500 BÀI TOÁN CƠ BẢN VÀ NÂNG CAO LỚP 5

["500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 86: C\u00f3 hai c\u00e1i \u0111\u1ed3ng h\u1ed3 c\u00e1t 4 ph\u00fat v\u00e0 7 ph\u00fat. C\u00f3 th\u1ec3 d\u00f9ng hai c\u00e1i \u0111\u1ed3ng h\u1ed3 n\u00e0y \u0111\u1ec3 \u0111o th\u1eddi gian 9 ph\u00fat \u0111\u01b0\u1ee3c kh\u00f4ng? B\u00e0i gi\u1ea3i: C\u00f3 nhi\u1ec1u c\u00e1ch \u0111\u1ec3 \u0111o \u0111\u01b0\u1ee3c 9 ph\u00fat: B\u1ea1n c\u00f3 th\u1ec3 cho c\u1ea3 2 c\u00e1i \u0111\u1ed3ng h\u1ed3 c\u00e1t c\u00f9ng ch\u1ea3y m\u1ed9t l\u00fac v\u00e0 ch\u1ea3y h\u1ebft c\u00e1t 3 l\u1ea7n. Khi \u0111\u1ed3ng h\u1ed3 4 ph\u00fat ch\u1ea3y h\u1ebft c\u00e1t 3 l\u1ea7n (4 x 3 = 12(ph\u00fat)) th\u00ec b\u1ea1n b\u1eaft \u0111\u1ea7u t\u00ednh th\u1eddi gian, t\u1eeb l\u00fac \u0111\u00f3 \u0111\u1ebfn khi \u0111\u1ed3ng h\u1ed3 7 ph\u00fat ch\u1ea3y h\u1ebft c\u00e1t 3 l\u1ea7n th\u00ec v\u1eeba \u0111\u00fang \u0111\u01b0\u1ee3c 9 ph\u00fat (7 x 3 - 12 = 9(ph\u00fat)); ho\u1eb7c cho c\u1ea3 hai \u0111\u1ed3ng h\u1ed3 c\u00f9ng ch\u1ea3y m\u1ed9t l\u00fac, \u0111\u1ed3ng h\u1ed3 7 ph\u00fat ch\u1ea3y h\u1ebft c\u00e1t m\u1ed9t l\u1ea7n (7 ph\u00fat), \u0111\u1ed3ng h\u1ed3 4 ph\u00fat ch\u1ea3y h\u1ebft c\u00e1t 4 l\u1ea7n (16 ph\u00fat). Khi \u0111\u1ed3ng h\u1ed3 7 ph\u00fat ch\u1ea3y h\u1ebft c\u00e1t ta b\u1eaft \u0111\u1ea7u t\u00ednh th\u1eddi gian, t\u1eeb l\u00fac \u0111\u00f3 \u0111\u1ebfn l\u00fac \u0111\u1ed3ng h\u1ed3 4 ph\u00fat ch\u1ea3y h\u1ebft c\u00e1t 4 l\u1ea7n l\u00e0 v\u1eeba \u0111\u00fang 9 ph\u00fat (16 - 7 = 9 (ph\u00fat)); ... B\u00e0i 87: Vui xu\u00e2n m\u1edbi, c\u00e1c b\u1ea1n c\u00f9ng l\u00e0m ph\u00e9p to\u00e1n sau, nh\u1edb r\u1eb1ng c\u00e1c ch\u1eef c\u00e1i kh\u00e1c nhau c\u1ea7n thay b\u1eb1ng c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau, c\u00e1c ch\u1eef c\u00e1i gi\u1ed1ng nhau thay b\u1eb1ng c\u00e1c ch\u1eef s\u1ed1 gi\u1ed1ng nhau. NHAM + NGO = 2002 B\u00e0i gi\u1ea3i: - V\u00ec A\u2260G m\u00e0 ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c c\u1ee7a t\u1ed5ng l\u00e0 