SGV TOAN 8-CTST

["V\u1eadn d\u1ee5ng 2. Ng\u01b0\u1eddi ta c\u00f3 th\u1ec3 d\u00f9ng hai s\u1ed1 \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh v\u1ecb tr\u00ed c\u1ee7a m\u1ed9t \u0111i\u1ec3m tr\u00ean m\u1eb7t \u0111\u1ea5t ho\u1eb7c \u0111\u1ecba c\u1ea7u, ch\u1eb3ng h\u1ea1n L\u00fd S\u01a1n l\u00e0 m\u1ed9t huy\u1ec7n \u0111\u1ea3o n\u1ed5i ti\u1ebfng c\u1ee7a Vi\u1ec7t Nam, n\u1eb1m \u1edf v\u1ecb tr\u00ed 109\u00b007'3''\u0110, 15\u00b022'51''B. Em h\u00e3y l\u1ea5y m\u1ed9t b\u1ea3n \u0111\u1ed3 \u0111\u1ecba l\u00ed Vi\u1ec7t Nam v\u00e0 x\u00e1c \u0111\u1ecbnh v\u1ecb tr\u00ed c\u1ee7a \u0111\u1ea3o L\u00fd S\u01a1n theo kinh \u0111\u1ed9 v\u00e0 v\u0129 \u0111\u1ed9. \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a V\u1eadn d\u1ee5ng 2: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf \u0111\u1ecdc to\u1ea1 \u0111\u1ed9 \u0111\u1ecba l\u00ed tr\u00ean b\u1ea3n \u0111\u1ed3. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. 3. \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 H\u0110KP 3 \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a H\u0110KP 3: H\u01b0\u01a1\u0301ng d\u00e2\u0303n HS x\u00e2y d\u01b0\u0323ng kh\u00e1i ni\u1ec7m \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 3: Y\u00eau c\u00e2\u0300u HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 3. V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = f(x) cho b\u1eb1ng b\u1ea3ng sau: x \u20132 \u20131 0 1 2 y 2 1 0 \u20131 \u20132 Mu\u0323c \u0111i\u0301ch c\u1ee7a Th\u1ef1c h\u00e0nh 3: HS th\u01b0\u0323c ha\u0300nh v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00ea\u0309 re\u0300n luy\u00ea\u0323n ki\u0303 n\u0103ng theo y\u00eau c\u00e2\u0300u c\u00e2\u0300n \u0111a\u0323t. V\u1eadn d\u1ee5ng 3. Cho h\u00e0m s\u1ed1 y = f(x) c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 H\u00ecnh 10. y H\u00e3y ho\u00e0n th\u00e0nh b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 sau \u0111\u00e2y: x \u20132 \u20131 0 1 2 4 y????? \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a V\u1eadn d\u1ee5ng 3: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng 1 3 x ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf, v\u1eadn d\u1ee5ng t\u1ed5ng h\u1ee3p c\u00e1c k\u0129 n\u0103ng \u20133 \u20132 \u20131 O 1 2 th\u00f4ng qua vi\u1ec7c \u0111\u1ecdc to\u1ea1 \u0111\u1ed9 m\u1ed9t \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1. Hi\u0300nh 10 \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 3: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: x \u20132 \u20131 0 1 2 y41014 150","IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) C\u00e1c \u0111i\u1ec3m A, B, C thu\u1ed9c tr\u1ee5c ho\u00e0nh. b) M\u1ed9t \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean tr\u1ee5c ho\u00e0nh c\u00f3 tung \u0111\u1ed9 b\u1eb1ng 0. 2. \ta) C\u00e1c \u0111i\u1ec3m M, N, P thu\u1ed9c tr\u1ee5c tung. b) M\u1ed9t \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean tr\u1ee5c tung c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng 0. 3. T\u1ee9 gi\u00e1c ABCD c\u00f3 b\u1ed1n c\u1ea1nh b\u1eb1ng nhau v\u00e0 4. \u0110\u1ed3 th\u1ecb \u0111\u01b0\u1ee3c v\u1ebd nh\u01b0 h\u00ecnh d\u01b0\u1edbi \u0111\u00e2y. b\u1ed1n g\u00f3c b\u1eb1ng nhau. y y B 4 4 3 A3 2 1 2 1 \u20134 \u20133 \u20132 \u20131 O 1 2 3 4x \u20131 \u20134 \u20133 \u20132 \u20131 O 1 2 3 4x \u20132 \u20131 C \u20133 \u20132 \u20134 \u20133 D \u20135 \u20136 \u20134 5. \t\u0110i\u1ec3m M(\u20131; \u20134) v\u00e0 P \uf8ed\uf8eb\uf8ec 1 ; 1\uf8f8\uf8f6\uf8f7 thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = 4x. 4 6. \ta) HS t\u1ef1 v\u1ebd. b) C\u00e1c \u0111i\u1ec3m \u0111\u00e3 cho trong c\u00e2u a) th\u1eb3ng h\u00e0ng. 7. \ta) H(3; 9), D(4; 12), M(2; 6). b) D\u0169ng mua nhi\u1ec1u quy\u1ec3n v\u1edf nh\u1ea5t. 8. \t\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 S theo bi\u1ebfn s\u1ed1 t \u0111\u01b0\u1ee3c v\u1ebd nh\u01b0 h\u00ecnh sau: y 60 50 40 30 20 10 \u201310 O 10 20 30 x \u201310 151","B\u00e0i 3. H\u00c0M S\u1ed0 B\u1eacC NH\u1ea4T y = ax + b (a \u2260 0) I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u00e2\u0300n \u0111a\u0323t: \u2013 Nh\u1eadn bi\u1ebft \u0111\u01b0\u1ee3c kh\u00e1i ni\u1ec7m h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t. \u2013 Thi\u00ea\u0301t l\u1eadp \u0111\u01b0\u01a1\u0323c b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b (a \u2260 0). \u2013 V\u1ebd \u0111\u01b0\u01a1\u0323c \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b (a \u2260 0). \u2013 V\u00e2\u0323n du\u0323ng \u0111\u01b0\u1ee3c h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 \u0111\u1ed3 th\u1ecb va\u0300o gia\u0309i quy\u00ea\u0301t m\u1ed9t s\u1ed1 b\u00e0i to\u00e1n th\u1ef1c ti\u1ec5n. 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc, m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc, giao ti\u00ea\u0301p toa\u0301n ho\u0323c. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng, t\u00edch h\u1ee3p c\u00e1c m\u00f4n h\u1ecdc kh\u00e1c. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd 1. GV c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng c\u00e1c b\u00e0i toa\u0301n v\u1ec1 chuy\u1ec3n \u0111\u1ed9ng \u0111\u1ec1u trong V\u1eadt l\u00ed, v\u00f2i n\u01b0\u1edbc ch\u1ea3y, ... \u0111\u1ec3 gi\u00fap HS l\u00e0m quen v\u1edbi h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t. 2. HS c\u1ea7n \u0111\u01b0\u1ee3c luy\u1ec7n t\u1eadp k\u0129 n\u0103ng l\u1eadp b\u1ea3ng gi\u00e1 tr\u1ecb v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 C\u00f3 m\u1ed9t c\u00e1i b\u1ec3 \u0111\u00e3 ch\u1ee9a s\u1eb5n 5 m3 n\u01b0\u1edbc. 2 m3\/h Ng\u01b0\u1eddi ta b\u1eaft \u0111\u1ea7u m\u1edf m\u1ed9t v\u00f2i n\u01b0\u1edbc 5 m3 cho ch\u1ea3y v\u00e0o b\u1ec3, m\u1ed7i gi\u1edd ch\u1ea3y \u0111\u01b0\u1ee3c 2 m3. H\u00e3y t\u00ednh: a) L\u01b0\u1ee3ng n\u01b0\u1edbc ch\u1ea3y v\u00e0o b\u1ec3 sau 1 gi\u1edd. b) L\u01b0\u1ee3ng n\u01b0\u1edbc ch\u1ea3y v\u00e0o b\u1ec3 sau x gi\u1edd. c) L\u01b0\u1ee3ng n\u01b0\u1edbc y c\u00f3 trong b\u1ec3 sau x gi\u1edd. \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t th\u00f4ng qua vi\u1ec7c t\u00ednh to\u00e1n l\u01b0\u1ee3ng n\u01b0\u1edbc ch\u1ea3y v\u00e0o b\u1ec3 b\u1eb1ng m\u1ed9t v\u00f2i ch\u1ea3y c\u00f3 t\u1ed1c \u0111\u1ed9 ch\u1ea3y \u0111\u1ec1u. C\u00e1ch \u0111\u1eb7t v\u1ea5n \u0111\u1ec1 n\u00e0y c\u00f3 kh\u1ea3 n\u0103ng thu h\u00fat HS v\u00e0o b\u00e0i h\u1ecdc. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110K\u0110: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV s\u1eed d\u1ee5ng c\u01a1 h\u1ed9i \u0111\u1ec3 gi\u1edbi thi\u1ec7u b\u00e0i. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c chia t\u1ed5 \u0111\u1ec3 th\u1ea3o lu\u1eadn. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: a) 2 m3;\t b) 2x (m3); \t c) y = 2x + 5 (m3). 152","1. H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t H\u0110KP 1 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 1: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 c\u00e1ch m\u00f4 t\u1ea3 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng y = ax + b (a \u2260 0). \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 1: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 1. T\u00ecm c\u00e1c h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t trong c\u00e1c h\u00e0m s\u1ed1 sau \u0111\u00e2y v\u00e0 ch\u1ec9 ra c\u00e1c h\u1ec7 s\u1ed1 a, b c\u1ee7a c\u00e1c h\u00e0m s\u1ed1 \u0111\u00f3: y = 4x \u2013 7; y = x2; y = \u2013 6x \u2013 4; y = 4x; y = 3 ; s = 5v + 8; m = 30n \u2013 25. x M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: HS th\u01b0\u0323c ha\u0300nh nh\u1eadn d\u1ea1ng h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. V\u1eadn d\u1ee5ng 1. M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc l\u00e0 2m 3m x 2 m v\u00e0 3 m. G\u1ecdi y l\u00e0 chu vi c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt n\u00e0y sau khi x Hi\u0300nh 1 t\u0103ng chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng th\u00eam x (m). H\u00e3y ch\u1ee9ng t\u1ecf y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t theo bi\u1ebfn s\u1ed1 x. T\u00ecm c\u00e1c h\u1ec7 s\u1ed1 a, b c\u1ee7a h\u00e0m s\u1ed1 n\u00e0y. \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 1: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf, s\u01b0\u0309 du\u0323ng m\u00f4 h\u00ecnh h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t \u0111\u1ec3 bi\u1ec3u di\u1ec5n m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 1: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: y = 4x + 10; a = 4; b = 10. 2. B\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t 2H.\u0110BK\u1ea2PN2G GI\u00c1 TR\u1eca C\u1ee6A H\u00c0M S\u1ed0 B\u1eacC NH\u1ea4T 2 L\u01b0\u1ee3ng n\u01b0\u1edbc y (t\u00ednh theo m3) c\u00f3 trong m\u1ed9t b\u1ec3 n\u01b0\u1edbc sau x gi\u1edd m\u01a1\u0309 vo\u0300i c\u00e2\u0301p n\u01b0\u1edbc \u0111\u01b0\u1ee3c cho b\u01a1\u0309i h\u00e0m s\u1ed1 y = 2x + 3. T\u00ednh l\u01b0\u1ee3ng n\u01b0\u1edbc c\u00f3 trong b\u1ec3 sau 0 gi\u1edd; 1 gi\u1edd; 2 gi\u1edd; 3 gi\u1edd; 10 gi\u1edd v\u00e0 ho\u00e0n th\u00e0nh b\u1ea3ng gi\u00e1 tr\u1ecb sau: x 0 1 2 3 10 y = f(x) = 2x + 3 ? ? ? ? ? \u0110\u1ec3 l\u1eadp b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u00e2\u0301t y = ax + b ta l\u1ea7n l\u01b0\u1ee3t cho x nh\u1eadn c\u00e1c gi\u00e1 tr\u1ecb x1, x2, x3, \u2026 (x1, x2, x3, \u2026 t\u0103ng d\u1ea7n) v\u00e0 t\u00ednh c\u00e1c gi\u00e1 tr\u1ecb t\u01b0\u01a1ng \u1ee9ng c\u1ee7a y r\u1ed3i ghi v\u00e0o b\u1ea3ng153 c\u00f3 d\u1ea1ng sau: x x x x\u2026","\u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a H\u0110KP 2: Gi\u00fap HS l\u00e0m quen v\u1edbi vi\u1ec7c x\u00e1c \u0111\u1ecbnh b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t qua th\u1ef1c t\u1ebf nh\u1eadn bi\u1ebft l\u01b0\u1ee3ng n\u01b0\u1edbc c\u00f3 trong b\u1ec3 t\u1ea1i m\u1ed7i th\u1eddi \u0111i\u1ec3m. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 2: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 2. L\u1eadp b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a m\u1ed7i h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t sau: \t\t\t y = f(x) = 4x \u2013 1 v\u00e0 y = h(x) = \u2013 0,5x + 8 v\u1edbi x l\u1ea7n l\u01b0\u1ee3t b\u1eb1ng \u20133; \u20132; \u20131; 0; 1; 2; 3. \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a Th\u1ef1c h\u00e0nh 2: HS th\u01b0\u0323c ha\u0300nh l\u1eadp b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a m\u1ed9t h\u00e0m b\u1eadc nh\u1ea5t cho tr\u01b0\u1edbc \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c Th\u1ef1c h\u00e0nh 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: x \u20133 \u20132 \u20131 0 1 23 3 7 11 y = f(x) = 4x \u2013 1 \u201313 \u20139 \u20135 \u20131 x \u20133 \u20132 \u20131 0 1 2 3 y = h(x) = \u20130,5x + 8 9,5 9 8,5 8 7,5 7 6,5 V\u1eadn d\u1ee5ng 2. M\u1ed9t xe kh\u00e1ch kh\u1edfi h\u00e0nh t\u1eeb b\u1ebfn xe ph\u00eda B\u1eafc b\u01b0u \u0111i\u1ec7n th\u00e0nh ph\u1ed1 Nha Trang \u0111\u1ec3 \u0111i ra th\u00e0nh ph\u1ed1 \u0110\u00e0 N\u1eb5ng v\u1edbi t\u1ed1c \u0111\u00f4\u0323 40 km\/h (H\u00ecnh 2). B\u01b0u \u0111i\u00ea\u0323n B\u00ea\u0301n xe \u0110a\u0300 N\u0103\u0303ng Nha Trang Hi\u0300nh 2 a) Bi\u1ebft r\u1eb1ng b\u1ebfn xe c\u00e1ch b\u01b0u \u0111i\u1ec7n th\u00e0nh ph\u1ed1 Nha Trang 6 km. Sau x gi\u1edd, xe kh\u00e1ch c\u00e1ch b\u01b0u \u0111i\u1ec7n th\u00e0nh ph\u1ed1 Nha Trang y km. T\u00ednh y theo x. b) Ch\u1ee9ng minh r\u1eb1ng y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t theo bi\u1ebfn s\u1ed1 x. c) Ho\u00e0n th\u00e0nh b\u1ea3ng gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u1edf c\u00e2u b) v\u00e0 gi\u1ea3i th\u00edch \u00fd ngh\u0129a c\u1ee7a b\u1ea3ng gi\u00e1 tr\u1ecb n\u00e0y: x 0123 y ???? \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a V\u1eadn d\u1ee5ng 2: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf s\u01b0\u0309 du\u0323ng m\u00f4 h\u00ecnh h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t \u0111\u1ec3 bi\u1ec3u di\u1ec5n v\u1ecb tr\u00ed c\u1ee7a xe t\u1ea1i m\u1ed7i th\u1eddi \u0111i\u1ec3m. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: a) y = 40x + 6; 154","b) HS t\u1ef1 ch\u1ee9ng minh. c) x 0 123 y = 40x + 6 6 46 86 126 B\u1ea3ng gi\u00e1 tr\u1ecb cho bi\u1ebft kho\u1ea3ng c\u00e1ch t\u1eeb xe kh\u00e1ch \u0111\u1ebfn b\u01b0u \u0111i\u1ec7n th\u00e0nh ph\u1ed1 Nha Trang \u1edf th\u1eddi \u0111i\u1ec3m tr\u01b0\u1edbc khi kh\u1edfi h\u00e0nh v\u00e0 1; 2; 3 gi\u1edd sau khi kh\u1edfi h\u00e0nh. 3. \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax (a \u2260 0) H\u0110KP 3 \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a H\u0110KP 3: H\u01b0\u01a1\u0301ng d\u00e2\u0303n HS kh\u00e1m ph\u00e1 t\u00ednh ch\u1ea5t c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax (a \u2260 0). \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 3: Y\u00eau c\u00e2\u0300u HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 3. a) V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1: y = 0,5x; y = \u20133x; y = x. b) C\u00e1c \u0111\u1ed3 th\u1ecb sau \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 n\u00e0o? yy y 3 3 3 2A B2 2 1 1 1 \u20133 \u20132 \u20131\u20131 O 1 2 3 x \u20133 \u20132 \u20131\u20131O 123 x \u20133 \u20132 \u20131 O 123 x \u20132 \u20132 \u20131 C \u20133 \u20133 \u20132 \u20133 a) b) c) Hi\u0300nh 6 155","3 3 3 2A B2 2 1 1 1 \u20133 \u20132 \u20131 O 1 2 3 x \u20133 \u20132 \u20131 O 1 2 3 x \u20133 \u20132 \u20131 O 1 2 3 x \u20131 \u20131 \u20131 C \u20132 \u20132 \u20132 Mu\u0323c \u0111i\u0301ch c\u1ee7\u20133a Th\u1ef1c h\u00e0nh 3: HS th\u01b0\u0323c ha\u0300nh\u20133v\u1ebd v\u00e0 nh\u1eadn d\u1ea1ng \u0111\u1ed3 th\u1ecb c\u1ee7a h\u2013\u00e03 m s\u1ed1 y = ax \u0111\u00ea\u0309 re\u0300n luy\u00ea\u0323n ki\u0303 n\u0103nag) theo y\u00eau c\u00e2\u0300u c\u00e2\u0300n \u0111a\u0323t. b) c) \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax + b (a \u2260 0, b H\u2260i\u0300n0h)6 H\u0110K\u0110P\u1ed34th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax + b (a \u2260 0, b \u2260 0) 4 Cho hai h\u00e0m s\u1ed1 y = f(x) = x v\u00e0 y = g(x) = x + 3. a) Thay d\u1ea5u ? b\u1eb1ng s\u1ed1 th\u00edch h\u1ee3p. x \u20132 \u20131 0 1 2 y = f(x) = x ????? y = g(x) = x + 3 ? ? ? ? ? b) Tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng to\u1ea1 \u0111\u1ed9, v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = f(x) v\u00e0 bi\u1ec3u di\u1ec5n c\u00e1c \u0111i\u1ec3m c\u00f3 to\u1ea1 \u0111\u1ed9 tho\u1ea3 m\u00e3n h\u00e0m s\u1ed1 y = g(x) c\u00f3 trong b\u1ea3ng tr\u00ean. c) Ki\u1ec3m tra xem c\u00e1c \u0111i\u1ec3m thu\u1ed9c \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = g(x) v\u1ebd \u1edf c\u00e2u b c\u00f3 th\u1eb3ng h\u00e0ng kh\u00f4ng. V\u00e0 d\u1ef1 \u0111o\u00e1n c\u00e1ch v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = g(x). \u2013 MTua\u0323csu\u0111yi\u0301crha tc\u00ed\u1ee7nha cHh\u0110\u1ea5tKc\u1ee7Pa4\u0111:\u1ed3Gthi\u1ecb\u00fahp\u00e0HmSs\u1ed1kbh\u1ead\u00e1cmnhp\u1ea5ht\u00e1nth\u00ed\u01b0nhsacuh: \u1ea5t c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b (a \u2260 0, b \u2260 0) v\u00e0 c\u00e1ch v\u1ebd \u0111\u1ed3 th\u1ecb. \u2013 G\u0110\u01a1\u0323i\u1ed3y\u0301tht\u00f4\u1ecb\u0309cc\u1ee7ha\u01b0\u0301hc\u00e0Hm\u0110s\u1ed1KyP=4:aHx S+ tbr\u1ea3(al\u1edd\u2260i 0y,\u00eabu \u2260c\u1ea70u)cl\u00e0\u1ee7amh\u1ed9ot\u1ea1\u0111t\u01b0\u0111\u1edd\u1ed9nnggthv\u1eb3\u00e0nogv:\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. G\u2013VCc\u1eaf\u00f3t thr\u1ee5\u1ec3ccthuongHtS\u1ea1il\u0111\u00e0im\u1ec3mvic\u1ec7\u00f3ctuthnego\u0111n\u1ed9hb\u00f3\u1eb1mng. b; Th\u1ef1\u2013cSho\u00e0nnghso4n.gVv\u1ebd\u1edb\u0111i \u0111\u1ed3\u01b0t\u1eddhn\u1ecb gc\u1ee7tha\u1eb3cn\u00e1gcyh=\u00e0maxs.\u1ed1 sau: a) y = 5x + 2; \t\t b) y = \u20132x \u2013 6. C\u00e1ch v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax + b (a \u2260 0, b \u2260 0) \u0111\u1ec3 \u2013r\u00e8MntTlruua\u00ea\u0323cny\u0111\u1ec7\u0111\u00e3tnai\u0301cbckhih\u1ebf\u0129ct\u1ec9n\u1ee7c\u0111\u0103a\u1ea7\u1ed3nngTthxht\u1ecb\u00e1h\u1ef1ccec\u1ee7o\u0111ah\u1ecbyn\u00e0h\u00eahn\u00e0uhm\u0111c\u01b04\u1ea7s\u1ee3:\u1ed1ucHychS\u1ea7a=nith\u0111a\u0111\u1ef1ix\u1ea1\u1ec3ctm+.hbp\u00e0nhl\u00e0\u00e2hnmvb\u1ebd\u1ed9i\u1ec7t\u0111t\u0111\u1ed3t\u01b0ut\u1edd\u1ef3hn\u1ecb\u00fdgct\u1ee7thhau\u1eb3\u1ed9hnc\u00e0gm.\u0111\u0110\u1ed3s\u1ec3t\u1ed1hv\u1ecbb\u1ebdr\u1ead\u1ed3\u0111ci\u1ed3nvth\u1ebdh\u1ea5\u0111\u1ecbt\u01b0hy\u1edd\u00e0mn=gast\u1ed1xh\u1eb3n+n\u00f3gbi \u2013 G\u0111\u1ee3iiq\u00fduat\u1ed5hcahi \u1ee9\u0111ic\u1ec3mTh\u0111\u1ef1\u00f3c. Th\u00e0hn\u00f4hng4t:hH\u01b0\u1eddSntgh\u1ef1tacxh\u00e1ic\u1ec7n\u0111\u1ecbvn\u00e0hohvai\u1edf\u0111, iG\u1ec3mV\u0111n\u1eb7hc\u1eadbni\u1ec7xt\u00e9lt\u00e0vg\u00e0ia\u0111o\u00e1n\u0111ih\u1ec3mgi\u00e1c.\u1ee7a \u0111\u1ed3 th\u1ecb V\u1eadvn\u1edbdi \u1ee5hnaigtr3\u1ee5.cMto\u1ed9\u1ea1t\u0111l\u1ed9\u00f2.xo c\u00f3 chi\u1ec1u d\u00e0i ban \u0111\u1ea7u khi ch\u01b0a treo v\u1eadt n\u1eb7ng l\u00e0 10 cm. Cho bi\u1ebft khi treo th\u00eam v\u00e0o l\u00f2 xo m\u1ed9t v\u1eadt n\u1eb7ng 1 kg th\u00ec chi\u1ec1u d\u00e0i l\u00f2 xo t\u0103ng th\u00eam 3 cm. 20a) T\u00ednh chi\u1ec1u d\u00e0i y (cm) c\u1ee7a l\u00f2 xo theo kh\u1ed1i l\u01b0\u1ee3ng x (kg) c\u1ee7a v\u1eadt. b) V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y theo bi\u1ebfn s\u1ed1 x. \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a V\u1eadn d\u1ee5ng 3: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng Hi\u0300nh 10 ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf, \u00e1p d\u1ee5ng ki\u1ebfn th\u1ee9c li\u00ean m\u00f4n, v\u1eadn d\u1ee5ng t\u1ed5ng h\u1ee3p c\u00e1c k\u0129 n\u0103ng th\u00f4ng qua vi\u1ec7c t\u00ednh chi\u1ec1u d\u00e0i c\u1ee7a m\u1ed9t l\u00f2 xo theo kh\u1ed1i l\u01b0\u1ee3ng \u0111\u01b0\u1ee3c treo v\u00e0o. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 3: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m. 156","H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: y a) y = 3x + 10. 15 b) \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u01b0\u1ee3c v\u1ebd nh\u01b0 h\u00ecnh b\u00ean. 10 5 \u201310 \u20135 O 5 10 x \u20135 IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) y = 4x + 2 l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u1edbi a = 4; b = 2. b) y = 5 \u2013 3x l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u1edbi a = \u20133; b = 5. c) y = 2 + x2 kh\u00f4ng ph\u1ea3i l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00ec c\u00f3 ch\u1ee9a x2. d) y = \u20130,2x l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u1edbi a = \u20130,2; b = 0. e) y = 5 x \u2013 1 l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u1edbi a = 5 ; b = \u20131. 2. \ta) m \u2260 1; b) m \u2260 0. y 3 3. \ta) \u0110\u1ed3 th\u1ecb c\u00e1c h\u00e0m s\u1ed1 y = x; y = x + 2; y = \u2013x; y = \u2013x + 2 y=x+2 y = \u2013x + 2 \u0111\u01b0\u1ee3c v\u1ebd tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng to\u1ea1 \u0111\u1ed9 nh\u01b0 h\u00ecnh b\u00ean. y = \u2013x 2B y=x A 1 C b) T\u1ee9 gi\u00e1c OABC c\u00f3 b\u1ed1n c\u1ea1nh b\u1eb1ng nhau v\u00e0 c\u00f3 b\u1ed1n g\u00f3c vu\u00f4ng n\u00ean l\u00e0 h\u00ecnh vu\u00f4ng. 4. \ta) C = 5 (F \u2013 32) suy ra C = 5 F \u2212 160 . \u20132 \u20131 O 1 2x 9 99 \u20131 Do \u0111\u00f3 C l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t theo bi\u1ebfn s\u1ed1 F. \u20132 s = 4t 1 2t b) V\u1edbi F = 32 th\u00ec C = 0; v\u1edbi C = 100 th\u00ec F = 212. s 4 5. \tTa c\u00f3 C = 2\u03c0r, v\u1eady C l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t theo 3 bi\u1ebfn s\u1ed1 r v\u1edbi a = 2\u03c0 v\u00e0 b = 0. 2 1 6. \ta) s = vt. b) Khi v = 4 ta c\u00f3 s = 4t. \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 s = 4t \u20132 \u20131 O \u0111\u01b0\u1ee3c v\u1ebd nh\u01b0 h\u00ecnh b\u00ean. \u20131 \u20132 \u20133 \u20134 157","B\u00e0i 4. H\u1ec6 S\u1ed0 G\u00d3C C\u1ee6A \u0110\u01af\u1edcNG TH\u1eb2NG I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u00e2\u0300n \u0111a\u0323t: \u2013 Nh\u1eadn bi\u1ebft \u0111\u01b0\u1ee3c kh\u00e1i ni\u1ec7m h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b (a \u2260 0). \u2013 S\u1eed d\u1ee5ng \u0111\u01b0\u1ee3c h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ec3 nh\u1eadn bi\u1ebft v\u00e0 gi\u1ea3i th\u00edch \u0111\u01b0\u01a1\u0323c s\u1ef1 c\u1eaft nhau ho\u1eb7c song song c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng cho tr\u01b0\u1edbc. 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc, m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc, giao ti\u00ea\u0301p toa\u0301n ho\u0323c. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng, t\u00edch h\u1ee3p c\u00e1c m\u00f4n h\u1ecdc kh\u00e1c. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd 1. B\u00e0i n\u00e0y th\u1ec3 hi\u1ec7n s\u1ef1 t\u00edch h\u1ee3p n\u1ed9i m\u00f4n gi\u1eefa \u0111\u1ea1i s\u1ed1 v\u00e0 h\u00ecnh h\u1ecdc. 2. GV c\u00f3 th\u1ec3 t\u00ecm c\u00e1c t\u00ecnh hu\u1ed1ng th\u1ef1c t\u1ebf \u0111\u1ec3 gi\u00fap HS chuy\u1ec3n t\u1eeb \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t sang \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 Khi n\u00e0o th\u00ec hai \u0111\u01b0\u1eddng th\u1eb3ng y y = ax + b (a \u2260 0) v\u00e0 y = a'x + b' (a' \u2260 0) 6 song song v\u1edbi nhau, 5 tr\u00f9ng nhau, c\u1eaft nhau? 4 3 2 1 \u20133 \u20132 \u20131 O 1 2 3 4 5 x \u20131 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 d\u1ea5u hi\u1ec7u song song, tr\u00f9ng nhau ho\u1eb7c c\u1eaft nhau c\u1ee7a hai \u0111\u1ed3 th\u1ecb h\u00e0m b\u1eadc nh\u1ea5t. C\u00e1ch \u0111\u1eb7t v\u1ea5n \u0111\u1ec1 n\u00e0y c\u00f3 kh\u1ea3 n\u0103ng thu h\u00fat HS v\u00e0o b\u00e0i h\u1ecdc. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110K\u0110: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV s\u1eed d\u1ee5ng c\u01a1 h\u1ed9i \u0111\u1ec3 gi\u1edbi thi\u1ec7u b\u00e0i. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: \u2013 Hai \u0111\u01b0\u1eddng th\u1eb3ng song song khi a = a\u2032; b \u2260 b\u2032. \u2013 Hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau khi a \u2260 a\u2032. \u2013 Hai \u0111\u01b0\u1eddng th\u1eb3ng tr\u00f9ng nhau khi a = a\u2032; b = b\u2032. 158","L\u01b0u \u00fd: \u0110\u00e2y l\u00e0 c\u00e2u h\u1ecfi m\u1edf, t\u1ea1o s\u1ef1 ch\u00fa \u00fd v\u00e0 k\u1ebft n\u1ed1i. GV kh\u00f4ng c\u1ea7n \u0111\u00e1nh gi\u00e1 c\u00e2u tr\u1ea3 l\u1eddi c\u1ee7a HS \u0111\u00fang hay sai. Tinh th\u1ea7n chung c\u1ee7a H\u0110K\u0110 l\u00e0: \u201cM\u1ecdi c\u00e2u tr\u1ea3 l\u1eddi \u0111\u1ec1u \u0111\u01b0\u1ee3c ghi nh\u1eadn, mu\u1ed1n bi\u1ebft \u0111\u00fang \u2013 sai, h\u1ecdc xong b\u00e0i n\u00e0y s\u1ebd r\u00f5!\u201d. 1. H\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b (a \u2260 0) H\u0110KP 1 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 1: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 kh\u00e1i ni\u1ec7m h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng y = ax + b (a \u2260 0) v\u00e0 s\u1ef1 li\u00ean quan c\u1ee7a a \u0111\u1ebfn \u0111\u1ed9 d\u1ed1c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00f3. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 1: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 1. T\u00ecm h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng sau \u0111\u00e2y: a) y = \u20135x \u2013 5;\t\t\t\t\t\t\t b) y = 3 x + 3;\t\t\t\t\t c) y = 11x + 7. M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: HS th\u01b0\u0323c ha\u0300nh t\u00ecm h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. V\u1eadn d\u1ee5ng 1. Trong ca\u0301c \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng sau \u0111\u00e2y, \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng na\u0300o ta\u0323o v\u01a1\u0301i Ox m\u00f4\u0323t go\u0301c nho\u0323n, \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng na\u0300o ta\u0323o v\u01a1\u0301i Ox m\u00f4\u0323t go\u0301c tu\u0300? a) y = 3x + 6;\t\t\t\t\t\t\t\t b) y = \u2013 4x + 1;\t\t\t\t\t\t c) y = \u20133x \u2013 6. 159","\u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 1: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf x\u00e1c \u0111\u1ecbnh g\u00f3c d\u1ef1a theo h\u1ec7 s\u1ed1 g\u00f3c. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 1: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: a) G\u00f3c nh\u1ecdn; \t b) G\u00f3c t\u00f9; \t c) G\u00f3c t\u00f9. 2. Hai \u0111\u01b0\u1eddng th\u1eb3ng song song, hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau Nh\u1eadn bi\u1ebft hai \u0111\u01b0\u1eddng th\u1eb3ng song song H\u0110KP 2 V\u00ed d\u1ee5 2. a) N\u00eau nh\u1eadn x\u00e9t v\u1ec1 v\u1ecb tr\u00ed gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng d1: y = \u2013x + 1 v\u00e0 d2: y = \u2013x \u2013 2. b) X\u00e1c \u0111\u1ecbnh h\u00e0m s\u1ed1 bi\u1ebft \u0111\u1ed3 th\u1ecb c\u1ee7a n\u00f3 l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d3 song song v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng d1 v\u00e0 c\u1eaft tr\u1ee5c Oy t\u1ea1i \u0111i\u1ec3m (0; 3). Gi\u1ea3i a) Hai \u0111\u01b0\u1eddng th\u1eb3ng d1 v\u00e0 d2 ph\u00e2n bi\u1ec7t (c\u1eaft Oy t\u1ea1i hai \u0111i\u1ec3m kh\u00e1c nhau) v\u00e0 c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c b\u1eb1ng nhau (c\u00f9ng b\u1eb1ng \u20131), suy ra d1 \/\/ d2. khi\u2013bMi\u1ebfqb\u1ee5tu)cpa\u0110h\u0111\u0111\u01b0\u01b0\u00edci\u1edd\u01a1\u1ec3hnmncgg\u1ee7(t0athr;\u1eb3H\u00ec3nn\u0110)hg. KVdc\u1ee73P\u1eadasy2oh:dnaG3gicis\u0111\u00fa\u00f3o\u01b0pnp\u1eddgHhn\u01b0vSg\u01a1\u1edbnntihhgd\u1ead\u1eb31tnn,r\u00ecsgbnuih\u0111y\u1ebf\u00f3tyr.\u0111a=\u01b0d\u1ee3\u20133cxpdh+\u1ea5\u1ea33ui .ch\u00f3i\u1ec7hu\u1ec7sso\u1ed1ngg\u00f3sconbg\u1eb1ncg\u1ee7a\u20131h.aTi \u0111a\u01b0l\u1ea1\u1eddincg\u00f3thd\u1eb33 n\u0111gi \u2013 GC\u1ee3hi \u00fa\u00fd\u00fdt:\u1ed5Hcahi\u1ee9\u0111c\u01b0H\u1eddn\u0110gKthP\u1eb3n2g: Gy =Vanx\u00ea+u bc\u00e2vu\u00e0 hy\u1ecf=i,aH\u2032xS+tbr\u1ea3\u2032 trl\u1edd\u00f9ni,gl\u1edbnhpanuhk\u1eadhni xv\u00e0\u00e9tc,hG\u1ec9 Vkh\u0111i a\u00e1n=hag\u2032,ib\u00e1.= b\u2032. Nh\u1eadNnh\u1eadbni\u1ebfbt ih\u1ebftaih\u0111ai\u01b0\u0111\u1edd\u01b0n\u1eddgntght\u1eb3hn\u1eb3gngc\u1eafct\u1eaftnnhhaauu 3 Quan s\u00e1t H\u00ecnh 4. y = 2x y y=x 3 a) T\u00ecm giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng 2 1 d: y = 2x v\u00e0 d\u2032: y = x. \u20133 \u20132 \u20131 O 1 2 3 x b) N\u00eau nh\u1eadn x\u00e9t v\u1ec1 hai \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 \u20131 h\u1ec7 s\u1ed1 g\u00f3c kh\u00e1c nhau. \u20132 \u20133 c) Cho \u0111\u01b0\u1eddng th\u1eb3ng d\u2032\u2032: y = ax + b v\u00e0 cho bi\u1ebft d\u2032\u2032 c\u1eaft d. H\u1ec7 s\u1ed1 g\u00f3c a c\u1ee7a d\u2032\u2032 c\u00f3 th\u1ec3 nh\u1eadn c\u00e1c gi\u00e1 tr\u1ecb n\u00e0o? Hi\u0300nh 4 160 Hai \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c kh\u00e1c nhau th\u00ec c\u1eaft nhau v\u00e0 ng\u01b0\u1ee3c l\u1ea1i, hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau th\u00ec c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c kh\u00e1c nhau.","\u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a H\u0110KP 3: Gi\u00fap HS nh\u1eadn bi\u1ebft d\u1ea5u hi\u1ec7u c\u1eaft nhau c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng khi bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00f3. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 3: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 2. H\u00e3y ch\u1ec9 ra ba c\u1eb7p \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau v\u00e0 c\u00e1c c\u1eb7p \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi nhau trong c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng sau: \t d1: y = 3x;\t\t\t\t\t\t d2: y = \u20137x + 9;\t\t\t\t\t d3: y = 3x \u2013 0,8; \t d4: y = \u20137x \u2013 1;\t\t\t\t d5: y = 2x + 10\u00a0;\t\t\t\t d6: y = 2x + 10. \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a Th\u1ef1c h\u00e0nh 2: HS th\u01b0\u0323c ha\u0300nh t\u00ecm c\u00e1c c\u1eb7p \u0111\u01b0\u1eddng th\u1eb3ng song song ho\u1eb7c c\u1eaft nhau d\u1ef1a theo h\u1ec7 s\u1ed1 g\u00f3c \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c Th\u1ef1c h\u00e0nh 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. V\u1eadn d\u1ee5ng 2. A 3 km B 2 km C D y = f(x) y = g(x) H\u00ecnh 5 Hai \u00f4 t\u00f4 kh\u1edfi h\u00e0nh c\u00f9ng l\u00fac v\u00e0 c\u00f9ng v\u1edbi t\u1ed1c \u0111\u1ed9 50 km\/h, m\u1ed9t \u00f4 t\u00f4 b\u1eaft \u0111\u1ea7u t\u1eeb B, m\u1ed9t \u00f4 t\u00f4 b\u1eaft \u0111\u1ea7u t\u1eeb C v\u00e0 c\u00f9ng \u0111i v\u1ec1 ph\u00eda D (H\u00ecnh 5). a) Vi\u1ebft c\u00f4ng th\u1ee9c c\u1ee7a hai h\u00e0m s\u1ed1 bi\u1ec3u th\u1ecb kho\u1ea3ng c\u00e1ch t\u1eeb A \u0111\u1ebfn m\u1ed7i xe sau x gi\u1edd. b) Ch\u1ee9ng t\u1ecf \u0111\u1ed3 th\u1ecb c\u1ee7a hai h\u00e0m s\u1ed1 tr\u00ean l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng song song. \u2013 Mu\u0323c \u0111i\u0301ch c\u1ee7a V\u1eadn d\u1ee5ng 2: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf, \u00e1p d\u1ee5ng ki\u1ebfn th\u1ee9c li\u00ean m\u00f4n, v\u1eadn d\u1ee5ng t\u1ed5ng h\u1ee3p c\u00e1c k\u0129 n\u0103ng th\u00f4ng qua vi\u1ec7c d\u00f9ng m\u00f4 h\u00ecnh h\u00e0m s\u1ed1 \u0111\u1ec3 bi\u1ec3u di\u1ec5n hai chuy\u1ec3n \u0111\u1ed9ng th\u1eb3ng \u0111\u1ec1u v\u00e0 x\u00e1c \u0111\u1ecbnh t\u00ednh song song c\u1ee7a hai \u0111\u1ed3 th\u1ecb c\u1ee7a hai h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u c\u1ee7a ho\u1ea1t \u0111\u1ed9ng v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: a) y = f(x) = 50x + 3; y = g(x) = 50x + 5. b) \u0110\u1ed3 th\u1ecb c\u1ee7a hai h\u00e0m s\u1ed1 tr\u00ean l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng ph\u00e2n bi\u1ec7t v\u00e0 c\u00f3 c\u00f9ng h\u1ec7 s\u1ed1 g\u00f3c n\u00ean \u0111\u00f3 l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng song song. IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) Thay x = 1 v\u00e0 y = \u20132 v\u00e0o y = ax \u2013 4 ta \u0111\u01b0\u1ee3c \u20132 = a \u2013 4, suy ra a = 2. b) HS t\u1ef1 v\u1ebd. 2. \ta) HS t\u1ef1 v\u1ebd; b) \u03b1 = 45o. 3. \tBa c\u1eb7p \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau: d1 v\u00e0 d2; d1 v\u00e0 d4; d1 v\u00e0 d5. C\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng song song: d1 v\u00e0 d3; d2 v\u00e0 d4; d5 v\u00e0 d6. 