0 n\u00ean ph\u00e9p c\u1ed9ng c\u00f3 nh\u1edb 1 sang h\u00e0ng tr\u0103m n\u00ean \u1edf h\u00e0ng tr\u0103m: H + N + 1 (nh\u1edb) = 10; nh\u1edb 1 sang h\u00e0ng ngh\u00ecn. Do \u0111\u00f3 H + N = 10 - 1 = 9. - Ph\u00e9p c\u1ed9ng \u1edf h\u00e0ng ngh\u00ecn: N + 1 (nh\u1edb) = 2 n\u00ean N = 2 - 1 = 1. Thay N = 1 ta c\u00f3: H + 1 = 9 n\u00ean H = 9 - 1 = 8 - Ph\u00e9p c\u1ed9ng \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb: C\u00f3 2 tr\u01b0\u1eddng h\u1ee3p x\u1ea3y ra: * Tr\u01b0\u1eddng h\u1ee3p 1: Ph\u00e9p c\u1ed9ng \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb kh\u00f4ng nh\u1edb sang h\u00e0ng ch\u1ee5c. Khi \u0111\u00f3: M + O = 0 v\u00e0 A + G = 10. Ta c\u00f3 b\u1ea3ng: (L\u01b0u \u00fd 4 ch\u1eef M, O, A, G ph\u1ea3i kh\u00e1c nhau v\u00e0 kh\u00e1c 1; 8) * Tr\u01b0\u1eddng h\u1ee3p 2: Ph\u00e9p c\u1ed9ng \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb c\u00f3 nh\u1edb 1 sang h\u00e0ng ch\u1ee5c. Khi \u0111\u00f3: M + O = 12 v\u00e0 A + G = 9. Ta c\u00f3 b\u1ea3ng: 49","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i V\u1eady b\u00e0i to\u00e1n c\u00f3 24 \u0111\u00e1p s\u1ed1 nh\u01b0 tr\u00ean. B\u00e0i 88: H\u00e3y x\u1ebfp 8 qu\u00e2n \u0111\u00f4min\u00f4 v\u00e0o m\u1ed9t h\u00ecnh vu\u00f4ng 4x4 sao cho t\u1ed5ng s\u1ed1 ch\u1ea5m tr\u00ean c\u00e1c h\u00e0ng ngang, d\u1ecdc, ch\u00e9o c\u1ee7a h\u00ecnh vu\u00f4ng \u0111\u1ec1u b\u1eb1ng 11. L\u1eddi gi\u1ea3i: C\u00f3 ba c\u00e1ch gi\u1ea3i c\u01a1 b\u1ea3n sau: T\u1eeb ba c\u00e1ch gi\u1ea3i c\u01a1 b\u1ea3n n\u00e0y c\u00f3 th\u1ec3 t\u1ea1o n\u00ean nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n kh\u00e1c, ch\u1eb3ng h\u1ea1n: B\u00e0i 89: S\u1eed d\u1ee5ng c\u00e1c con s\u1ed1 trong m\u1ed7i bi\u1ec3n s\u1ed1 xe \u00f4 t\u00f4 39A 0452, 38B 0088, 52N 8233 c\u00f9ng c\u00e1c d\u1ea5u +, -, x, : v\u00e0 d\u1ea5u ngo\u1eb7c ( ), [ ] \u0111\u1ec3 l\u00e0m th\u00e0nh m\u1ed9t ph\u00e9p t\u00ednh \u0111\u00fang. L\u1eddi gi\u1ea3i: * Bi\u1ec3n s\u1ed1 39A 0452. C\u00f3 m\u1ed9t s\u1ed1 c\u00e1ch: 