161","4. \ta = 9. 5. \ta) m = 1; b) m \u2260 1. 6. \ty = x + 2 024; y = x + 2 025. 7. \ty = x; y = 2x. 8. a) To\u1ea1 \u0111\u1ed9 (x; y) c\u1ee7a c\u00e1c \u0111i\u1ec3m A, B, C, D, E, F tr\u00f9ng v\u1edbi c\u00e1c gi\u00e1 tr\u1ecb t\u01b0\u01a1ng \u1ee9ng trong b\u1ea3ng \u0111\u00e3 cho. b) m = 1 . 2 9. \ta) y = 50x + 4; \t\t b) H\u1ec7 s\u1ed1 g\u00f3c a = 50. 10. a) y = x + 3; \t \t b) HS t\u1ef1 v\u1ebd. B\u00c0I T\u1eacP CU\u1ed0I CH\u01af\u01a0NG 5 C\u00c2U H\u1eceI TR\u1eaeC NGHI\u1ec6M 1. A 2. A 3. B 4. D 5. D 6. C 7. C 8. D 9. C B\u00c0I T\u1eacP T\u1ef0 LU\u1eacN 10. a) f \uf8eb 1 \uf8f6 = 25 ; f(\u20135) =\u22121; f \uf8eb 4 \uf8f6 = 25 . \uf8ec\uf8ed 5 \uf8f7\uf8f8 4 4 \uf8ec\uf8ed 5 \uf8f7\uf8f8 16 b) x \u20133 \u20132 \u20131 \u2212 1 1 1 2 24 y = f(x) = 5 \u22125 \u20135 \u22125 \u22125 5 5 5 4x 12 8 4 2 48 11. f(\u20133) = \u20138; f(\u20132) = \u20133; f(\u20131) = 0; f(0) = 1; f(1) = 0. 12. ABCD l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. 162","13. \ta) a = \u2212 4 ; 5 \t b) V\u1ebd \u0111i\u1ec3m c\u00f3 to\u1ea1 \u0111\u1ed9 (\u20135; 4); \t c) V\u1ebd \u0111i\u1ec3m c\u00f3 to\u1ea1 \u0111\u1ed9 \uf8ed\uf8eb\uf8ec \u2212 5 ; 2 \uf8f6\uf8f7\uf8f8. 2 14. \ty = \u20132x + b v\u1edbi b \u2260 10. 15. \ta) s = 3t; b) HS t\u1ef1 v\u1ebd. 16. m = 3. 17. n = 2. y = \u2013x + 3 y 18. k \u2260 4. 6 5 y=x+3 19. a) \u0110\u1ed3 th\u1ecb hai h\u00e0m s\u1ed1 \u0111\u00e3 cho \u0111\u01b0\u1ee3c v\u1ebd nh\u01b0 h\u00ecnh b\u00ean. 4 T\u1eeb \u0111\u1ed3 th\u1ecb ta c\u00f3 A(0; 3), B(\u20133; 0), C(3; 0). 3A b) 45o v\u00e0 135o. 2 c) Chu vi tam gi\u00e1c ABC b\u1eb1ng 6 + 6 2. 1 C Di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 9. B 1 2 3 4x \u20134 \u20133 \u20132 \u20131 O 1 163","Ch\u01b0\u01a1ng 6 PH\u01af\u01a0NG TR\u00ccNH A. M\u1ee4C TI\u00caU 1. N\u0103ng l\u01b0\u0323c chuy\u00ean m\u00f4n Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n \u2013 Hi\u1ec3u \u0111\u01b0\u1ee3c kh\u00e1i ni\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n. \u2013 Hi\u1ec3u \u0111\u01b0\u1ee3c c\u00e1ch gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n. Gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t\t \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t. V\u00ed d\u1ee5: C\u00e1c b\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn chuy\u1ec3n \u0111\u1ed9ng trong V\u1eadt l\u00ed, c\u00e1c b\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn Ho\u00e1 h\u1ecdc, ...\t\t\t\t\t\t\t\t 2. N\u0103ng l\u01b0\u0323c chung \u2013 N\u0103ng l\u01b0\u0323c t\u01b0\u0323 chu\u0309 va\u0300 t\u01b0\u0323 ho\u0323c trong ti\u0300m to\u0300i, kha\u0301m pha\u0301. \u2013 N\u0103ng l\u01b0\u0323c giao ti\u00ea\u0301p va\u0300 h\u01a1\u0323p ta\u0301c trong tri\u0300nh ba\u0300y, tha\u0309o lu\u00e2\u0323n va\u0300 la\u0300m vi\u00ea\u0323c nho\u0301m. \u2013 N\u0103ng l\u01b0\u0323c gia\u0309i quy\u00ea\u0301t v\u00e2\u0301n \u0111\u00ea\u0300 va\u0300 sa\u0301ng ta\u0323o trong th\u01b0\u0323c ha\u0300nh va\u0300 v\u00e2\u0323n du\u0323ng. 3. Hi\u0300nh tha\u0300nh ca\u0301c ph\u00e2\u0309m ch\u00e2\u0301t \u2013 Y\u00eau n\u01b0\u01a1\u0301c, nh\u00e2n \u00e1i. \u2013 Ch\u0103m chi\u0309, trung th\u01b0\u0323c, tra\u0301ch nhi\u00ea\u0323m. B. H\u01af\u1edaNG D\u1eaaN D\u1ea0Y H\u1eccC B\u00e0i 1. PH\u01af\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T M\u1ed8T \u1ea8N I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t: \u2013 Hi\u1ec3u \u0111\u01b0\u1ee3c kh\u00e1i ni\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n v\u00e0 c\u00e1ch gi\u1ea3i. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t. 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc; m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc; s\u1eed d\u1ee5ng c\u00f4ng c\u1ee5, c\u00e1c ph\u01b0\u01a1ng ti\u1ec7n h\u1ecdc to\u00e1n. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd \u2013 Kh\u00f4ng gi\u1edbi thi\u1ec7u ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng. \u2013 Kh\u00f4ng d\u00f9ng d\u1ea5u \u201c\u21d4\u201d trong qu\u00e1 tr\u00ecnh gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh. 164","\u2013 Vi\u1ec7c gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh s\u1eed d\u1ee5ng c\u00e1c quy t\u1eafc (quy t\u1eafc chuy\u1ec3n v\u1ebf, quy t\u1eafc nh\u00e2n v\u1edbi m\u1ed9t s\u1ed1, quy t\u1eafc chia cho m\u1ed9t s\u1ed1) \u0111\u01b0\u1ee3c h\u00ecnh th\u00e0nh qua H\u0110K\u0110 2, do \u0111\u00f3 GV c\u1ea7n ph\u00e2n t\u00edch k\u0129, gi\u1ea3ng ch\u1eadm t\u1eebng b\u01b0\u1edbc \u0111\u1ec3 HS n\u1eafm \u0111\u01b0\u1ee3c kh\u00e1i ni\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n v\u00e0 c\u00e1ch gi\u1ea3i. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 Quan s\u00e1t h\u00ecnh b\u00ean. Bi\u1ebft r\u1eb1ng xg xg xg xg 1 PH\u01af\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T M\u1ed8T \u1ea8NB\u00e0i c\u00e2n th\u0103ng b\u1eb1ng, c\u00f3 th\u1ec3 t\u00ecm 600 g x g \u0111\u01b0\u1ee3c kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a qu\u1ea3 c\u00e2n x g kh\u00f4ng? T\u00ecm b\u1eb1ng c\u00e1ch n\u00e0o? Quan s\u00e1t h\u00ecnh b\u00ean. Bi\u1ebft r\u1eb1ng xg xg \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110:cK\u00e2n\u00edcthh\u0103thn\u00edgchbH\u1eb1nSg,t\u01b0c\u00f3duthy\u1ec3st\u00e1\u00ecnmg t\u1ea1o, tx\u00ecgm hxig\u1ec3u v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh60b0 g\u1eadc xngh\u1ea5t m\u1ed9t \u1ea9n v\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u0111\u01a1\u01b0n\u1ee3gc tkr\u00echn\u1ed1hi bl\u01b0\u1ead\u1ee3cnngh\u1ea5ct\u1ee7ma \u1ed9qtu\u1ea9\u1ea3n. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110K\u0110cc:\u00e2\u00e1GnchVnx\u00e0gno\u00ea?ukhc\u00f4\u00e2nug?h\u1ecfTi\u00ec,mHbS\u1eb1tnrga\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV s\u1eed d\u1ee5ng c\u01a1 h\u1ed9i gi\u1edbi thi\u1ec7u b\u00e0i. \u0111\u1ec3 1. Ph\u01b0\u01a1ng tr\u00ecnh m\u1ed9t \u1ea9n H1.\u0110PKHP \u01af1 \u01a0NG TR\u00ccNH M\u1ed8T \u1ea8N 1 a) \u1ede tr\u00ean, vi\u1ebft c\u00e1c bi\u1ec3u th\u1ee9c bi\u1ec3u th\u1ecb t\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a c\u00e1c v\u1eadt tr\u00ean m\u1ed7i \u0111\u0129a c\u00e2n. T\u1eeb \u0111i\u1ec1u ki\u1ec7n c\u00e2n th\u0103ng b\u1eb1ng, hai bi\u1ec3u th\u1ee9c c\u00f3 m\u1ed1i quan h\u1ec7 nh\u01b0 th\u1ebf n\u00e0o? b) N\u1ebfu x = 200 th\u00ec c\u00e2n c\u00f3 th\u0103ng b\u1eb1ng kh\u00f4ng? T\u1ea1i sao? N\u1ebfu x = 100 th\u00ec c\u00e2n c\u00f3 th\u0103ng b\u1eb1ng kh\u00f4ng? T\u1ea1i sao? ngh\u2013i\u1ec7MmnT\u1ee5hcrco\u1ee7anua\u0111g,\u00edpcth\u1eebh\u01b0\u0111c1\u01a1\u00f3\u1ee7tnratg\u00eaanHtn,r\u00ec\u0110hdn\u1eadoKhn,Pc\u0111t\u00e2h\u01b01n\u1ef1\u1ee3:ctcGhc\u0103ihn\u00fa\u1ea5gpt bH\u1eb1Sngc\u00f3n\u00eacn\u01a1t\u1ed5hn\u1ed9gi tkr\u1ea3hi\u1ed1ingl\u01b0h\u1ee3i\u1ec7nmg ,c\u00e1thc\u1ea3vo\u1eadltut\u1eadr\u00eannvh\u1ec1api h\u0111\u01b0\u0129a\u01a1nc\u00e2gntrb\u00ec\u1eb1nnhg, l\u00e0 c\u00e1c b\u00e0i to\u00e1n t\u00ecm x m\u00e0 HS \u0111\u00e3 l\u00e0m quen \u1edf c\u00e1c l\u1edbp d\u01b0\u1edbi. GV\u2013mG\u1edd\u01a1\u0323iiH\u00fdSt\u1ed5tr\u1ea3c4hlx\u1edd\u1ee9=icc6H\u00e10c0\u0110c+K\u00e2uxP.h1\u1ecf:i Gtr(oV1n)gy\u00eaHu\u0110cK\u1ea7uPH1,Sviq\u1ebfutacn\u00e1cs\u00e1bti\u1ec3hu\u00ecnthh\u1ee9hcabi i\u0111\u1ec3\u0129uathc\u00e2\u1ecb nt\u1ed5tnrgonkgh\u1ed1Hi \u0110l\u01b0K\u1ee3n\u0110g. c\u1ee7a c\u00e1cTav\u1eadgt\u1ecdtir(\u00ea1n) ml\u00e0\u1ed7mi \u1ed9\u0111t\u0129aphc\u01b0\u00e2n\u01a1,ngl\u1edbtpr\u00ecnh\u1eadvn\u1edbix\u1ea9\u00e9nt, sG\u1ed1Vx (\u0111h\u00e1anyh\u1ea9gni\u00e1x,).ch\u1ed1t l\u1ea1i ki\u1ebfn th\u1ee9c. Ch\u00faK\u00fdh:i x = 200, hai v\u1ebf c\u1ee7a (1) c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng nhau, \u0111\u1ec1u b\u1eb1ng 800. Ta n\u00f3i s\u1ed1 200 tho\u1ea3 m\u00e3n \u2013 G(Vhoc\u1eb7\u1ea7cnnnghh\u1ea5i\u1ec7nmm\u0111\u1ea1\u00fannhg:) ph\u01b0\u01a1ng tr\u00ecnh (1). Ta c\u0169ng n\u00f3i s\u1ed1 200 (hay x = 200) l\u00e0 m\u1ed9t nghi\u1ec7m + Phc\u01b0\u1ee7a\u01a1npgh\u01b0t\u01a1r\u00ecnnghtvr\u00ec\u1edbnih\u1ea9(n1)x. c\u00f3 d\u1ea1ng A(x) = B(x), trong \u0111\u00f3 v\u1ebf tr\u00e1i A(x) v\u00e0 v\u1ebf ph\u1ea3i B(x) l\u00e0 hai bi\u1ec3Tu\u1ed5tnhg\u1ee9cquc\u00e1\u1ee7ta, pch\u00f9\u01b0n\u01a1gnmg \u1ed9trt\u00ecnbhi\u1ebfvn\u1edbxi .\u1ea9n x c\u00f3 d\u1ea1ng A(x) = B(x), trong \u0111\u00f3 v\u1ebf tr\u00e1i A(x) v\u00e0 v\u1ebf ph\u1ea3i + GBi\u00e1(xt)r\u1ecbl\u00e0ch\u1ee7aai bbii\u1ec3\u1ebfun tlh\u00e0\u1ee9mc cc\u1ee7haoc\u00f9hnagi vm\u1ebf\u1ed9tc\u1ee7bia\u1ebfnphx\u01b0. N\u01a1ngg\u01b0\u1eddtir\u00ectnahthb\u01b0\u1eb1\u1eddnngg nd\u00f9hnagu pgh\u1ecd\u01b0i\u01a1ln\u00e0gntgr\u00echnih\u1ec7mkhicn\u1ee7\u00f3ai ph\u01b0\u01a1ngv\u1ec1trv\u00ecni\u1ec7hc\u0111t\u00ec\u00f3m. x0 \u0111\u1ec3 A(x0) = B(x0). \u2013 \u1edeGVi\u00e1\u00ed tdr\u1ecb\u1ee5c1\u1ee7,a GbiV\u1ebfncl\u1ea7\u00e0nmpch\u00e2onhat\u00edicvh\u1ebfkc\u0129\u1ee7a\u0111\u1ec3phH\u01b0S\u01a1nhgi\u1ec3tru\u00ecnrh\u00f5ck\u00f3hg\u00e1ii\u00e1ntri\u1ecb\u1ec7mb\u1eb1npgh\u01b0nh\u01a1anuggt\u1ecdr\u00ecinlh\u00e0 nmg\u1ed9hit\u1ec7\u1ea9mncv\u1ee7\u00e0a nghi\u1ec7mphc\u01b0\u1ee7\u01a1anpght\u01b0r\u00ec\u01a1nnhg\u0111t\u00f3r.\u00ecnh. Ngo\u00e0i ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi \u1ea9n x, ta c\u00f3 th\u1ec3 l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi \u1ea9n y, \u1ea9n t, \u2026 Ch\u1eb3ng h\u1ea1n, 7y \u2013 4 = 2(y + 3) l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi \u1ea9n y; 3t + 5 = 2t l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi \u1ea9n t. 165 V\u00ed d\u1ee5 1. N\u0103m nay m\u1eb9 39 tu\u1ed5i, g\u1ea5p 3 l\u1ea7n tu\u1ed5i c\u1ee7a Lan n\u0103m ngo\u00e1i.","Th\u1ef1c h\u00e0nh 1. Cho ph\u01b0\u01a1ng tr\u00ecnh 4t \u2013 3 = 12 \u2013 t. Trong hai s\u1ed1 3 v\u00e0 5, c\u00f3 s\u1ed1 n\u00e0o l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho kh\u00f4ng? \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: HS tr\u1ea3i nghi\u1ec7m nh\u1eadn ra nghi\u1ec7m c\u1ee7a m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c Th\u1ef1c h\u00e0nh 1: GV cho HS ph\u00e1t bi\u1ec3u, l\u1edbp nh\u1eadn x\u00e9t, GV \u0111\u00e1nh gi\u00e1. V\u1eadn d\u1ee5ng 1. \u0110\u1eb7t l\u00ean hai \u0111\u0129a nh\u1eefng qu\u1ea3 cG\u00e2in\u1ea3inh\u01b0 H\u00ecnh 1. a) Bai)\u1ebfTt ru\u1eb1\u1ed5nigcc\u1ee7\u00e2anLtahn\u0103nng\u0103mb\u1eb1nnggo, \u00e1hi\u00e3ly\u00e0vxi\u1ebf\u2013t 1p.hT\u01b0h\u01a1enog\u0111tr\u1ec1\u00ecnbh\u00e0i, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 3(x \u2013 1) = 39. x g bi\u1ec3u th\u1ecb s\u1ef1 th\u0103ng b\u1eb1ng n\u00e0y. b) Nb)\u1ebfVu \u1edbxi x==10130, tvh\u1ebf\u00ec tcr\u00e1\u00e2inc\u1ee7ca\u00f3 pthh\u01b0\u0103n\u01a1gngbt\u1eb1r\u00ecnngh ktrh\u00ea\u00f4nncg\u00f3?gi\u00e1 tr\u1ecbx3g(11003g \u2013x1g ) = 3 . 12 = 36 \u2260 34090.g x g V\u00ec sVa\u1eadoy?13 kh\u00f4ng tho\u1ea3 m\u00e3n ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean. N\u1ebfuVv\u1ebfx\u1edbp=i hx1\u1ea35=i.01Dt4ho,\u00ec\u0111vc\u00f3\u1ebf\u00e2,nt1rc\u00e14\u00f3i lct\u00e0h\u1ee7n\u0103angpghhib\u1ec7\u01b0\u1eb1m\u01a1nngcg\u1ee7ktahr\u00ec\u00f4pnnhhg\u01b0t?\u01a1r\u00eanVng\u00ecctsr\u00f3a\u00econg?hi\u00e1trt\u00earn\u1ecb .3(14 \u2013 1) = 3 . 13 = 39, b\u1eb1ng gi\u00e1 tr\u1ecb T\u1eeb V\u0111\u1ead\u00f3y, tcuh\u1ed5\u1ec9i cr\u1ee7aa mLa\u1ed9nt nn\u0103gmhin\u1ec7amy lc\u00e0\u1ee71a4.pBh\u1ea1\u01b0n\u01a1Mngaitnr\u00ec\u00f3nih\u0111\u00fang. x g Hi\u0300nh 1 \u1edf c\u00e2u a. 200 g xg xg 200 g 200 g \u2013 MT\u1ee5hc\u1ef1\u0111c\u00edhc\u00e0hnch\u1ee7a1.VC\u1eadhnodp\u1ee5hn\u01b0g\u01a1n1g: tHr\u00ecSnhc4\u00f3tc\u2013\u01a13h=\u1ed9i12v\u1ead\u2013nt.d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c ti\u00ea\u0303n th\u00f4ng Tqruoangvih\u1ec7aci vs\u1ed1i\u1ebf3t pvh\u00e0\u01b05\u01a1, nc\u00f3g st\u1ed1r\u00ecnh\u00e0obli\u00e0\u1ec3ungthhi\u1ecb\u1ec7ms\u1ef1cc\u1ee7\u00e2anpbh\u1eb1\u01b0n\u01a1gngc\u1ee7tra\u00ecnhha\u0111i\u00e3\u0111c\u0129ahoc\u00e2knh,\u00f4ncg\u1ee7?ng c\u1ed1 th\u00eam v\u1ec1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. \u2013 GV\u1ee3\u1eadin\u00fddt\u1ed5\u1ee5nchg\u1ee91c. V\u0110\u1ead\u1eb7tnl\u00eadn\u1ee5nhgai1\u0111:\u0129aHnSh\u1eefvni\u1ebfgt qvu\u00e0\u1ea3trc\u00ec\u00e2nnhnbh\u00e0\u01b0yHk\u00ec\u1ebfnthq1u.\u1ea3 theo y\u00eau c\u1ea7u. L\u1edbp nh\u1eadn x\u00e9t, GV s\u1eedaa)bB\u00e0ii\u1ebfcthru\u1eb1nngg ctr\u00e2\u01b0n\u1edbtch\u0103l\u1edbnpg.b\u1eb1ng, h\u00e3y vi\u1ebft ph\u01b0\u01a1ng xg tr\u00ecnh bi\u1ec3u th\u1ecb s\u1ef1 th\u0103ng b\u1eb1ng n\u00e0y. xg 200 g x g 100 g x g 400 g x g H\u01b0\u1edbb)ngNd\u1ebfu\u1eabnx\u2013=\u01111\u00e10p0\u00e1tnh:\u00ec c\u00e2n c\u00f3 th\u0103ng b\u1eb1ng kh\u00f4ng? a) PVh\u00ec\u01b0s\u01a1anog? tr\u00ecnh 3x + 100 = 400 + x. b) xNT=\u1eeb\u1ebfu1\u01110x\u00f30=,, 1cc5h\u00e20n\u1ec9 trkhah\u00ec c\u00f4m\u00e2nn\u1ed9gtc\u00f3tnhtg\u0103hhn\u0103ing\u1ec7gmbb\u1eb1\u1eb1ncn\u1ee7gga(k4ph0\u00f4h0n\u01b0g\u01a1\u2260?nV5g0\u00ect0sra\u00ec)no. h? Hi\u0300nh 1 c) x\u1edf=c\u00e21u5a0., c\u00e2n th\u0103ng b\u1eb1ng (550 = 550). 2. Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n v\u00e0 c\u00e1ch gi\u1ea3i xg 200 g xg xg 200 g 200 g 2H.\u0110PKHP\u01af2\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T M\u1ed8T \u1ea8N V\u00c0x gC\u00c1CH GI\u1ea2I 400 g x g x g 100 g x g 2 X\u00e9t c\u00e2n th\u0103ng b\u1eb1ng \u1edf . x g 200 g a) Gi\u1ea3i th\u00edch t\u1ea1i sao n\u1ebfu b\u1ecf ra kh\u1ecfi m\u1ed7i \u0111\u0129a c\u00e2n m\u1ed9t qu\u1ea3 c\u00e2n x g th\u00ec c\u00e2n v\u1eabn th\u0103ng b\u1eb1ng. b) Nx \u1ebfg u10x0tghg axyg qua c\u00e2n 600 g b\u1eb14n0g0 gbaxqg u\u1ea3 c\u00e2n 200 g xg 2x )g th\u00ec c\u00e2n c\u00f2n th\u010320n0gg 2b0\u1eb10 ng g200kgh\u00f4ng? (xHg \u00ecnh T\u1ea1i sao? c) Ti\u1ebfp theo, chia c\u00e1c qu\u1ea3 c\u00e2n tr\u00ean m\u1ed7i \u0111\u0129a c\u00e2n th\u00e0nh ba ph\u1ea7n b\u1eb1ng nhau, r\u1ed3i b\u1ecf \u0111i hai ph\u1ea7n (H\u00ecnh 3). Khi \u0111\u00f3, c\u00e2n c\u00f2n th\u0103ng b\u1eb1ng kh\u00f4ng? T\u1ea1i sao? xg 200 g xg 200 g xg xg 200 g 200 g Hi\u0300nh 2 Hi\u0300nh 3 xg 200 g 166 32","\u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 2: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m, hi\u1ec3u \u0111\u01b0\u1ee3c c\u00e1ch gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1eadt m\u1ed9t \u1ea9n. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110KP 2: T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m, l\u1edbp nh\u1eadn x\u00e9t k\u1ebft qu\u1ea3 c\u1ee7a c\u00e1c nh\u00f3m, GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. Th\u1ef1c h\u00e0nh 2. Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau: a) 2 x + 1 1 = 0; \t\t\t\t\t\t\t\t\t\t\t b) 2 1 \u2212 0, 75x = 0. 32 2 Th\u1ef1c ha\u0300nh 3. Gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau: a) 15 \u2013 4x = x \u2013 5;\t\t\t\t\t\t\t\t\t\t b) 5x + 2 + 3x \u2212 2= 3 \u22c5 4 32 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 2, 3: Gi\u00fap HS r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. Ph\u01b0\u01a1ng tr\u00ecnh quy v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n \u0111\u01b0\u1ee3c \u0111\u01b0a v\u00e0o d\u01b0\u1edbi d\u1ea1ng Ch\u00fa \u00fd th\u00f4ng qua V\u00ed d\u1ee5 3, do \u0111\u00f3 GV c\u1ea7n khai th\u00e1c k\u0129 V\u00ed d\u1ee5 3 \u0111\u1ec3 gi\u00fap HS n\u1eafm v\u1eefng c\u00e1c b\u01b0\u1edbc gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh. GV ch\u00fa \u00fd trong qu\u00e1 tr\u00ecnh bi\u1ebfn \u0111\u1ed5i c\u00f3 th\u1ec3 d\u1eabn \u0111\u1ebfn ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m ho\u1eb7c nghi\u1ec7m \u0111\u00fang v\u1edbi m\u1ecdi x. Qua V\u00ed d\u1ee5 4 v\u00e0 V\u00ed d\u1ee5 5, GV c\u1ea7n ph\u00e2n t\u00edch k\u0129 \u0111\u1ec3 HS hi\u1ec3u r\u00f5 khi n\u00e0o ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m ho\u1eb7c nghi\u1ec7m \u0111\u00fang v\u1edbi m\u1ecdi x. V\u1eadn d\u1ee5ng 2. Hai b\u1ea1n An v\u00e0 Mai gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh x = 2x nh\u01b0 sau: An: \t\t \t\t x = 2x \t \t \t \t \t \t \t 1 = 2. \t (chia hai v\u1ebf cho x) V\u00e2\u0323y ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. Mai: \t\t\t \t x = 2x \t \t \t \t x \u2013 2x = 0\t\t (chuy\u1ec3n 2x sang v\u1ebf tr\u00e1i) \t\t\t\t\t\t \u2013x = 0\t\t (r\u00fat g\u1ecdn) \t\t\t\t\t\t\t x =0.\t\t (nh\u00e2n hai v\u1ebf v\u01a1\u0301i \u20131) V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m x = 0. Em h\u00e3y cho bi\u1ebft b\u1ea1n n\u00e0o gi\u1ea3i \u0111\u00fang. M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 2: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c ti\u1ec5n khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, c\u1ee7ng c\u1ed1 th\u00eam v\u1ec1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh, r\u00e8n luy\u1ec7n ki\u1ebfn th\u1ee9c theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \t5x + 450 = 700. 2. \ta) 7x + 4 = 0 l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n v\u1edbi a = 7 v\u00e0 b = 4 .\t 7 7 b) 3 y\u22125 = 4, chuy\u1ec3n v\u1ebf ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh 3 y\u22129 = 0 l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t 2 2 m\u1ed9t \u1ea9n v\u1edbi a = 3 v\u00e0 b = \u20139.\t 2 167","c) 0t + 6 = 0 v\u00e0 d) x2 + 3 = 0 kh\u00f4ng l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n. 3. a) 5x \u2013 30 = 0 b) 4 \u2013 3x = 11 5x = 30 \u20133x = 7 x = 6; x= \u22127 ; 3 c) 3x + x + 20 = 0 4x + 20 = 0 d) 1 x + 1 = x + 2 4x = \u201320 32 x = \u2013 5; \u22122 x = 3 3 2 x= \u22129 . 4 4. a) 8 \u2013 (x \u2013 15) = 2(3 \u2013 2x) b) \u2013 6(1,5 \u2013 2u) = 3(\u201315 + 2u) \u20139 + 12u = \u2013 45 + 6u 8 \u2013 x + 15 = 6 \u2013 4x 6u = \u201336 u = \u2013 6; 3x = \u201317 x= \u221217 ; 3 c) (x + 3)2 \u2013 x(x + 4) = 13 d) (y + 5)(y \u2013 5) \u2013 (y \u2013 2)2 = \u20135 x2 + 6x + 9 \u2013 x2 \u2013 4x = 13 y2 \u2013 25 \u2013 y2 + 4y \u2013 4 = \u20135 2x = 4 4y = 24 x = 2; y = 6. 5. a) 5x \u2212 3 = x+2 b) 9x + 5 = 1 \u2212 6 + 3x 4 3 6 8 3(5x \u2013 3) = 4(x + 2) 4(9x + 5) = 24 \u2013 3(6 + 3x) 15x \u2013 9 = 4x + 8 36x + 20 = 24 \u2013 18 \u2013 9x 11x = 17 45x = \u201314 x = 17 ; \t x = \u2212 14 ; 11 45 c) 2(x + 1) \u2212 1 = 1 + 3x d) x+3 \u2212 2 x= 3 3 2 4 5 3 10 8(x + 1) \u2013 6 = 3(1 + 3x) 6(x + 3) \u2013 20x = 9 8x + 8 \u2013 6 = 3 + 9x 6x + 18 \u2013 20x = 9 \u2013x = 1 \u221214x = \u2212 9 x = \u20131; 9 x = 14 . 168","6. \t\uf8eb\uf8ed\uf8ec x \u2212 1 \uf8f6 \u22c5 1 = 1 2 \uf8f7\uf8f8 2 8 x\u2013 1 = 1 2 4 x = 3. 4 B\u00e0i 2. GI\u1ea2I B\u00c0I TO\u00c1N B\u1eb0NG C\u00c1CH L\u1eacP PH\u01af\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t: \u2212 Bi\u1ec3u di\u1ec5n \u0111\u01b0\u1ee3c m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng b\u1edfi bi\u1ec3u th\u1ee9c ch\u1ee9a \u1ea9n. \u2212 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t. 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc; m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc; s\u1eed d\u1ee5ng c\u00f4ng c\u1ee5, ph\u01b0\u01a1ng ti\u1ec7n h\u1ecdc to\u00e1n. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd GV c\u1ea7n gi\u1ea3ng ch\u1eadm, kh\u1eafc s\u00e2u ph\u1ea7n \u201c1. Bi\u1ec3u di\u1ec5n m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng b\u1edfi bi\u1ec3u th\u1ee9c ch\u1ee9a \u1ea9n\u201d. T\u1ed5 ch\u1ee9c l\u1edbp th\u1ea3o lu\u1eadn nh\u00f3m, k\u1ebft h\u1ee3p l\u00e0m vi\u1ec7c c\u00e1 nh\u00e2n gi\u00fap HS t\u00ecm ra m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng, thi\u1ebft l\u1eadp \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 Sau khi gi\u1ea3m gi\u00e1 15% th\u00ec \u0111\u00f4i gi\u00e0y Gia\u0309m gia\u0301 15% th\u1ec3 thao c\u00f3 gi\u00e1 l\u00e0 1 275 000 \u0111\u1ed3ng. H\u1ecfi l\u00fac ch\u01b0a gi\u1ea3m gi\u00e1 th\u00ec \u0111\u00f4i gi\u00e0y c\u00f3 gi\u00e1 l\u00e0 bao nhi\u00eau? \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110: K\u00edch th\u00edch HS t\u00f2 m\u00f2, t\u01b0 duy, t\u00ecm hi\u1ec3u v\u1ec1 vi\u1ec7c gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110K\u0110: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV s\u1eed d\u1ee5ng c\u01a1 h\u1ed9i \u0111\u1ec3 gi\u1edbi thi\u1ec7u b\u00e0i. 169","H\u01b0\u1edbng d\u1eabn \u2212 \u0111\u00e1p \u00e1n: x . (100% \u2013 15%) = 1 275 000 \t\t\t \t x = 1 500 000. 1. Bi\u1ec3u di\u1ec5n m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng b\u1edfi bi\u1ec3u th\u1ee9c ch\u1ee9a \u1ea9n H\u0110KP 1 \u2212 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 1: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m, th\u1ea3o lu\u1eadn \u0111\u1ec3 hi\u1ec3u \u0111\u01b0\u1ee3c c\u00e1ch bi\u1ec3u di\u1ec5n m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng b\u1edfi bi\u1ec3u th\u1ee9c ch\u1ee9a \u1ea9n. \u2212 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110KP 1: + T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m. + L\u1edbp nh\u1eadn x\u00e9t k\u1ebft qu\u1ea3 c\u1ee7a c\u00e1c nh\u00f3m. + GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. V\u00ed d\u1ee5 1. M\u1ed9t \u00f4 t\u00f4 kh\u1edfi h\u00e0nh t\u1eeb th\u00e0nh ph\u1ed1 A \u0111\u1ebfn th\u00e0nh ph\u1ed1 B v\u1edbi t\u1ed1c \u0111\u00f4\u0323 40 km\/h. Khi t\u1eeb B quay v\u1ec1 A xe ch\u1ea1y v\u1edbi t\u1ed1c \u0111\u00f4\u0323 50 km\/h. G\u1ecdi x (km) l\u00e0 chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng AB. Vi\u1ebft bi\u1ec3u th\u1ee9c bi\u1ec3u th\u1ecb: a) Th\u1eddi gian \u00f4 t\u00f4 \u0111i t\u1eeb A \u0111\u1ebfn B; b) T\u1ed5ng th\u1eddi gian \u00f4 t\u00f4 \u0111i t\u1eeb A \u0111\u1ebfn B v\u00e0 t\u1eeb B v\u1ec1 A. \u2212 M\u1ee5c \u0111\u00edch c\u1ee7a V\u00ed d\u1ee5 1: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m c\u00e1ch bi\u1ec3u di\u1ec5n m\u1ed9t \u0111\u1ea1i l\u01b0\u1ee3ng b\u1edfi bi\u1ec3u th\u1ee9c ch\u1ee9a \u1ea9n. \u2212 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c V\u00ed d\u1ee5 1: GV cho HS nh\u1eafc l\u1ea1i c\u00e1c c\u00f4ng th\u1ee9c t\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng, v\u1eadn t\u1ed1c, th\u1eddi gian. Tr\u00ean c\u01a1 s\u1edf c\u00e1c c\u00f4ng th\u1ee9c \u0111\u00e3 bi\u1ebft, GV g\u1ee3i \u00fd HS thi\u1ebft l\u1eadp bi\u1ec3u th\u1ee9c li\u00ean quan \u0111\u1ebfn c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng c\u1ee7a b\u00e0i to\u00e1n. Th\u1ef1c h\u00e0nh 1. Ti\u1ec1n l\u01b0\u01a1ng c\u01a1 b\u1ea3n c\u1ee7a anh Minh m\u1ed7i th\u00e1ng l\u00e0 x (tri\u1ec7u \u0111\u1ed3ng). Ti\u1ec1n ph\u1ee5 c\u1ea5p m\u1ed7i th\u00e1ng l\u00e0 3 500 000 \u0111\u1ed3ng. a) Vi\u1ebft bi\u1ec3u th\u1ee9c bi\u1ec3u th\u1ecb ti\u1ec1n l\u01b0\u01a1ng m\u1ed7i th\u00e1ng c\u1ee7a anh Minh. Bi\u1ebft ti\u1ec1n l\u01b0\u01a1ng m\u1ed7i th\u00e1ng b\u1eb1ng t\u1ed5ng ti\u1ec1n l\u01b0\u01a1ng c\u01a1 b\u1ea3n v\u00e0 ti\u1ec1n ph\u1ee5 c\u1ea5p. b) Th\u00e1ng T\u1ebft, anh Minh \u0111\u01b0\u1ee3c th\u01b0\u1edfng 1 th\u00e1ng l\u01b0\u01a1ng c\u00f9ng v\u1edbi 60% ti\u1ec1n ph\u1ee5 c\u1ea5p. Vi\u1ebft bi\u1ec3u th\u1ee9c ch\u1ec9 s\u1ed1 ti\u1ec1n anh Minh \u0111\u01b0\u1ee3c nh\u1eadn \u1edf th\u00e1ng T\u1ebft. \u2212 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: Qua th\u1ef1c h\u00e0nh gi\u00fap HS r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. 170","\u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c Th\u1ef1c h\u00e0nh 1: + T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m. + L\u1edbp nh\u1eadn x\u00e9t k\u1ebft qu\u1ea3 c\u1ee7a c\u00e1c nh\u00f3m. + GV \u0111\u00e1nh gi\u00e1. H\u01b0\u1edbng d\u1eabn \u2212 \u0111\u00e1p \u00e1n: a) x + 3 500 000. b) x + 3 500 000 + x + 3 500 000 + 60% . 3 500 000 = 2x + 9 100 000. 2. Gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t 2H.\u0110GKIP\u1ea22I B\u00c0I TO\u00c1N B\u1eb0NG C\u00c1CH L\u1eacP PH\u01af\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T 2 Thay d\u1ea5u ? b\u1eb1ng c\u00e1c d\u1eef li\u1ec7u th\u00edch h\u1ee3p \u0111\u1ec3 ho\u00e0n th\u00e0nh l\u1eddi gi\u1ea3i b\u00e0i to\u00e1n. M\u1ed9t ng\u01b0\u1eddi \u0111i xe g\u1eafn m\u00e1y t\u1eeb A \u0111\u1ebfn B v\u1edbi t\u1ed1c \u0111\u1ed9 40 km\/h. L\u00fac v\u1ec1 ng\u01b0\u1eddi \u0111\u00f3 \u0111i v\u1edbi t\u1ed1c \u0111\u1ed9 50 km\/h n\u00ean th\u1eddi gian v\u1ec1 \u00edt h\u01a1n th\u1eddi gian \u0111i l\u00e0 30 ph\u00fat. T\u00ecm chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng AB. Gi\u1ea3i G\u1ecdi chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0 x (km). \u0110i\u1ec1u ki\u1ec7n x > ? . Th\u1eddi gian \u0111i l\u00e0: x gi\u1edd. 40 Th\u1eddi gian v\u1ec1 l\u00e0: ? . Ta c\u00f3: 30 ph\u00fat = 1 gi\u1edd. 2 V\u00ec th\u1eddi gian v\u1ec1 \u00edt h\u01a1n th\u1eddi gian \u0111i l\u00e0 1 gi\u1edd n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x \u2013?=1 2 40 2 Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = ? tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > ? . V\u1eady chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0 ? . \u2013 MT\u1ee5\u00f3cm\u0111t\u00ed\u1eafcthcc\u00e1\u1ee7cab\u01b0H\u1edb\u0110cKgiP\u1ea3i2b:\u00e0Gi ito\u00fa\u00e1pnHbS\u1eb1ncg\u00f3cc\u00e1\u01a1chh\u1ed9l\u1eadiptrp\u1ea3hi\u01b0n\u01a1gnhgi\u1ec7trm\u00ecn,ht:h\u1ea3o lu\u1eadn v\u1ec1 vi\u1ec7c bi\u1ec3u di\u1ec5n m\u1ed9t \u0111\u1ea1Bi\u01b0l\u1edb\u01b0c\u1ee31n.gLb\u1eadp\u1edfipbh\u01b0i\u1ec3\u01a1ungtht\u1ee9r\u00eccnhc.h\u1ee9a \u1ea9n, l\u1eadp \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh, gi\u1ea3i \u0111\u01b0\u1ee3c b\u00e0i to\u00e1n theo y\u00eau c\u1ea7\u2013u.Ch\u1ecdn \u1ea9n s\u1ed1 v\u00e0 \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n th\u00edch h\u1ee3p cho \u1ea9n s\u1ed1. \u2013 G\u2013\u01a1\u0323iB\u00fdi\u1ec3tu\u1ed5dcih\u1ec5n\u1ee9c\u00e1Hc \u0110\u0111\u1ea1Ki Pl\u01b02\u1ee3:ng ch\u01b0a bi\u1ebft theo \u1ea9n v\u00e0 theo c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng \u0111\u00e3 bi\u1ebft. + T\u2013\u1ed5 Lch\u1eadp\u1ee9cphth\u01b0\u1ea3\u01a1onglut\u1eadr\u00ecnnhnhb\u00f3i\u1ec3mu.di\u1ec5n m\u1ed1i quan h\u1ec7 gi\u1eefa c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng. B\u01b0\u1edbc 2. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh. + LB\u1edb\u01b0p\u1edbnch3\u1ead.nTxr\u1ea3\u00e9tl\u1eddki\u1ebf.t qu\u1ea3 c\u1ee7a c\u00e1c nh\u00f3m. + G\u2013VK\u0111i\u00e1\u1ec3nmhtgrai\u00e1x, ecmh\u1ed1ttrokni\u1ebfgnct\u00e1hc\u1ee9ncg. hi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh, nghi\u1ec7m n\u00e0o tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n + Tc\u00f3\u1ee7ma \u1ea9t\u1eafnt, cn\u00e1gchib\u1ec7\u01b0m\u1edbnc\u00e0\u0111o\u1ec3kghi\u00f4\u1ea3nigb.\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t. \u2013 K\u1ebft lu\u1eadn. V\u00ed d\u1ee5 2. B\u00e1c Thanh g\u1eedi 300 000 000 \u0111\u1ed3ng v\u00e0o m\u1ed9t ng\u00e2n h\u00e0ng v\u1edbi k\u00ec h\u1ea1n m\u00f4\u0323t n\u0103m. SauVm\u00ed \u00f4d\u0323t\u1ee5n2\u0103.mBb\u00e1\u00e1ccTrh\u00faatnvh\u1ec1 g\u0111\u1eed\u01b0i\u1ee33c0c0\u1ea30v0\u1ed10n0l0\u1eab0n \u0111l\u00e3\u1ed3inlg\u00e0 v3\u00e01o8 6m0\u1ed90t0n0g0\u00e2n\u0111\u1ed3hn\u00e0gn.gTv\u00edn\u1edbhi kl\u00e3\u00ecihs\u1ea1un\u1ea5tmm\u1ed9\u1ed9t tnn\u0103\u0103mm. c\u1ee7aSkahuom\u1ea3n\u1ed9ttin\u1ec1\u0103nmbb\u00e1\u00e1ccTrh\u00faat nvh\u1ec1 \u0111g\u01b0\u1eed\u1ee3i c\u1edfcn\u1ea3gv\u00e2\u1ed1nnhl\u00e0\u1eabnngl\u00e3\u0111i \u00f3l\u00e0. 318 600 000 \u0111\u1ed3ng. T\u00ednh l\u00e3i su\u1ea5t m\u1ed9t n\u0103m c\u1ee7a kho\u1ea3n ti\u1ec1n b\u00e1c Thanh g\u1eedi \u1edf ng\u00e2n h\u00e0ng \u0111\u00f3. Gi\u1ea3i G\u1ecdi l\u00e3i su\u1ea5t ti\u1ec1n g\u1eedi m\u1ed9t n\u0103m l\u00e0 x. \u0110i\u1ec1u ki\u1ec7n: x > 0. 171 Ti\u1ec1n l\u00e3i sau m\u1ed9t n\u0103m g\u1eedi l\u00e0: 300 000 000 . x (\u0111\u1ed3ng). V\u00ec c\u1ea3 v\u1ed1n l\u1eabn l\u00e3i sau m\u1ed9t n\u0103m l\u00e0 318 600 000 \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:","\u2212 M\u1ee5c \u0111\u00edch c\u1ee7a V\u00ed d\u1ee5 2: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m gi\u1ea3i b\u00e0i to\u00e1n th\u1ef1c t\u1ebf b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t. \u2212 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c V\u00ed d\u1ee5 2: GV cho HS l\u00e0m vi\u1ec7c c\u00e1 nh\u00e2n t\u1ef1 thi\u1ebft l\u1eadp c\u00e1c b\u01b0\u1edbc gi\u1ea3i b\u00e0i to\u00e1n. L\u1edbp nh\u1eadn x\u00e9t, GV \u0111\u00e1nh gi\u00e1. Th\u1ef1c h\u00e0nh 2. M\u1ed9t ng\u01b0\u1eddi mua 36 b\u00f4ng hoa h\u1ed3ng v\u00e0 b\u00f4ng hoa c\u1ea9m ch\u01b0\u1edbng h\u1ebft t\u1ea5t c\u1ea3 136 800 \u0111\u1ed3ng. Gi\u00e1 m\u1ed7i b\u00f4ng hoa h\u1ed3ng l\u00e0 3 000 \u0111\u1ed3ng, gi\u00e1 m\u1ed7i b\u00f4ng hoa c\u1ea9m ch\u01b0\u1edbng l\u00e0 4 800 \u0111\u1ed3ng. T\u00ednh s\u1ed1 b\u00f4ng hoa m\u1ed7i lo\u1ea1i. Hoa h\u00f4\u0300ng Hoa c\u00e2\u0309m ch\u01b0\u01a1\u0301ng 3 000 \u0111\u00f4\u0300ng\/b\u00f4ng 4 800 \u0111\u00f4\u0300ng\/b\u00f4ng \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 2: Qua th\u1ef1c h\u00e0nh gi\u00fap HS r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c Th\u1ef1c h\u00e0nh 2: + T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m. + L\u1edbp nh\u1eadn x\u00e9t k\u1ebft qu\u1ea3 c\u1ee7a c\u00e1c nh\u00f3m. + GV \u0111\u00e1nh gi\u00e1. H\u01b0\u1edbng d\u1eabn \u2212 \u0111\u00e1p \u00e1n: G\u1ecdi x (b\u00f4ng) l\u00e0 s\u1ed1 b\u00f4ng hoa h\u1ed3ng (0 < x < 36, x \u2208 \uf0a5). S\u1ed1 b\u00f4ng hoa c\u1ea9m ch\u01b0\u1edbng l\u00e0: 36 \u2013 x (b\u00f4ng). Ph\u01b0\u01a1ng tr\u00ecnh: 3 000x + 4 800. (36 \u2013 x) = 136 800. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 20 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n. V\u1eady s\u1ed1 b\u00f4ng hoa h\u1ed3ng l\u00e0 20 b\u00f4ng, s\u1ed1 b\u00f4ng hoa c\u1ea9m ch\u01b0\u1edbng l\u00e0 16 b\u00f4ng. V\u1eadn d\u1ee5ng. Gi\u1ea3i b\u00e0i to\u00e1n \u0111\u00e3 cho trong (trang 37). \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c ti\u00ea\u0303n th\u00f4ng qua vi\u1ec7c t\u00ecm gi\u00e1 ch\u01b0a gi\u1ea3m c\u1ee7a \u0111\u00f4i gi\u00e0y b\u1eb1ng c\u00e1ch l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c V\u1eadn d\u1ee5ng: + HS vi\u1ebft v\u00e0 tr\u00ecnh b\u00e0y k\u1ebft qu\u1ea3 theo y\u00eau c\u1ea7u. + L\u1edbp nh\u1eadn x\u00e9t, GV s\u1eeda b\u00e0i chung tr\u01b0\u1edbc l\u1edbp. H\u01b0\u1edbng d\u1eabn \u2212 \u0111\u00e1p \u00e1n: x . (100% \u2212 15%) = 1 275 000. Suy ra x = 1 500 000. 172","IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \tG\u1ecdi x l\u00e0 s\u1ed1 \u0111\u01a1n h\u00e0ng \u0111\u01b0\u1ee3c giao trong ng\u00e0y th\u1ee9 nh\u1ea5t. \u0110i\u1ec1u ki\u1ec7n: 0 < x < 95, x \u2208 \uf0a5. S\u1ed1 \u0111\u01a1n h\u00e0ng giao trong ng\u00e0y th\u1ee9 hai l\u00e0 x + 15 (\u0111\u01a1n h\u00e0ng). Ph\u01b0\u01a1ng tr\u00ecnh: x + (x + 15) = 95. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 40 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n 0 < x < 95, x \u2208 \uf0a5. V\u1eady s\u1ed1 \u0111\u01a1n h\u00e0ng \u0111\u01b0\u1ee3c giao trong ng\u00e0y th\u1ee9 nh\u1ea5t l\u00e0 40 \u0111\u01a1n h\u00e0ng. 2. \tG\u1ecdi x (ph\u00fat) l\u00e0 th\u1eddi gian ch\u1ea1y b\u1ed9 c\u1ee7a anh B\u00ecnh. \u0110i\u1ec1u ki\u1ec7n: 0 < x < 40. Th\u1eddi gian b\u01a1i c\u1ee7a anh B\u00ecnh l\u00e0 40 \u2013 x (ph\u00fat). Ph\u01b0\u01a1ng tr\u00ecnh: 10x + (40 \u2013 x) . 14 = 500. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 15 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n 0 < x < 40. V\u1eady th\u1eddi gian ch\u1ea1y b\u1ed9 c\u1ee7a anh B\u00ecnh l\u00e0 15 ph\u00fat. 3. \tG\u1ecdi x (kg) l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng g\u1ea1o b\u00e1n \u0111\u01b0\u1ee3c trong ng\u00e0y th\u1ee9 nh\u1ea5t. \u0110i\u1ec1u ki\u1ec7n: x > 560. Kh\u1ed1i l\u01b0\u1ee3ng g\u1ea1o b\u00e1n trong ng\u00e0y th\u1ee9 hai l\u00e0 x \u2013 560 (kg). Ph\u01b0\u01a1ng tr\u00ecnh: x + 60 = 1,5 . (x \u2013 560). Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 1 800 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 560. S\u1ed1 g\u1ea1o b\u00e1n \u0111\u01b0\u1ee3c trong ng\u00e0y th\u1ee9 nh\u1ea5t l\u00e0 1 800 kg. 4. \tG\u1ecdi chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0 x (km). \u0110i\u1ec1u ki\u1ec7n: x > 0. Ph\u01b0\u01a1ng tr\u00ecnh: x + x = 27 . 50 40 5 Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 120 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng AB d\u00e0i 120 km. 5. G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n b\u00e1c N\u0103m g\u1eedi ti\u1ebft ki\u1ec7m. \u0110i\u1ec1u ki\u1ec7n: x > 0. Ph\u01b0\u01a1ng tr\u00ecnh: (106,2%)2 . x = 225 568 800. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 200 000 000 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady B\u00e1c N\u0103m g\u1eedi ti\u1ebft ki\u1ec7m 200 000 000 \u0111\u1ed3ng. 6. \tG\u1ecdi x l\u00e0 s\u1ed1 h\u1ecdc sinh kh\u1ed1i 8. \u0110i\u1ec1u ki\u1ec7n: 0 < x < 580, x \u2208 \uf0a5. S\u1ed1 h\u1ecdc sinh kh\u1ed1i 9 l\u00e0 580 \u2013 x (h\u1ecdc sinh). Ph\u01b0\u01a1ng tr\u00ecnh: 40% . x + (580 \u2013 x) . 48% = 256. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 280 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n 0 < x < 580, x \u2208 \uf0a5. V\u1eady s\u1ed1 h\u1ecdc sinh kh\u1ed1i 8 l\u00e0 280 h\u1ecdc sinh, s\u1ed1 h\u1ecdc sinh kh\u1ed1i 9 l\u00e0 300 h\u1ecdc sinh. 173","7. \tG\u1ecdi x l\u00e0 s\u1ed1 gam dung d\u1ecbch trong l\u1ecd l\u00fac \u0111\u1ea7u. \u0110i\u1ec1u ki\u1ec7n: x > 0. Kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch l\u00fac sau l\u00e0 x + 350 (g). Ph\u01b0\u01a1ng tr\u00ecnh: (x + 350) . 5% = x . 12%. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 250 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch trong l\u1ecd l\u00fac \u0111\u1ea7u l\u00e0 250 g. 8. \tG\u1ecdi x (\u0111\u1ed3ng) l\u00e0 gi\u00e1 \u0111i\u1ec7n m\u1ee9c 1. \u0110i\u1ec1u ki\u1ec7n: x > 0. S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 \u1edf m\u1ee9c 1 l\u00e0 50x (\u0111\u1ed3ng). S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 \u1edf m\u1ee9c 2 l\u00e0 50(x + 56) (\u0111\u1ed3ng). S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 \u1edf m\u1ee9c 3 l\u00e0 85(x + 56 + 280) = 85(x + 336) (\u0111\u1ed3ng). S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 ch\u01b0a t\u00ednh thu\u1ebf VAT l\u00e0 375 969 : 110% = 341 790 (\u0111\u1ed3ng). Ph\u01b0\u01a1ng tr\u00ecnh: 50x + 50(x + 56) + 85(x + 336) = 341 790. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 1 678 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady gi\u00e1 m\u1ed7i s\u1ed1 \u0111i\u1ec7n \u1edf m\u1ee9c th\u1ee9 3 l\u00e0 1 678 + 56 + 280 = 2 014 (\u0111\u1ed3ng). B\u00c0I T\u1eacP CU\u1ed0I CH\u01af\u01a0NG 6 C\u00c2U H\u1eceI TR\u1eaeC NGHI\u1ec6M 1. D 2. B 3. B 4. C 5. C 6. C B\u00c0I T\u1eacP T\u1ef0 LU\u1eacN 7. \ta) x = 3;\t \t \t \t \t b) y = \u20135; \t \t \t \t \t \t c) x = 13; \t \t \t \t \t \t d) x = \u2013 2 . 3 8. \ta) x = \u20135; \t \t \t \t b) x = \u22125 ;\t \t \t c) x = \u221216 ;\t\t \t \t \t \t d) x = 20 . 2 7 3 9. \ta) 3x \u22121 = 3 + 2x \t \t \t \t \t \t \t \t \t \t \t b) x+5 =1\u2212 x\u22122 6 3 3 4 3x \u2013 1 = 2(3 + 2x)\t\t \t \t \t \t \t \t \t 4(x + 5) = 12 \u2013 3(x \u2013 2) x = \u20137; \t \t \t \t \t \t \t \t \t \t \t \t \t \t \t\t x = \u22122 ; \t\t\t\t 7 c) 3x \u2212 2 + 3 = 4 \u2212 x \t \t \t \t \t \t \t \t d) x + 2x +1 = 4(x \u2212 2) 5 2 10 36 5 2(3x \u2013 2) + 15 = 4 \u2013 x \t \t \t\t \t 10x + 5(2x + 1) = 24(x \u2013 2) x = \u20131;\t \t \t \t \t \t \t \t\t \t \tx = 53 . 4 174","10. G\u1ecdi x l\u00e0 s\u1ed1 \u00e1o ph\u1ea3i may theo k\u1ebf ho\u1ea1ch. \u0110i\u1ec1u ki\u1ec7n: x \u2208 \uf0a5*. Ph\u01b0\u01a1ng tr\u00ecnh: x \u2212 x + 20 = 3. 30 40 Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 420 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u2208 \uf0a5*. V\u1eady s\u1ed1 \u00e1o ph\u1ea3i may theo k\u1ebf ho\u1ea1ch l\u00e0 420 chi\u1ebfc. 11. G\u1ecdi x l\u00e0 s\u1ed1 c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang. \u0110i\u1ec1u ki\u1ec7n: 0 < x < 50, x \u2208 \uf0a5. S\u1ed1 c\u00e2u tr\u1ea3 l\u1eddi sai (ho\u1eb7c kh\u00f4ng tr\u1ea3 l\u1eddi) l\u00e0 50 \u2013 x (c\u00e2u). Ph\u01b0\u01a1ng tr\u00ecnh: 5x + (50 \u2013 x ).(\u2013 2) = 194. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 42 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n 0 < x < 50, x \u2208 \uf0a5. V\u1eady s\u1ed1 c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang l\u00e0 42 c\u00e2u. 12. G\u1ecdi x l\u00e0 s\u1ed1 gam n\u01b0\u1edbc c\u1ea7n th\u00eam. \u0110i\u1ec1u ki\u1ec7n: x > 0. Kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch l\u00fac sau l\u00e0 x + 500 (g). Ph\u01b0\u01a1ng tr\u00ecnh: (x + 500) . 20% = 150. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 250 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady l\u01b0\u1ee3ng n\u01b0\u1edbc c\u1ea7n th\u00eam v\u00e0o l\u00e0 250 g. 13. G\u1ecdi x (km) l\u00e0 chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng AB. \u0110i\u1ec1u ki\u1ec7n: x > 0. Th\u1eddi gian d\u1ef1 \u0111\u1ecbnh \u0111i h\u1ebft qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0 x (gi\u1edd). 50 Th\u1eddi gian th\u1ef1c t\u1ebf \u0111i 2 qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0 2 x : 50 = 1 x (gi\u1edd). 3 3 75 Th\u1eddi gian th\u1ef1c t\u1ebf \u0111i qu\u00e3ng \u0111\u01b0\u1eddng c\u00f2n l\u1ea1i l\u00e0 1 x : 40 = 1 x (gi\u1edd). 3 120 Ph\u01b0\u01a1ng tr\u00ecnh: x + x \u2212 x = 1 . 75 120 50 2 Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 300 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng AB d\u00e0i 300 km. 14. G\u1ecdi x (m) l\u00e0 chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt. \u0110i\u1ec1u ki\u1ec7n: x > 2. Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 3x (m). Ph\u01b0\u01a1ng tr\u00ecnh: 3x2 \u2013 (3x + 3)(x \u2013 2) = 90. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 28 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 2. V\u1eady chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 28 m, chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 84 m. 175","15. G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 ti\u1ec1n l\u01b0\u01a1ng c\u1ee7a m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng. \u0110i\u1ec1u ki\u1ec7n: x > 0. Ti\u1ec1n l\u01b0\u01a1ng c\u1ee7a m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c t\u0103ng ca l\u00e0 x + 200 000 (\u0111\u1ed3ng). Ph\u01b0\u01a1ng tr\u00ecnh: 24x + 4(x + 200 000) = 7 800 000. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 250 000 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 0. V\u1eady ti\u1ec1n l\u01b0\u01a1ng c\u1ee7a m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng l\u00e0 250 000 \u0111\u1ed3ng. 16. G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 gi\u00e1 ti\u1ec1n t\u1ee7 l\u1ea1nh l\u00fac ch\u01b0a gi\u1ea3m gi\u00e1. \u0110i\u1ec1u ki\u1ec7n: x > 12 800 000. Ph\u01b0\u01a1ng tr\u00ecnh: x . 80% . 80% = 12 800 000. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c x = 20 000 000 tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n x > 12 800 000. V\u1eady t\u1ee7 l\u1ea1nh l\u00fac ch\u01b0a gi\u1ea3m c\u00f3 gi\u00e1 l\u00e0 20 000 000 \u0111\u1ed3ng. 176","Ph\u1ea7n HI\u0300NH HO\u0323C VA\u0300 \u0110O L\u01af\u01a0\u0300NG \t H\u00ccNH H\u1eccC PH\u1eb2NG Ch\u01b0\u01a1ng 7 \u0110\u1ecaNH L\u00cd THAL\u00c8S A. M\u1ee4C TI\u00caU 1. N\u0103ng l\u01b0\u0323c chuy\u00ean m\u00f4n \u0110\u1ecbnh l\u00ed Thal\u00e8s \u2212 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c \u0111\u1ecbnh l\u00ed Thal\u00e8s trong tam gi\u00e1c (\u0111\u1ecbnh l\u00ed thu\u1eadn v\u00e0 \u0111\u1ea3o). \u2212 T\u00ednh \u0111\u01b0\u1ee3c \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng b\u1eb1ng c\u00e1ch s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s. \u2212 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi vi\u1ec7c v\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s. \u0110\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c \u2212 M\u00f4 t\u1ea3 \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c. \u2212 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a tam gi\u00e1c \u2212 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a tam gi\u00e1c. 2. N\u0103ng l\u01b0\u0323c chung \u2013 N\u0103ng l\u01b0\u0323c t\u01b0\u0323 chu\u0309 va\u0300 t\u01b0\u0323 ho\u0323c trong ti\u0300m to\u0300i, kha\u0301m pha\u0301. \u2013 N\u0103ng l\u01b0\u0323c giao ti\u00ea\u0301p va\u0300 h\u01a1\u0323p ta\u0301c trong tri\u0300nh ba\u0300y, tha\u0309o lu\u00e2\u0323n va\u0300 la\u0300m vi\u00ea\u0323c nho\u0301m. \u2013 N\u0103ng l\u01b0\u0323c gia\u0309i quy\u00ea\u0301t v\u00e2\u0301n \u0111\u00ea\u0300 va\u0300 sa\u0301ng ta\u0323o trong th\u01b0\u0323c ha\u0300nh va\u0300 v\u00e2\u0323n du\u0323ng. 3. Hi\u0300nh tha\u0300nh ca\u0301c ph\u00e2\u0309m ch\u00e2\u0301t \u2013 Y\u00eau n\u01b0\u01a1\u0301c, nh\u00e2n \u00e1i. \u2013 Ch\u0103m chi\u0309, trung th\u01b0\u0323c, tra\u0301ch nhi\u00ea\u0323m. B. H\u01af\u1edaNG D\u1eaaN D\u1ea0Y H\u1eccC B\u00e0i 1. \u0110\u1ecaNH L\u00cd THAL\u00c8S TRONG TAM GI\u00c1C I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u00e2\u0300n \u0111a\u0323t: \u2013 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c \u0111\u1ecbnh l\u00ed Thal\u00e8s thu\u1eadn v\u00e0 \u0111\u1ea3o trong tam gi\u00e1c. \u2013 T\u00ednh \u0111\u01b0\u1ee3c \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng b\u1eb1ng c\u00e1ch s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi vi\u1ec7c v\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s (v\u00ed d\u1ee5: t\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai v\u1ecb tr\u00ed, ...). 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc, m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc, giao ti\u00ea\u0301p toa\u0301n ho\u0323c. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng, t\u00edch h\u1ee3p c\u00e1c m\u00f4n h\u1ecdc kh\u00e1c. 177","II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd 1. GV gi\u1edbi thi\u1ec7u c\u00e1c t\u00ecnh hu\u1ed1ng c\u1ee5 th\u1ec3 gi\u00fap HS l\u00e0m quen v\u1edbi kh\u00e1i ni\u1ec7m t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng v\u00e0 kh\u00e1i ni\u1ec7m c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7. 2. GV c\u1ea7n t\u1eadp trung v\u00e0o c\u00e1c \u00e1p d\u1ee5ng c\u1ee7a \u0111\u1ecbnh l\u00ed Thal\u00e8s trong th\u1ef1c t\u1ebf nh\u01b0 t\u00ednh \u0111\u01b0\u1ee3c \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng b\u1eb1ng c\u00e1ch s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s. 3. Ch\u01b0\u01a1ng tr\u00ecnh To\u00e1n 8 kh\u00f4ng y\u00eau c\u1ea7u ch\u1ee9ng minh \u0111\u1ecbnh l\u00ed Thal\u00e8s, ch\u1ec9 n\u00eau nh\u1eadn x\u00e9t v\u00e0 r\u00fat ra \u0111\u1ecbnh l\u00ed. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 Nh\u1eefng s\u1ee3i c\u00e1p treo c\u1ee7a c\u1ea7u thu\u1eadn Ph\u01b0\u1edbc (thu\u1ed9c tha\u0300nh ph\u00f4\u0301 \u0110\u00e0 N\u1eb5ng) cho ta h\u00ecnh \u1ea3nh nh\u1eefng \u0111o\u1ea1n th\u1eb3ng song song. C\u00e1c \u0111o\u1ea1n th\u1eb3ng AA\u2032, BB\u2032, CC\u2032 th\u1ec3 hi\u1ec7n ba s\u1ee3i c\u00e1p c\u1ee7a c\u1ea7u. N\u1ebfu bi\u1ebft \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n AB, BC, A\u2032B\u2032, c\u00f3 th\u1ec3 t\u00ednh \u0111\u1ed9 d\u00e0i B\u2032C\u2032 kh\u00f4ng? C' B\u00e0i B' 1 \u0110\u1ecaNH L\u00cd THAL\u00c8S TRONG TAMA'GI\u00c1C Nh\u1eefng s\u1ee3i c\u00e1p treo c\u1ee7a c\u1ea7u ThAu\u1eadn Ph\u01b0\u1edbc (tBhu\u1ed9c thaC\u0300nh ph\u00f4\u0301 \u0110a\u0300 N\u1eb5ng) cho ta h\u00ecnh \u1ea3nh nh\u1eefng \u0111o\u1ea1n th\u1eb3ng song song. C\u00e1c \u0111o\u1ea1n th\u1eb3ng AA\u2032, BB\u2032, CC\u2032 th\u1ec3 hi\u1ec7n ba s\u1ee3i c\u00e1p c\u1ee7a c\u1ea7u. N\u1ebfu bi\u1ebft \u0111\u1ed9 da\u0300i c\u00e1c \u0111o\u1ea1n \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KA\u0110B:, BGCi,\u00faAp\u2032B\u2032H, cS\u00f3 tcho\u1ec3\u0301 tc\u00ed\u01a1nhh\u0111\u00f4\u0323\u1ed9i dtar\u0300\u1ea3i Bi \u2032nC\u2032gkhhi\u00f4\u1ec7nmg,? tha\u0309o lu\u00e2\u0323n v\u1ec1 t\u00ecnh hu\u1ed1ng th\u1ef1c t\u1ebf d\u1eabn \u0111\u1ebfn \u0111\u1ecbnh l\u00ed Thal\u00e8s th\u00f4ng qua t\u00ecnh hu\u1ed1ng quan s\u00e1t c\u00e1c \u0111o\u1ea1n c\u00e1p treo song song. C\u00e1ch \u0111\u1eb7t v\u1ea5n \u0111\u1ec1 n\u00e0y c\u00f3 kh\u1ea3 n\u0103ng thu h\u00fat HS v\u00e0o b\u00e0i h\u1ecdc. C' \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110K\u0110: GV n\u00eau c\u00e2u B's\u1eed d\u1ee5ng c\u01a1 gi\u1edbi thi\u1ec7u b\u00e0i. h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nhA\u1ead'n x\u00e9t, GV h\u1ed9i \u0111\u1ec3 L\u01b0u \u00fd: \u0110\u00e2y l\u00e0 c\u00e2u h\u1ecfi m\u1edf, t\u1ea1o s\u1ef1 ch\u00fa \u00fd v\u00e0 k\u1ebft n\u1ed1i. GV kh\u00f4ng c\u1ea7n \u0111\u00e1nh gi\u00e1 c\u00e2u tr\u1ea3 l\u1eddi c\u1ee7a HS \u0111\u00fang hay sai. Tinh th\u1ea7n chung c\u1ee7a H\u0110K\u0110 l\u00e0: M\u1ecdi c\u00e2u tAr\u1ea3 l\u1eddi \u0111\u1ec1u B\u0111\u01b0\u1ee3c ghCi nh\u1eadn, mu\u1ed1n bi\u1ebft \u0111\u00fang \u2013 sai, h\u1ecdc xong b\u00e0i n\u00e0y s\u1ebd r\u00f5! 1. \u0110o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 1T.\u1ec9 s\u0110\u1ed1Oc\u1ee7\u1ea0aNhaTiH\u0111\u1eb2o\u1ea1NnGthT\u1eb3\u1ec8nLg\u1ec6 H\u0110TK\u1ec9Ps\u1ed11c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng 1 a) Cho hai s\u1ed1 5 v\u00e0 8. H\u00e3y t\u00ednh t\u1ec9 s\u1ed1 gi\u1eefa hai s\u1ed1 \u0111\u00e3 cho. A B D C Hi\u0300nh 1 b) H\u00e3y \u0111o v\u00e0 t\u00ednh t\u1ec9 s\u1ed1 gi\u1eefa hai \u0111\u1ed9 d\u00e0i (theo mm) c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng AB v\u00e0 CD trong H\u00ecnh 1. T\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng l\u00e0 t\u1ec9 s\u1ed1 \u0111\u1ed9 d\u00e0i c\u1ee7a ch\u00fang theo c\u00f9ng m\u1ed9t \u0111\u01a1n v\u1ecb \u0111o. 178 T\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng AB v\u00e0 CD \u0111\u01b0\u1ee3c k\u00ed hi\u1ec7u l\u00e0: AB . CD","\u2013 M\u1ee5c \u0111\u00edch cu\u0309a H\u0110KP 1: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 kh\u00e1i ni\u1ec7m t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 1: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 1. H\u00e3y t\u00ednh t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng AB v\u00e0 CD trong c\u00e1c tr\u01b0\u1eddng h\u1ee3p sau: a) AB = 6 cm; CD = 8 cm;\t\t\t\t\t\t\t\t\t\t b) AB = 1,2 m; CD = 42 cm. M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: HS th\u1ef1c h\u00e0nh t\u00ednh t\u1ec9 s\u1ed1 c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. \u0110o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 H\u0110KP 2 \u2013 M\u1ee5c \u0111\u00edch cu\u0309a H\u0110KP 2: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 kh\u00e1i ni\u1ec7m \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 2: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 2. Trong H\u00ecnh 3, ch\u1ee9ng minh r\u1eb1ng: C A A\u2032 B B\u2032 a) AB v\u00e0 BC t\u1ec9 l\u1ec7 v\u1edbi A\u2032B\u2032 v\u00e0 B\u2032C\u2032; C\u2032 Hi\u0300nh 3 b) AC v\u00e0 A\u2032C\u2032 t\u1ec9 l\u1ec7 v\u1edbi AB v\u00e0 A\u2032B\u2032. C 6m M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 2: HS th\u01b0\u0323c ha\u0300nh t\u00ecm c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. V\u1eadn d\u1ee5ng 1. H\u00e3y t\u00ecm c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 trong h\u00ecnh v\u1ebd s\u01a1 \u0111\u1ed3 m\u1ed9t g\u00f3c c\u00f4ng vi\u00ean \u01a1\u0309 H\u00ecnh 4. \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 1: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng B ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf nh\u1eadn bi\u1ebft c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 trong s\u01a1 \u0111\u1ed3 m\u1ed9t c\u00f4ng g\u00f3c vi\u00ean h\u00ecnh tam gi\u00e1c. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c V\u1eadn d\u1ee5ng 1: HS tr\u1ea3 l\u1eddi y\u00eau c\u1ea7u v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS 3 m E l\u00e0m vi\u1ec7c theo nh\u00f3m. 3m H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: A=E A=C EC . \t D AD AB DB 1,5 m A Hi\u0300nh 4 179","2. \u0110\u1ecbnh l\u00ed Thal\u00e8s trong tam gi\u00e1c H\u0110KP 3 \u2013 Mu\u0323c \u0111i\u0301ch cu\u0309a H\u0110KP 3: Gi\u00fap HS kh\u00e1m ph\u00e1 \u0111\u1ecbnh l\u00ed Thal\u00e8s qua vi\u1ec7c nh\u1eadn bi\u1ebft c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 t\u1ea1o b\u1edfi m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi m\u1ed9t c\u1ea1nh c\u1ee7a tam gi\u00e1c. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 3: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 3. T\u00ednh \u0111\u1ed9 d\u00e0i x, y trong H\u00ecnh 8. M A x 3 5,5 y dE F 2 1,5 RS 2,5 B d \/\/ BC C NP a) 6 H\u00ecnh 8 b) \u2013 Mu\u0323c \u0111i\u0301ch cu\u0309a Th\u1ef1c h\u00e0nh 3: Gi\u00fap HS th\u1ef1c h\u00e0nh s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s trong vi\u1ec7c t\u00ecm \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n th\u1eb3ng \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. 180","H\u0110KP 4 \u2013 Mu\u0323c \u0111i\u0301ch cu\u0309a H\u0110KP 4: Gi\u00fap HS kh\u00e1m ph\u00e1 h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Thal\u00e8s qua vi\u1ec7c nh\u1eadn bi\u1ebft c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 t\u1ea1o b\u1edfi m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi m\u1ed9t c\u1ea1nh c\u1ee7a tam gi\u00e1c. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 4: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 4. T\u00ecm \u0111\u1ed9 d\u00e0i x tr\u00ean H\u00ecnh 13. \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 4: HS th\u01b0\u0323c ha\u0300nh s\u1eed d\u1ee5ng O \u0111\u1ecbnh l\u00ed Thal\u00e8s v\u00e0 h\u1ec7 qu\u1ea3 v\u00e0o vi\u1ec7c t\u00ecm c\u00e1c \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. 3,6 \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c Th\u1ef1c h\u00e0nh 4: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u C xD v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp.\t A 1,8 V\u1eadn d\u1ee5ng 2. V\u1edbi s\u1ed1 li\u1ec7u \u0111o \u0111\u1ea1c \u0111\u01b0\u1ee3c ghi tr\u00ean 7,8 B H\u00ecnh 14, h\u00e3y t\u00ednh b\u1ec1 r\u1ed9ng CD c\u1ee7a con k\u00eanh. CD \/\/ AB H\u00ecnh 13 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 2: HS c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng D ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o th\u1ef1c t\u1ebf t\u00ednh b\u1ec1 r\u1ed9ng c\u1ee7a m\u1ed9t con k\u00eanh. ?E \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u C 3m v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS 8m B 8m l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. A H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: H\u00ecnh 14 BE = AB suy ra 3 = 8 , suy ra CD = 6 m. CD CA CD 16 181","H\u0110KP 5 Mu\u0323c \u0111i\u0301ch cu\u0309a H\u0110KP 5: H\u01b0\u01a1\u0301ng d\u00e2\u0303n HS kh\u00e1m ph\u00e1 \u0111\u1ecbnh l\u00ed Thal\u00e8s \u0111\u1ea3o b\u1eb1ng c\u00e1ch v\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 5: Y\u00eau c\u00e2\u0300u HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 5. Ha\u0303y ch\u1ec9 ra c\u00e1c c\u1eb7p \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi nhau trong m\u00f4\u0303i hi\u0300nh d\u01b0\u01a1\u0301i \u0111\u00e2y. A 4 B\u2032\u2032 A\u2032\u2032 12 MN 2 O A\u2032 3 2 3 B\u2032 4,5 B 2P 3 C A B H\u00ecnh 18 b) a) M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 5: HS th\u01b0\u0323c ha\u0300nh s\u01b0\u0309 du\u0323ng \u0111\u1ecbnh l\u00ed Thal\u00e8s \u0111\u1ea3o trong vi\u1ec7c ki\u1ec3m tra t\u00ednh song song c\u1ee7a c\u00e1c c\u1eb7p \u0111o\u1ea1n th\u1eb3ng \u0111\u00ea\u0309 re\u0300n luy\u00ea\u0323n ki\u0303 n\u0103ng theo y\u00eau c\u00e2\u0300u c\u00e2\u0300n \u0111a\u0323t. V\u1eadn d\u1ee5ng 3. \u0110o chi\u1ec1u cao A AB c\u1ee7a m\u1ed9t to\u00e0 nh\u00e0 b\u1eb1ng hai c\u00e2y c\u1ecdc FE, DK, m\u1ed9t s\u1ee3i d\u00e2y v\u00e0 m\u1ed9t th\u01b0\u1edbc cu\u1ed9n nh\u01b0 sau: \u2013 \u0110\u1eb7t c\u1ecdc FE c\u1ed1 \u0111\u1ecbnh, di F chuy\u1ec3n c\u1ecdc DK sao cho nh\u00ecn K th\u1ea5y K, F, A th\u1eb3ng h\u00e0ng. \u2013 C\u0103ng th\u1eb3ng d\u00e2y FC \u0111i qua K va\u0300 c\u1eaft m\u1eb7t \u0111\u1ea5t t\u1ea1i C. \u2013 \u0110o kho\u1ea3ng c\u00e1ch BC v\u00e0 DC B E DC tr\u00ean m\u1eb7t \u0111\u1ea5t. H\u00ecnh 19 Cho bi\u1ebft DK = 1 m, BC = 24 m, DC = 1,2 m. T\u00ednh chi\u1ec1u cao AB c\u1ee7a to\u00e0 nh\u00e0. 182","\u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 3: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc v\u00e0o vi\u1ec7c t\u00ednh chi\u1ec1u cao c\u1ee7a to\u00e0 nh\u00e0 trong th\u1ef1c t\u1ebf. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 3: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: DK = DC suy ra 1 = 1, 2 , suy ra AB = 20 m. AB BC AB 24 IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) HS \u0111o chi\u1ec1u d\u00e0i a v\u00e0 chi\u1ec1u r\u1ed9ng b b\u00e0n h\u1ecdc theo c\u00f9ng m\u1ed9t \u0111\u01a1n v\u1ecb \u0111o v\u00e0 t\u00ednh t\u1ec9 s\u1ed1 a. b b) 70 = 1 . \t\t\t\t\t\t\t c) AB = 3 suy ra C=D 5.A=B 5=.6 10 (cm). 350 5 CD 5 33 2. \ta) AM = AN suy ra x = 4, 5 , suy ra x = 3; MB NC 2 3 b) CE = CB suy ra x = 2, 4 , suy ra x = 7,2; CD CA 9 3 c) DM = PM suy ra x = 5 , suy ra x = 2. EN PN 2, 6 6, 5 3. \tTa c\u00f3 B\uf0b5 = C\uf0b5 , suy ra EB \/\/ DC, suy ra CD = AC , EB AB suy ra CD = AC \u22c5 EB = 600 \u22c5120 = 360 (m). AB 200 4. Ta c\u00f3 AM= 3, 6= 3= 4, 5= AN , suy ra MN \/\/ BC. MB 2, 4 2 3 NC 5. \ta) HK = AK suy ra x = 3 , suy ra x = 4. BC AC 6 4,5 b) MQ = PQ suy ra x = 3,8 , suy ra 6,4x = 3,8x + 3,8 . 1,8, MH NH x +1,8 6, 4 suy ra x = 3,8.1,8 \u2248 2,6. 6, 4 \u2212 3,8 c) EC = ED2 + DC2 = 82 + 62 = 10. A=B C=B CA suy ra =x y= 5 , suy ra x \u2248 6,7; y \u2248 8,3. DE CE CD 8 10 6 6. a) Ta c\u00f3 I=M J=M 1 suy ra IJ \/\/ NP; I=M K=P 1 suy ra IK \/\/ MP; IN JP IN KN P=J P=K 1 suy ra JK \/\/ MN. JM KN 183","b) A=M A=N 2 suy ra MN \/\/ BC; C=N C=P 5 suy ra NP \/\/ AB. MB NC 5 AN PB 2 7. \tTa c\u00f3 AB \/\/ CD, suy ra OA = OB , suy ra OA . OD = OB . OC. A B OC OD NP M Q C 8. \tTa c\u00f3 M=N D=N C=Q PQ , suy ra MN = PQ.\t D AB DB CB AB 9. \tTa c\u00f3 BC \/\/ B\u2032C\u2032 suy ra AB = BC , suy ra x x h = a , suy ra xa\u2032 = a(x + h), AB\u2032 B\u2032C\u2032 + a\u2032 suy ra x = ah . a\u2032\u2212 a B\u00e0i 2. \u0110\u01af\u1edcNG TRUNG B\u00ccNH C\u1ee6A TAM GI\u00c1C I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u00e2\u0300n \u0111a\u0323t: \u2013 M\u00f4 t\u1ea3 \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c. \u2013 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c (\u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c th\u00ec song song v\u1edbi c\u1ea1nh th\u1ee9 ba v\u00e0 b\u1eb1ng n\u1eeda c\u1ea1nh \u0111\u00f3). \u2013 Bi\u1ebft v\u1eadn d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ee7a \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c trong gi\u1ea3i to\u00e1n v\u00e0 gi\u1ea3i quy\u1ebft m\u1ed9t s\u1ed1 v\u1ea5n \u0111\u1ec1 th\u1ef1c t\u1ebf. 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc, m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc, giao ti\u00ea\u0301p toa\u0301n ho\u0323c. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng, t\u00edch h\u1ee3p c\u00e1c m\u00f4n h\u1ecdc kh\u00e1c. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd GV h\u01b0\u1edbng d\u1eabn HS s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s \u0111\u1ec3 ch\u1ee9ng minh t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 Gi\u1eefa hai \u0111i\u1ec3m B v\u00e0 C c\u00f3 m\u1ed9t h\u1ed3 n\u01b0\u1edbc (xem B C h\u00ecnh b\u00ean). Bi\u1ebft DE = 45 m. L\u00e0m th\u1ebf n\u00e0o \u0111\u1ec3 D E t\u00ednh \u0111\u01b0\u1ee3c kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m B v\u00e0 C? A 184","\u2013 M\u1ee5c \u0111\u00edch cu\u0309a H\u0110K\u0110: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 t\u00ecnh hu\u1ed1ng xu\u1ea5t hi\u1ec7n \u0111\u01b0\u1eddng trung b\u00ecnh khi t\u00ednh kho\u1ea3ng c\u00e1ch trong th\u1ef1c t\u1ebf. C\u00e1ch \u0111\u1eb7t v\u1ea5n \u0111\u1ec1 n\u00e0y c\u00f3 kh\u1ea3 n\u0103ng thu h\u00fat HS v\u00e0o b\u00e0i h\u1ecdc. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110K\u0110: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV s\u1eed d\u1ee5ng c\u01a1 h\u1ed9i \u0111\u1ec3 gi\u1edbi thi\u1ec7u b\u00e0i. L\u01b0u \u00fd: \u0110\u00e2y l\u00e0 c\u00e2u h\u1ecfi m\u1edf, t\u1ea1o s\u1ef1 ch\u00fa \u00fd v\u00e0 k\u1ebft n\u1ed1i. GV kh\u00f4ng c\u1ea7n \u0111\u00e1nh gi\u00e1 c\u00e2u tr\u1ea3 l\u1eddi c\u1ee7a HS \u0111\u00fang hay sai. Tinh th\u1ea7n chung c\u1ee7a H\u0110K\u0110 l\u00e0: M\u1ecdi c\u00e2u tr\u1ea3 l\u1eddi \u0111\u1ec1u \u0111\u01b0\u1ee3c ghi nh\u1eadn, mu\u1ed1n bi\u1ebft \u0111\u00fang \u2013 sai, h\u1ecdc xong b\u00e0i n\u00e0y s\u1ebd r\u00f5! 1. \u0110\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c H\u0110KP 1 \u2013 M\u1ee5c \u0111\u00edch cu\u0309a H\u0110KP 1: Gi\u00fap HS c\u00f3 c\u01a1 h\u00f4\u0323i kh\u00e1m ph\u00e1 \u0111\u1ecbnh ngh\u0129a \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 1: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 1. T\u00ecm \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u0103\u0309ng NQ trong H\u00ecnh 4. P Q M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: HS th\u01b0\u0323c ha\u0300nh t\u00ecm \u0111\u1ed9 d\u00e0i 5 ? \u0111\u01b0\u1eddng trung b\u00ecnh \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. M N 54 V\u1eadn d\u1ee5ng 1. Trong H\u00ecnh 5, ch\u1ee9ng minh MN l\u00e0 \u0111\u01b0\u1eddng O trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC. H\u00ecnh 4 C \u2013 M\u1ee5c \u0111\u00edch V\u1eadn d\u1ee5ng 1: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1eeba h\u1ecdc \u0111\u1ec3 nh\u1eadn bi\u1ebft \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a N tam gi\u00e1c. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 1: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u AM B v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS H\u00ecnh 5 l\u00e0m vi\u1ec7c theo nh\u00f3m. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: Ta c\u00f3 MN \u22a5 AB, AC \u22a5 AB, suy ra MN \/\/ AC. Theo \u0111\u1ecbnh l\u00ed Thal\u00e8s ta c\u00f3 B=N B=M 1, suy ra NB = NC, hay N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC. NC AM V\u00ec M v\u00e0 N l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB v\u00e0 BC n\u00ean MN l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC. 185","2. T\u00ednh ch\u1ea5t c\u1ee7a \u0111\u01b0\u1eddng trung b\u00ecnh H\u0110KP 2 \u2013 Mu\u0323c \u0111i\u0301ch cu\u0309a H\u0110KP 2: Gi\u00fap HS kh\u00e1m ph\u00e1 t\u00ednh ch\u1ea5t c\u1ee7a \u0111\u01b0\u1eddng trung b\u00ecnh b\u1eb1ng c\u00e1ch \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110KP 2: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. Th\u1ef1c h\u00e0nh 2. Trong Hi\u0300nh 8, cho bi\u00ea\u0301t JK = 10 cm, J DE = 6,5 cm, EL = 3,7 cm. T\u00ednh DJ, EF, DF, KL. DE \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 2: HS th\u01b0\u0323c ha\u0300nh v\u1eadn d\u1ee5ng K FL t\u00ednh ch\u1ea5t c\u1ee7a \u0111\u01b0\u1eddng trung b\u00ecnh \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo H\u00ecnh 8 y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c Th\u1ef1c h\u00e0nh 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp.\t V\u1eadn d\u1ee5ng 2. H\u00e3y t\u00ednh kho\u1ea3ng c\u00e1ch BC trong ph\u00e2\u0300n (trang 52). \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng 2: HS c\u00f3 c\u01a1 h\u1ed9i v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1ec1 \u0111\u01b0\u1eddng trung b\u00ecnh \u0111\u1ec3 t\u00ednh kho\u1ea3ng c\u00e1ch th\u1ef1c t\u1ebf trong t\u00ecnh hu\u1ed1ng kh\u00f4ng th\u1ec3 \u0111o \u0111\u1ea1c tr\u1ef1c ti\u1ebfp. \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c V\u1eadn d\u1ee5ng 2: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: BC = 2DE = 90 m. IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) x = 12; \t b) x = 11 ; \t c) x = 6. 2 2.\t P=Q B=C 9 (cm). 2 2 3. \t\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pythagore, ta c\u00f3: AC = AB = 42 + 22 =2 5 (cm), \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t \t BC = 62 + 22 =2 10 (cm). \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC, ta c\u00f3: P=Q A=B 5 (cm); =PR A=C 5 (cm); R=Q B=C 10 (cm). 2 2 2 186","4. \ta) \u0394FBA = \u0394FCK (g.c.g) b) Ta c\u00f3 \u0394FBA = \u0394FCK, suy ra AF = FK v\u00e0 AB = CK. EF l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a \u0394ADK, suy ra EF \/\/ DK v\u00e0 EF = DK , suy ra EF \/\/ CD \/\/ AB. 2 c) Ta c\u00f3=EF D=K CK + C=D AB + CD . 2 2 2 5. Ta c\u00f3 MN l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC, A suy ra MN \/\/ BC, suy ra MNPH l\u00e0 h\u00ecnh thang. (1) N Trong tam gi\u00e1c vu\u00f4ng AHB, ta c\u00f3 HM l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng M v\u1edbi c\u1ea1nh huy\u1ec1n AB, suy ra M=H A=B MB. B HP C 2 Trong tam gi\u00e1c MBH c\u00e2n t\u1ea1i M, ta c\u00f3 M\uf0b7HB = M\uf0b7BH. Ta l\u1ea1i c\u00f3 NP l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC, suy ra NP \/\/ AB, suy ra N\uf0b7PC = M\uf0b7BH (hai g\u00f3c \u0111\u1ed3ng v\u1ecb). Suy ra M\uf0b7HB = N\uf0b7PC , suy ra M\uf0b7HP = N\uf0b7PH. (2) T\u1eeb (1) v\u00e0 (2) suy ra MNPH l\u00e0 h\u00ecnh thang c\u00e2n. 6. \tx = 1,4 m. 7. \tDE = 2BC = 464 m. B\u00e0i 3. T\u00cdNH CH\u1ea4T \u0110\u01af\u1edcNG PH\u00c2N GI\u00c1C C\u1ee6A TAM GI\u00c1C I. M\u1ee5c ti\u00eau 1.Y\u00eau c\u1ea7u c\u00e2\u0300n \u0111a\u0323t: \u2013 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a tam gi\u00e1c. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 v\u1ea5n \u0111\u1ec1 th\u1ef1c ti\u1ec5n g\u1eafn v\u1edbi t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c. 2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc, m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc, giao ti\u00ea\u0301p toa\u0301n ho\u0323c. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng, t\u00edch h\u1ee3p c\u00e1c m\u00f4n h\u1ecdc kh\u00e1c. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd 1. Ch\u01b0\u01a1ng tr\u00ecnh To\u00e1n 8 ch\u1ec9 gi\u1edbi thi\u1ec7u t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a tam gi\u00e1c, kh\u00f4ng y\u00eau c\u1ea7u gi\u1edbi thi\u1ec7u \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c ngo\u00e0i c\u1ee7a tam gi\u00e1c. 2. GV s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Thal\u00e8s \u0111\u1ec3 h\u01b0\u1edbng d\u1eabn HS kh\u00e1m ph\u00e1 t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a tam gi\u00e1c. 187","III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 A 6 3 H\u0110K\u0110 \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c AD c\u1ee7a tam gi\u00e1c ABC chia c\u1ea1nh \u0111\u1ed1i di\u1ec7n BC th\u00e0nh hai \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 v\u1edbi hai \u0111o\u1ea1n th\u1eb3ng n\u00e0o trong h\u00ecnh? B2 D 4C \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i tr\u1ea3i nghi\u1ec7m, tha\u0309o lu\u00e2\u0323n v\u1ec1 t\u00ednh chia t\u1ec9 l\u1ec7 c\u1ea1nh 3B\u0111v\u1ed1\u00e0\u00e0oiidbi\u00e0\u1ec7inhc\u1ecd\u1ee7Tca.\u00cd\u0111N\u01b0H\u1eddngCpHh\u00e2\u1ea4nTgi\u00e1\u0110c\u01afc\u1ee7\u1edcaNtaGm gPi\u00e1Hc.\u00c2CN\u00e1chG\u0111I\u1eb7\u00c1t vC\u1ea5nC\u0111\u1ee6\u1ec1 nA\u00e0yTcA\u00f3Mkh\u1ea3GnI\u0103\u00c1ngCthu h\u00fat HS \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c H\u0110K\u0110: GV n\u00eau c\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV s\u1eed d\u1ee5ng c\u01a1 h\u1ed9i \u0111\u1ec3 gi\u1edbi thi\u1ec7u b\u00e0i. mc\u1ee7uaL\u1ed1H\u01b0nuSbi\u00fd\u0111\u1ebf:\u00fat\u0110n\u0111g\u00e2\u00faynhgal\u00e0y\u2013css\u00e2aauii.,hTh\u1ecfi\u1ecdnichmAt\u0110hhxaB\u01b0t\u1edf\u1eb3oihC\u1eddn,\u0111n\u1ea7ngctongg\u1ea1h\u1ea1noicbpna\u00e0hh\u00e0osctui\u00e2\u1ef1h\u1ea1tnnnr\u1eb3nocg\u00e0nghhnygic\u00e1g\u00fa\u0111\u1ee7cst\u1ed1h\u1ec9\u00fd\u1ebdaAi\u00ecln\u1ec7DdrvHh\u00f5i\u00e0v\u1ec7c?!\u0110\u1edb\u1ee7nkiaK\u1ebfBhtt\u0110Caanmitl\u1ed1h\u0111\u00e0gi\u00e0o:.in\u1ea1\u00e1MGhnc V\u1ecdi kh\u00f4nAg c\u1ea7n \u0111\u00e1nh gi\u00e1 c\u00e2u tr\u1ea3 l\u1eddi c\u00e2u tr\u1ea3 l\u1eddi \u0111\u1ec1u \u0111\u01b0\u1ee3c ghi nh\u1eadn, 3 6 H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: DD=CB AB . AC B2 D 4C 1. T\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c H1.\u0110TK\u00cdPNH CH\u1ea4T \u0110\u01af\u1edcNG PH\u00c2N GI\u00c1C CU\u0309A TAM GI\u00c1C Cho tam gi\u00e1c ABC c\u00f3 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c AD. E V\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng qua B song song v\u1edbi AD v\u00e0 c\u1eaft A \u0111\u01b0\u1eddng th\u1eb3ng AC t\u1ea1i E (H\u00ecnh 1). Ha\u0303y gia\u0309i thi\u0301ch t\u1ea1i sao: a) Tam gi\u00e1c BAE c\u00e2n t\u1ea1i A. b) D=B A=E AB. B C DC AC AC D H\u00ecnh 1 \u2013 M\u0110\u1ee5c\u1ecbn\u0111h\u00edcl\u00edh c\u1ee7a H\u0110KP: Gi\u00fap HS co\u0301 c\u01a1 h\u00f4\u0323i kh\u00e1m ph\u00e1 t\u00ednh chia t\u1ec9 l\u1ec7 c\u1ee7a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c th\u00f4ng qua ho\u1ea1t \u0111\u1ed9ng tr\u1ea3i nghi\u1ec7m v\u1edbi thao t\u00e1c v\u1ebd th\u00eam \u0111\u01b0\u1eddng ph\u1ee5 BE qua B v\u00e0 song song v\u1edbi ADT.rong tam gi\u00e1c, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a m\u1ed9t g\u00f3c chia c\u1ea1nh \u0111\u1ed1i di\u1ec7n th\u00e0nh hai \u0111o\u1ea1n th\u1eb3ng \u2013 Gt\u01a1\u1ec9\u0323il\u1ec7y\u0301vt\u1edb\u00f4\u0309ichha\u01b0i\u0301c\u1ea1Hnh\u0110kK\u1ec1Ph:aGi \u0111Vo\u1ea1nn\u00eau\u1ea5yc.\u00e2u h\u1ecfi, HS tra\u0309 l\u01a1\u0300i, l\u01a1\u0301p nh\u1eadn x\u00e9t, GV \u0111a\u0301nh gia\u0301. 2. \u00c1p d\u1ee5ng t\u00ednh chia t\u1ec9 l\u1ec7 c\u1ee7a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c C Ph\u1ea7n nG\u00e0yTkh\u00f4AngDcl\u00f3\u00e0h\u0111o\u01b0\u1ea1\u1eddtn\u0111g\u1ed9nphg\u00e2knhg\u00e1im\u00e1cpch\u1ee7\u00e1a, cgh\u00f3\u1ec9ccA\u00f3 thraoinvg\u00ed d\u1ee5 minh ho\u1ea1 c\u00e1c \u00e1p d\u1ee5ng t\u00ednh chia l\u1ec7 c\u1ee7a \u0111\u01b0\u1eddng p\u2206hA\u00e2nBCgi,\u00e1Dc.\u2208BC t\u1ec9 GV c\u00f3KthL\u1ec3 gi\u1ea3Di tBh\u00ed=chAVB\u00ed d\u1ee5 2 D v\u00e0 cho ho\u1ea1t \u0111\u1ed9ng nh\u00f3m theo \u0111\u1ec1 b\u00e0i t\u01b0\u01a1ng t\u1ef1 V\u00ed d\u1ee5 3. DC AC 188 AB H\u00ecnh 2 V\u00ed d\u1ee5 1. Cho tam gi\u00e1c ABC c\u00f3 AB = 5 cm, AC = 8 cm. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1eaft","V\u00ed d\u1ee5 2. Cho tam gi\u00e1c OEF nh\u01b0 trong H\u00ecnh 4. E T\u00ednh t\u1ec9 s\u1ed1 hai \u0111o\u1ea1n th\u1eb3ng ME v\u00e0 MF. 16 M O 12 F H\u00ecnh 4 C V\u00ed d\u1ee5 3. T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n BZ, UC, UZ trong H\u00ecnh 5. 86 U Th\u1ef1c h\u00e0nh. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh MQ c\u1ee7a tam gi\u00e1c MPQ B H\u00ecnh 5 5 Z trong H\u00ecnh 6. M Q \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh: HS th\u01b0\u0323c ha\u0300nh s\u1eed d\u1ee5ng 7 C t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c \u0111\u1ec3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng theo y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t. P 4N 189 H\u00ecnh 6 \u2013 G\u01a1\u0323i y\u0301 t\u00f4\u0309 ch\u01b0\u0301c Th\u1ef1c h\u00e0nh: HS tra\u0309 l\u01a1\u0300i y\u00eau c\u1ea7u v\u00e0o v\u1edf, GV s\u1eeda chung tr\u01b0\u1edbc l\u1edbp. GV c\u00f3 th\u1ec3 cho HS l\u00e0m vi\u1ec7c theo nh\u00f3m ho\u1eb7c thuy\u1ebft tr\u00ecnh. IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) x = 4; \t\t\t\t\t\t\t\t b) x = 12;\t\t\t\t\t\t\t c) x = 12. 2. \ta) Ta c\u00f3 DB= AB= 6= 3, suy ra D=B D=C DB + D=C B=C 10 , DC AC 8 4 3 4 3+4 7 7 suy ra DB = 30 cm; DC = 40 cm. 77 b) Hai tam gi\u00e1c ADB v\u00e0 ADC c\u00f3 c\u00f9ng chi\u1ec1u cao, suy ra SA=DB D=B 3 . SADC DC 4 3. \ta) Ta c\u00f3 D=B A=B 1=5 3 , A DC AC 20 4 suy ra D=B D=C DB + D=C B=C 25 . E 3 4 3+4 7 7 V\u1eady DB = 75 cm; DC = 100 cm. B D 77 Ta c\u00f3 DE \/\/ AB, suy ra DE= DC= 4 , suy ra DE = 60 cm. AB BC 7 7","b) Tam gi\u00e1c ABC c\u00f3 AB2 + AC2 = BC2, suy ra tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. Do \u0111\u00f3 SABC = 1 AB . AC = 150 (cm2). 2 c) Hai tam gi\u00e1c ADB v\u00e0 ADC c\u00f3 c\u00f9ng chi\u1ec1u cao, suy ra SA=DB D=B 3 hay SA=DB SA=DC SA=BC 150 . SADC DC 4 3 4 77 V\u1eady SADB = 450 cm 2 ; SADC = 600 cm 2 . 7 7 12=\uf8ed\uf8ec\uf8eb 670 \uf8f6\uf8f8\uf8f72 Tam gi\u00e1c ADE vu\u00f4ng c\u00e2n t\u1ea1i E, suy ra SADE = 1 E=D . EA 1 800 (cm2 ). 2 49 SDCE = SADC \u2013 SADE =600 \u2212 1800 =2 400 (cm2 ). 7 49 49 A 4. \ta) Tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A, suy ra BC2 = AB2 + AC2 = 9 + 16 = 25, suy ra BC = 5 cm. Ta c\u00f3 AD l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c n\u00ean D=B A=B 3 , B HD C DC AC 4 suy ra D=B D=C DB + D=C B=C 5 . 3 4 3+4 7 7 Do \u0111\u00f3 DB = 15 cm; DC = 20 cm. 7 7 b) Trong tam gi\u00e1c vu\u00f4ng ABC, ta c\u00f3 2SABC = AB . AC = AH . BC, suy ra AH= AB.A=C 12 (cm). BC 5 Trong tam gi\u00e1c vu\u00f4ng ABH, ta c\u00f3 BH= AB2 \u2212 AH2 = 9 \u2212 144 =9 (cm). 25 5 Ta c\u00f3 HD = DB \u2013 BH = 15 \u2212 9 = 12 (cm), 7 5 35 suy ra AD = AH2 + DH2 = \uf8eb 12 \uf8f62 + \uf8eb 12 \uf8f62 = 12 2 (cm). \uf8ec\uf8ed 5 \uf8f7\uf8f8 \uf8ec\uf8ed 35 \uf8f7\uf8f8 7 5. \tMD l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c AMB, suy ra DA = MA . DB MB ME l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c AMC, suy ra EA = MA . EC MC Ta l\u1ea1i c\u00f3 MB = MC, suy ra DA = EA , suy ra DE \/\/ BC. DB EC 190","B\u00c0I T\u1eacP CU\u1ed0I CH\u01af\u01a0NG 7 C\u00c2U H\u1eceI TR\u1eaeC NGHI\u1ec6M 1. A 2. D 3. B 4. A 5. C 6. C 7. D 8. B 9. A B\u00c0I T\u1eacP T\u1ef0 LU\u1eacN A KH 10. V\u1ebd BH v\u00e0 DK l\u1ea7n l\u01b0\u1ee3t vu\u00f4ng g\u00f3c v\u1edbi AC 13,5 (H, K \u2208 AC), ta c\u00f3 BH \/\/ DK. Suy ra D=K A=D 13=,5 3 . \t D BH AB 18 4 4,5 BC 11. a) Ta c\u00f3 MN \/\/ CB, suy ra A=B A=C 5,3 , suy ra AB = 1, 5 . 5, 3 \u2248 3,3 m. AN AM 2,4 2, 4 b) Ta c\u00f3 36 \u2212 x = 1,6 , suy ra x = 33,6 m. 36 24 12. a) Ta c\u00f3 M=N A=K 1 , suy ra MN = 10 cm. A BC AH 3 MK N T\u01b0\u01a1ng t\u1ef1 EF = 20 cm. b) Ta c=\u00f3 AH 2=SABC 2.=1080 72 (cm), E I F BC 30 H C suy ra KI = 1 AH = 24 (cm). B D 3 H C =SMNFE KI(=MN + EF) 2=4(10 + 20) 360 (cm2) = 3,6 (dm2). 2 2 13. a) x = 3,5; \t\t\t\t\t\t b) x = 5,1; \t\t\t\t c) x = 5,2. 14. a) x = 3,125; \t\t\t\t\t b) x = 8,1.\t 15. a) Trong tam gi\u00e1c ABC, ta c\u00f3 OE \/\/ BC, A F suy ra AE = AO . \t\t\t\t\t\t\t\t\t\t\t (1) E O AB AC Trong tam gi\u00e1c ACD, ta c\u00f3 OF \/\/ CD, BG suy ra AF = AO . \t\t\t\t\t\t\t\t\t (2) AD AC T\u1eeb (1) v\u00e0 (2) ta c\u00f3 AE = AF , suy ra FE \/\/ BD. AB AD 191","b) Trong tam gi\u00e1c ABC, ta c\u00f3 OG \/\/ AB, suy ra CG = OC . \t\t\t\t\t\t (3) BG OA \tTrong tam gi\u00e1c ACD, ta c\u00f3 OH \/\/ AD, suy ra CH = OC . \t\t\t\t\t (4) DH OA \tT\u1eeb (3) v\u00e0 (4) ta c\u00f3 CG = CH , suy ra CG . DH = BG . CH. BG DH 16. \ta) Ta c\u00f3 AD \/\/ BK, suy ra EA = ED . Ta c\u00f3 AB \/\/ DG, suy ra EG = ED . EK EB EA EB \t Suy ra EA = EG , suy ra EA2 = EK . EG hay AE2 = EK . EG. EK EA \t b) Ta c\u00f3 E=K E=B A=B D=C AK . EA ED DG DG AG \t Suy ra AK \u2212 AE = AK , suy ra AK . AG = AE . AG + AE . AK. \t\t (1) AE AG \t Chia hai v\u1ebf c\u1ee7a (1) cho AE . AK . AG ta \u0111\u01b0\u1ee3c 1 = 1 + 1 . AE AK AG 17. \ta) Ta c\u00f3 AK \/\/ BD (v\u00ec c\u00f9ng vu\u00f4ng g\u00f3c v\u1edbi AH). Suy ra K=B A=D AB , suy ra AK l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A trong tam gi\u00e1c ABC. KC AC AC \tb) C\u00e1ch v\u1ebd \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A c\u1ee7a tam gi\u00e1c ABC b\u1eb1ng th\u01b0\u1edbc k\u1ebb v\u00e0 \u00eake: \u2013 Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia AC l\u1ea5y \u0111i\u1ec3m D sao cho AD = AB. \u2013 V\u1ebd AH vu\u00f4ng g\u00f3c v\u1edbi BD (H \u2208 BD). \u2013 V\u1ebd AK vu\u00f4ng g\u00f3c v\u1edbi AH (K \u2208 BC). Ta c\u00f3 AK l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c A. 192","Ch\u01b0\u01a1ng 8 H\u00ccNH \u0110\u1ed2NG D\u1ea0NG A. M\u1ee4C TI\u00caU 1. N\u0103ng l\u01b0\u0323c chuy\u00ean m\u00f4n Hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u2013 M\u00f4 t\u1ea3 \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. C\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c \u2013 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c c\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi vi\u1ec7c v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. C\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c vu\u00f4ng \u2013 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c c\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c vu\u00f4ng. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u00f4\u0323t s\u00f4\u0301 v\u1ea5n \u0111\u1ec1 th\u01b0\u0323c ti\u00ea\u0303n g\u1eafn v\u1edbi vi\u1ec7c v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1ec1 hai tam gi\u00e1c vu\u00f4ng \u0111\u1ed3ng d\u1ea1ng. Hai h\u00ecnh \u0111\u1ed3ng d\u1ea1ng \u2013 Nh\u00e2\u0323n bi\u00ea\u0301t \u0111\u01b0\u01a1\u0323c hi\u0300nh \u0111\u00f4\u0300ng da\u0323ng ph\u00f4\u0301i ca\u0309nh (hi\u0300nh vi\u0323 t\u01b0\u0323), hi\u0300nh \u0111\u00f4\u0300ng da\u0323ng qua c\u00e1c h\u00ecnh \u1ea3nh c\u1ee5 th\u1ec3. \u2013 Nh\u1eadn bi\u1ebft \u0111\u01b0\u01a1\u0323c ve\u0309 \u0111e\u0323p trong t\u01b0\u0323 nhi\u00ean, ngh\u00ea\u0323 thu\u00e2\u0323t, ki\u00ea\u0301n tru\u0301c, c\u00f4ng ngh\u00ea\u0323 ch\u00ea\u0301 ta\u0323o, ... bi\u00ea\u0309u hi\u00ea\u0323n qua hi\u0300nh \u0111\u00f4\u0300ng da\u0323ng.