50","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i (4 x 2 - 5 + 0) x 3 = 9 5x2-4+3+0=9 45 : 9 - 3 - 2 = 0 (9 + 2 - 3) x 5 = 40 (4 + 5) : 9 + 2 + 0 = 3 9 : 3 - ( 5 - 4 + 2) = 0 3 - 9 : (4 + 5) - 0 = 2 9 : (4 + 5) + 2 + 0 = 3 (9 + 5) : 2 - 4 + 0 = 3 9 + 3 : (5 - 2) + 0 = 4 5+2-9:3-0=4 (9 : 3 + 0) + 4 - 2 = 5 (9 + 3) : 4 + 0 + 2 = 5 . . . . * Bi\u1ec3n s\u1ed1 38B 0088. C\u00f3 nhi\u1ec1u l\u1eddi gi\u1ea3i d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t \u201cnh\u00e2n m\u1ed9t s\u1ed1 v\u1edbi s\u1ed1 0\u201d 38 x 88 x 0 = 0 ho\u1eb7c t\u00ednh ch\u1ea5t \u201cchia s\u1ed1 0 cho m\u1ed9t s\u1ed1 kh\u00e1c 0\u201d M\u1ed9t v\u00e0i c\u00e1ch kh\u00e1c: 0 : (38 + 88) = 0 (9 - 8) + 0 - 8 : 8 = 0 8:8+8+0+0=9.... * Bi\u1ec3n s\u1ed1 52N 8233. C\u00f3 m\u1ed9t s\u1ed1 c\u00e1ch: 5x2-8+3-3=2 8 : (5 x 2 - 3 - 3) = 2 [(23 - 3) : 5] x 2 = 8 (5 + 2 + 2) - (3 : 3) = 8 (8 : 2 - 3) x (3 + 2) = 5 [(8 + 2) x 3 : 3] : 2 = 5 (5 x 2 + 3 + 3) : 2 = 8 3x3-5+2+2=8.... 51","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 90: M\u1ed9t chi\u1ebfc \u0111\u1ed3ng h\u1ed3 \u0111ang ho\u1ea1t \u0111\u1ed9ng b\u00ecnh th\u01b0\u1eddng, hi\u1ec7n t\u1ea1i kim gi\u1edd v\u00e0 kim ph\u00fat \u0111ang kh\u00f4ng tr\u00f9ng nhau. H\u1ecfi sau \u0111\u00fang 24 gi\u1edd (t\u1ee9c 1 ng\u00e0y \u0111\u00eam), hai kim \u0111\u00f3 tr\u00f9ng nhau bao nhi\u00eau l\u1ea7n? H\u00e3y l\u1eadp lu\u1eadn \u0111\u1ec3 l\u00e0m \u0111\u00fang s\u00e1ng t\u1ecf k\u1ebft qu \u0111\u00f3. L\u1eddi gi\u1ea3i: V\u1edbi m\u1ed9t chi\u1ebfc \u0111\u1ed3ng h\u1ed3 \u0111ang ho\u1ea1t \u0111\u1ed9ng b\u00ecnh th\u01b0\u1eddng, c\u1ee9 m\u1ed7i gi\u1edd tr\u00f4i qua th\u00ec kim ph\u00fat quay \u0111\u01b0\u1ee3c m\u1ed9t v\u00f2ng, c\u00f2n kim gi\u1edd quay \u0111\u01b0\u1ee3c 1\/12 v\u00f2ng. Hi\u1ec7u v\u1eadn t\u1ed1c c\u1ee7a kim ph\u00fat v\u00e0 kim gi\u1edd l\u00e0: 1 - 1\/12 = 11\/12 (v\u00f2ng\/gi\u1edd) Th\u1eddi gian \u0111\u1ec3 hai kim tr\u00f9ng nhau m\u1ed9t l\u1ea7n l\u00e0: 1 : 11\/12 = 12\/11 (gi\u1edd) V\u1eady sau 24 gi\u1edd hai kim s\u1ebd tr\u00f9ng nhau s\u1ed1 l\u1ea7n