\t\t\t\t\t\t 2. N\u0103ng l\u01b0\u0323c chung H\u00ecnh th\u00e0nh v\u00e0 ph\u00e1t tri\u1ec3n: \u2013 N\u0103ng l\u01b0\u0323c t\u01b0\u0323 chu\u0309 va\u0300 t\u01b0\u0323 ho\u0323c trong ti\u0300m to\u0300i, kha\u0301m pha\u0301. \u2013 N\u0103ng l\u01b0\u0323c giao ti\u00ea\u0301p va\u0300 h\u01a1\u0323p ta\u0301c trong tri\u0300nh ba\u0300y, tha\u0309o lu\u00e2\u0323n va\u0300 la\u0300m vi\u00ea\u0323c nho\u0301m. \u2013 N\u0103ng l\u01b0\u0323c gia\u0309i quy\u00ea\u0301t v\u00e2\u0301n \u0111\u00ea\u0300 va\u0300 sa\u0301ng ta\u0323o trong th\u01b0\u0323c ha\u0300nh va\u0300 v\u00e2\u0323n du\u0323ng. 3. Hi\u0300nh tha\u0300nh ca\u0301c ph\u00e2\u0309m ch\u00e2\u0301t \u2013 Y\u00eau n\u01b0\u01a1\u0301c, nh\u00e2n \u00e1i. \u2013 Ch\u0103m chi\u0309, trung th\u01b0\u0323c, tra\u0301ch nhi\u00ea\u0323m. B. H\u01af\u1edaNG D\u1eaaN D\u1ea0Y H\u1eccC B\u00e0i 1. HAI TAM GI\u00c1C \u0110\u1ed2NG D\u1ea0NG I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t: \u2013 M\u00f4 t\u1ea3 \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng, k\u00ed hi\u1ec7u, c\u00e1ch vi\u1ebft, t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 v\u1ea5n \u0111\u1ec1 th\u1ef1c ti\u1ec5n g\u1eafn v\u1edbi vi\u1ec7c v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. 193","2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc; m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc; s\u1eed d\u1ee5ng c\u00f4ng c\u1ee5, ph\u01b0\u01a1ng ti\u1ec7n h\u1ecdc to\u00e1n. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd V\u1edbi Th\u1ef1c h\u00e0nh 3 v\u00e0 V\u1eadn d\u1ee5ng, GV t\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m, nh\u1eadn x\u00e9t k\u0129, gi\u00fap HS d\u1ec5 d\u00e0ng ti\u1ebfp thu ki\u1ebfn th\u1ee9c v\u1ec1 c\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 Hai tam gi\u00e1c c\u00f3 ba c\u1ea1nh b\u1eb1ng nhau th\u00ec b\u1eb1ng nhau. C\u00f2n hai tam gia\u0301c c\u00f3 ba g\u00f3c b\u1eb1ng nhau thi\u0300 c\u00f3 b\u1eb1ng nhau kh\u00f4ng? M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110: K\u00edch th\u00edch kh\u1ea3 n\u0103ng t\u01b0 duy s\u00e1ng t\u1ea1o, t\u00ecm hi\u1ec3u v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng c\u1ee7a HS. 1. Tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng H\u0110KP 1 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 1: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m b\u01b0\u1edbc \u0111\u1ea7u nh\u1eadn bi\u1ebft hai h\u00ecnh \u0111\u1ed3ng d\u1ea1ng. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110KP 1: GV cho HS quan s\u00e1t c\u00e1c c\u1eb7p h\u00ecnh, nh\u1eadn x\u00e9t, n\u00eau \u00fd ki\u1ebfn. GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. 194","H\u0110KP 2 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 2: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m, th\u1ea3o lu\u1eadn \u0111\u1ec3 m\u00f4 t\u1ea3 \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng, vi\u1ebft \u0111\u00fang t\u1ec9 s\u1ed1 c\u00e1c c\u1ea1nh c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110KP 2: T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m, c\u00e1c nh\u00f3m n\u00eau \u00fd ki\u1ebfn, l\u1edbp nh\u1eadn x\u00e9t, GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. Th\u1ef1c h\u00e0nh 1. Quan s\u00e1t H\u00ecnh 3, cho bi\u1ebft \u2206AMN \u223d \u2206ABC. A a) H\u00e3y vi\u1ebft t\u1ec9 s\u1ed1 c\u1ee7a c\u00e1c c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng v\u00e0 t\u00ednh t\u1ec9 s\u1ed1 M? 4 N \u0111\u1ed3ng d\u1ea1ng. b) T\u00ednh A\uf0b7MN. 65o 12 C B Hi\u0300nh 3 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 1: HS r\u00e8n k\u0129 n\u0103ng c\u1ea7n \u0111\u1ea1t th\u00f4ng qua vi\u1ec7c \u00e1p d\u1ee5ng \u0111\u1ecbnh ngh\u0129a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u0111\u1ec3 vi\u1ebft \u0111\u00fang t\u1ec9 s\u1ed1 c\u00e1c c\u1ea1nh v\u00e0 c\u00e1c g\u00f3c b\u1eb1ng nhau t\u01b0\u01a1ng \u1ee9ng. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c Th\u1ef1c h\u00e0nh 1: HS l\u00e0m vi\u1ec7c c\u00e1 nh\u00e2n, GV \u0111\u00e1nh gi\u00e1 k\u1ebft qu\u1ea3. 2. T\u00ednh ch\u1ea5t H\u0110KP 3 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 3: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m, th\u1ea3o lu\u1eadn \u0111\u1ec3 hi\u1ec3u \u0111\u01b0\u1ee3c c\u00e1c t\u00ednh ch\u1ea5t c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110KP 3: GV n\u00eau c\u00e2u h\u1ecfi; HS tr\u1ea3 l\u1eddi, l\u1edbp nh\u1eadn x\u00e9t; GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. A Th\u1ef1c h\u00e0nh 2. Quan s\u00e1t H\u00ecnh 4, cho bi\u1ebft \u2206ADE \u223d \u2206AMN, D E \u2206AMN \u223d \u2206ABC, DE l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c M N AMN, MN l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC. Tam gi\u00e1c ADE \u0111\u1ed3ng d\u1ea1ng v\u1edbi tam gi\u00e1c ABC theo t\u1ec9 s\u1ed1 B Hi\u0300nh 4 \u0111\u1ed3ng d\u1ea1ng l\u00e0 bao nhi\u00eau? C 195","\u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 2: HS r\u00e8n k\u0129 n\u0103ng c\u1ea7n \u0111\u1ea1t th\u00f4ng qua vi\u1ec7c \u00e1p d\u1ee5ng c\u00e1c t\u00ednh ch\u1ea5t c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u0111\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp c\u1ee5 th\u1ec3. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c Th\u1ef1c h\u00e0nh 2: HS l\u00e0m vi\u1ec7c c\u00e1 nh\u00e2n, GV \u0111\u00e1nh gi\u00e1 k\u1ebft qu\u1ea3. 3. \u0110\u1ecbnh l\u00ed H\u0110KP 4 \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110KP 4: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m, th\u1ea3o lu\u1eadn \u0111\u1ec3 hi\u1ec3u \u0111\u01b0\u1ee3c \u0111\u1ecbnh l\u00ed v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. \u2013 G\u01a1\u0323i \u00fd t\u1ed5 ch\u1ee9c H\u0110KP 4: T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m; c\u00e1c nh\u00f3m n\u00eau \u00fd ki\u1ebfn, l\u1edbp nh\u1eadn x\u00e9t; GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. Th\u1ef1c h\u00e0nh 3. Quan s\u00e1t H\u00ecnh 8, cho bi\u1ebft DC \/\/ MP, P M Q EF \/\/ MQ. ED a) Ch\u1ee9ng minh r\u1eb1ng \u2206EPF \u223d \u2206DCQ. I b) \u2206ICF c\u00f3 \u0111\u1ed3ng d\u1ea1ng v\u1edbi \u2206MPQ kh\u00f4ng? T\u1ea1i sao? CF \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a Th\u1ef1c h\u00e0nh 3: HS th\u1ef1c h\u00e0nh \u00e1p d\u1ee5ng Hi\u0300nh 8 \u0111\u1ecbnh l\u00ed v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u0111\u1ec3 ch\u1ee9ng minh c\u00e1c tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. Qua \u0111\u00f3 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng c\u1ea7n \u0111\u1ea1t th\u00f4ng qua vi\u1ec7c \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u0111\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp c\u1ee5 th\u1ec3. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c Th\u1ef1c h\u00e0nh 3: HS l\u00e0m vi\u1ec7c c\u00e1 nh\u00e2n, GV \u0111\u00e1nh gi\u00e1 k\u1ebft qu\u1ea3. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: a) Ta c\u00f3 DC \/\/ MP, suy ra \u2206DCQ \u223d \u2206MPQ. EF \/\/ MQ, suy ra \u2206EPF \u223d \u2206MPQ. V\u1eady \u2206EPF \u223d \u2206DCQ. b) Ta c\u00f3 IF \/\/ DQ, suy ra \u2206ICF \u223d \u2206DCQ. Do \u0111\u00f3 \u2206ICF \u223d \u2206MPQ. 196","V\u1eadn d\u1ee5ng. Trong H\u00ecnh 10, cho bi\u1ebft ABCD l\u00e0 h\u00ecnh A E b\u00ecnh h\u00e0nh. B a) Ch\u1ee9ng minh r\u1eb1ng \u2206IEB \u223d \u2206IDA. I C b) Cho bi\u1ebft CB = 3BE v\u00e0 AI = 9 cm. T\u00ednh \u0111\u1ed9 d\u00e0i DC. \u2013 M\u1ee5c \u0111\u00edch c\u1ee7a V\u1eadn d\u1ee5ng: Gi\u00fap HS c\u1ee7ng c\u1ed1, kh\u1eafc s\u00e2u D Hi\u0300nh 10 ki\u1ebfn th\u1ee9c th\u00f4ng qua vi\u1ec7c \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u0111\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp c\u1ee5 th\u1ec3. \u2013 G\u1ee3i \u00fd t\u1ed5 ch\u1ee9c V\u1eadn d\u1ee5ng: T\u1ed5 ch\u1ee9c th\u1ea3o lu\u1eadn nh\u00f3m, GV \u0111\u00e1nh gi\u00e1 k\u1ebft qu\u1ea3. H\u01b0\u1edbng d\u1eabn \u2013 \u0111\u00e1p \u00e1n: b) \u2206IEB \u223d \u2206IDA, suy ra BE = IB , suy ra BE = IB , suy ra IB = 1. DA IA 3BE IA IA 3 Suy ra IB = 3 cm; DC = 12 cm. IV. H\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp 1. \ta) \u0110\u00fang. Hai tam gi\u00e1c b\u1eb1ng nhau th\u00ec \u0111\u1ed3ng d\u1ea1ng v\u1edbi nhau theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng k = 1. b) Sai. N\u1ebfu hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng k \u2260 1 th\u00ec \u0111\u1ed9 d\u00e0i hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng kh\u00f4ng b\u1eb1ng nhau, n\u00ean \u0111\u00f3 l\u00e0 hai tam gi\u00e1c kh\u00f4ng b\u1eb1ng nhau. 2. \tL\u1ea7n l\u01b0\u1ee3t l\u1ea5y c\u00e1c \u0111i\u1ec3m E, F tr\u00ean c\u00e1c c\u1ea1nh AB, AC sao cho AE = 1; AF =1. AB 2 AC 2 V\u1ebd \u0111o\u1ea1n th\u1eb3ng EF. Ta c\u00f3 \u2206ABC \u223d \u2206AEF theo t\u1ec9 s\u1ed1 A \u0111\u1ed3ng d\u1ea1ng k = 1 . E F 2 B C Ch\u1ee9ng minh: Trong tr\u01b0\u1eddng h\u1ee3p nh\u01b0 h\u00ecnh b\u00ean, ta c\u00f3 EF C' l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c ABC, do \u0111\u00f3 EF \/\/ BC. T\u1eeb \u0111\u00f3 suy ra \u2206AEF \u223d \u2206ABC theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng =k A=E 1 . \t A' AB 2 3. \ta) Ta c\u00f3 \u2206ABC \u223d \u2206A\u2032B\u2032C\u2032. A Suy ra A=B B=C AC , A\u2032B\u2032 B\u2032C\u2032 A\u2032C\u2032 =A\uf0b5 A\uf0b5=\u2032, B\uf0b5 B\uf0b5=\u2032, C\uf0b5 C\uf0b5\u2032. B C B' b) Ta c\u00f3 F\uf024 = 180o \u2013 (D\uf0b5 + E\uf0b5 ) D D' 78o ? \t\t\t\t\t\t = 180o \u2013 (78o + 57o) = 45o. 57o F E' ? V\u00ec \u2206DEF \u223d \u2206D\u2032E\u2032F\u2032 n\u00ean D\uf0b5\u2032 = D\uf0b5 = 78o, E F' F\uf0b5\u2032 = F\uf024 = 45o. \t 197","c) Ta c\u00f3 \u2206MNP \u223d \u2206M\u2032N\u2032P\u2032, do \u0111\u00f3 M' 15 M=N M=P NP . M\u2032N\u2032 M\u2032P\u2032 N\u2032P\u2032 N' 12 Suy ra MN= 10= 6= 1. M 10 P 15 M\u2032P\u2032 12 2 6 V\u1eady MN = 15 = 7,5; M\u2032P\u2032 = 10 . 2 = 20.\t N P' 2 C 4. \ta) Ta c\u00f3 AB \/\/ CD (gi\u1ea3 thi\u1ebft), do \u0111\u00f3 \u2206AEB \u223d \u2206DEC. b) \u2206AEB \u223d \u2206DEC suy ra AE = AB hay x \u2212 2 = 3 . Do \u0111\u00f3 x = 8. DE DC 10 5 5. \ta) \u2206ABC \u223d \u2206DEF theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng k = 2. 5 Do \u0111\u00f3 A=B A=C B=C 2 . DE DF EF 5 \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3 AB + AC + BC = 2 . V\u1eady P\u2206ABC = 2 . DE + DF + EF 5 P\u2206DEF 5 b) Ta c\u00f3 P\u2206ABC = 2 , suy ra P\u2206DEF = P\u2206ABC . P\u2206DEF 5 52 \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: P\u2206DE=F P\u2206AB=C P\u2206DEF \u2212 P\u2206AB=C 3=6 12. 52 5\u22122 3 A V\u1eady P\u2206DEF = 60 cm, P\u2206ABC = 24 cm. 16 m 30 m 6. \ta) DE \/\/ BC (gi\u1ea3 thi\u1ebft), suy ra \u2206ADE \u223d \u2206ABC. 22 m E D b) \u2206ADE \u223d \u2206ABC suy ra AD = DE hay 16 = 22 . B AB BC 30 BC Do \u0111\u00f3 BC = 30 . 22 = 41, 25 (m).\t 16 B\u00e0i 2. C\u00c1C TR\u01af\u1edcNG H\u1ee2P \u0110\u1ed2NG D\u1ea0NG C\u1ee6A HAI TAM GI\u00c1C I. M\u1ee5c ti\u00eau 1. Y\u00eau c\u1ea7u c\u1ea7n \u0111\u1ea1t: \u2013 Gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c c\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c. \u2013 V\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c \u0111\u00e3 h\u1ecdc \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. \u2013 Gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 v\u1ea5n \u0111\u1ec1 th\u1ef1c ti\u1ec5n g\u1eafn v\u1edbi vi\u1ec7c v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u1ec1 hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng. 198","2. N\u0103ng l\u1ef1c ch\u00fa tr\u1ecdng: t\u01b0 duy v\u00e0 l\u1eadp lu\u1eadn to\u00e1n h\u1ecdc; m\u00f4 h\u00ecnh ho\u00e1 to\u00e1n h\u1ecdc; s\u1eed d\u1ee5ng c\u00f4ng c\u1ee5, ph\u01b0\u01a1ng ti\u1ec7n h\u1ecdc to\u00e1n. 3. T\u00edch h\u1ee3p: To\u00e1n h\u1ecdc v\u00e0 cu\u1ed9c s\u1ed1ng. II. M\u1ed9t s\u1ed1 ch\u00fa \u00fd Sau m\u1ed7i tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c th\u00ec t\u1ec9 s\u1ed1 chu vi, t\u1ec9 s\u1ed1 hai \u0111\u01b0\u1eddng trung tuy\u1ebfn, t\u1ec9 s\u1ed1 hai \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng \u0111\u01b0\u1ee3c \u0111\u01b0a v\u00e0o d\u01b0\u1edbi d\u1ea1ng ch\u00fa \u00fd ho\u1eb7c nh\u1eadn x\u00e9t. GV ch\u1ec9 c\u1ea7n n\u00eau m\u00e0 kh\u00f4ng c\u1ea7n gi\u1ea3i th\u00edch. \u0110\u1ed1i v\u1edbi HS kh\u00e1 gi\u1ecfi th\u00ec xem nh\u01b0 l\u00e0 m\u1ed9t b\u00e0i t\u1eadp m\u1edf r\u1ed9ng. III. G\u1ee3i \u00fd c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u1ee5 th\u1ec3 H\u0110K\u0110 2B\u00e0i C\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng c\u1ee7a hai tam gi\u00e1c c\u00f3 \u0111i\u1ec1u g\u00ec kh\u00e1c v\u1edbi c\u00e1c Ct\u00c1r\u01b0C\u1eddngThR\u1ee3p\u01afb\u1edc\u1eb1nNgGnhaHu\u1ee2c\u1ee7Pa h\u0110ai \u1ed2tamNgGi\u00e1cD? \u1ea0NG C\u1ee6A HAI TAM GI\u00c1C M\u1ee5c \u0111\u00edch c\u1ee7a H\u0110K\u0110: HS c\u00f3 c\u01a1 h\u1ed9i ph\u00e1n \u0111o\u00e1n, suy lu\u1eadn. K\u00edch th\u00edch s\u1ef1 t\u00f2 m\u00f2, t\u00ecm hi\u1ec3u v\u1ec1 c\u00e1c tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1Cn\u00e1cgtcr\u01b0\u1ee7\u1eddanhgahi \u1ee3tapm\u0111\u1ed3gni\u00e1gcd.\u1ea1ng c\u1ee7a hai tam gi\u00e1c c\u00f3 \u0111i\u1ec1u g\u00ec kh\u00e1c v\u1edbi c\u00e1c tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau c\u1ee7a hai tam gi\u00e1c? 1. Tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng th\u1ee9 nh\u1ea5t (c.c.c) H1.\u0110TKRP\u01af1\u1edcNG H\u1ee2P \u0110\u1ed2NG D\u1ea0NG TH\u1ee8 NH\u1ea4T (c.c.c) 1 Cho tam gi\u00e1c ABC v\u00e0 tam gi\u00e1c A\u2032B\u2032C\u2032 c\u00f3 A c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 H\u00ecnh 1. Tr\u00ean c\u1ea1nh AB v\u00e0 AC c\u1ee7a tam gi\u00e1c ABC l\u1ea7n l\u01b0\u1ee3t l\u1ea5y hai 2 3 N9 \u0111i\u1ec3m M, N sao cho AM = 2 cm, AN = 3 cm. 6M a) So s\u00e1nh c\u00e1c t\u1ec9 s\u1ed1 A'B' , A'C' , B'C' . B 12 C AB AC BC a) b) T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng MN. A\u2032 c) Em c\u00f3 nh\u1eadn x\u00e9t g\u00ec v\u1ec1 m\u1ed1i quan h\u1ec7 gi\u1eefa 23 c\u00e1c tam gi\u00e1c ABC, AMN v\u00e0 A\u2032B\u2032C\u2032? B\u2032 4 C\u2032 b) Hi\u0300nh 1 \u2013 M\u0110\u1ee5\u1ecbcn\u0111h\u00edlc\u00edh: c\u1ee7a H\u0110KP 1: Gi\u00fap HS c\u00f3 c\u01a1 h\u1ed9i tr\u1ea3i nghi\u1ec7m, th\u1ea3o lu\u1eadn, gi\u1ea3i th\u00edch \u0111\u01b0\u1ee3c tr\u01b0\u1eddngNh\u1ebf\u1ee3upb\u0111a\u1ed3nc\u1ea1gndh\u1ea1cn\u1ee7gattha\u1ee9mnghi\u1ea5\u00e1tccn\u1ee7\u00e0ayhta\u1ec9il\u1ec7tavm\u1edbigbi\u00e1ac.c\u1ea1nh c\u1ee7a tam gi\u00e1c kia th\u00ec hai tam gi\u00e1c \u0111\u00f3 \u2013 G\u01a1\u0111\u0323i\u1ed3\u00fdngt\u1ed5d\u1ea1cnhg\u1ee9cv\u1edbHi\u0110nhKaPu.1: GV n\u00eau c\u00e2u h\u1ecfi; HS th\u1ef1c hi\u1ec7n v\u00e0o v\u1edf, n\u00eau \u00fd ki\u1ebfn; l\u1edbp nh\u1eadn x\u00e9t; GV \u0111\u00e1nh gi\u00e1, ch\u1ed1t ki\u1ebfn th\u1ee9c. A GT \u2206ABC v\u00e0 \u2206A\u2032B\u2032C\u2032, A\u2032 A='B' A=' C' B'C' AB AC BC 199 KL \u2206A\u2032B\u2032C\u2032 \u223d \u2206ABC"]