l\u00e0 : 24 : 12\/11 = 22 (l\u1ea7n). B\u00e0i 91: C\u00f3 ba ng\u01b0\u1eddi d\u00f9ng chung m\u1ed9t k\u00e9t ti\u1ec1n. H\u1ecfi ph\u1ea3i l\u00e0m cho c\u00e1i k\u00e9t \u00edt nh\u1ea5t bao nhi\u00eau \u1ed5 kho\u00e1 v\u00e0 bao nhi\u00eau ch\u00eca \u0111\u1ec3 k\u00e9t ch\u1ec9 m\u1edf \u0111\u01b0\u1ee3c n\u1ebfu c\u00f3 m\u1eb7t \u00edt nh\u1ea5t hai ng\u01b0\u1eddi? L\u1eddi gi\u1ea3i: V\u00ec k\u00e9t ch\u1ec9 m\u1edf \u0111\u01b0\u1ee3c n\u1ebfu c\u00f3 m\u1eb7t \u00edt nh\u1ea5t hai ng\u01b0\u1eddi, n\u00ean s\u1ed1 \u1ed5 kho\u00e1 ph\u1ea3i l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng 2. a) L\u00e0m 2 \u1ed5 kho\u00e1. + N\u1ebfu l\u00e0m 3 ch\u00eca th\u00ec s\u1ebd c\u00f3 hai ng\u01b0\u1eddi c\u00f3 c\u00f9ng m\u1ed9t lo\u1ea1i ch\u00eca; hai ng\u01b0\u1eddi n\u00e0y kh\u00f4ng m\u1edf \u0111\u01b0\u1ee3c k\u00e9t. + N\u1ebfu l\u00e0m nhi\u1ec1u h\u01a1n 3 ch\u00eca th\u00ec \u00edt nh\u1ea5t c\u00f3 m\u1ed9t ng\u01b0\u1eddi c\u1ea7m 2 ch\u00eca kh\u00e1c lo\u1ea1i; ch\u1ec9 c\u1ea7n m\u1ed9t ng\u01b0\u1eddi n\u00e0y \u0111\u00e3 m\u1edf \u0111\u01b0\u1ee3c k\u00e9t. V\u1eady kh\u00f4ng th\u1ec3 l\u00e0m 2 \u1ed5 kho\u00e1. b) L\u00e0m 3 \u1ed5 kho\u00e1 + N\u1ebfu l\u00e0m 3 ch\u00eca th\u00ec c\u1ea7n ph\u1ea3i c\u00f3 \u0111\u1ee7 ba ng\u01b0\u1eddi m\u1edbi m\u1edf \u0111\u01b0\u1ee3c k\u00e9t. + N\u1ebfu l\u00e0m 4 ch\u00eca ho\u1eb7c 5 ch\u00eca th\u00ec \u00edt nh\u1ea5t c\u00f3 hai ng\u01b0\u1eddi kh\u00f4ng m\u1edf \u0111\u01b0\u1ee3c k\u00e9t. + N\u1ebfu l\u00e0m 6 ch\u00eca (m\u1ed7i kho\u00e1 2 ch\u00eca) th\u00ec m\u1ed7i ng\u01b0\u1eddi c\u1ea7m hai ch\u00eca kh\u00e1c nhau th\u00ec ch\u1ec9 c\u1ea7n hai ng\u01b0\u1eddi b\u1ea5t k\u1ef3 l\u00e0 m\u1edf \u0111\u01b0\u1ee3c k\u00e9t. V\u1eady \u00edt nh\u1ea5t ph\u1ea3i l\u00e0m 3 \u1ed5 kho\u00e1 v\u00e0 m\u1ed7i \u1ed5 kho\u00e1 l\u00e0m 2 ch\u00eca. B\u00e0i 92 : C\u00f3 4 t\u1ea5m g\u1ed7 d\u00e0i v\u00e0 4 t\u1ea5m g\u1ed7 h\u00ecnh cung tr\u00f2n. N\u1ebfu s\u1eafp x\u1ebfp nh\u01b0 h\u00ecnh b\u00ean th\u00ec \u0111\u01b0\u1ee3c 4 chu\u1ed3ng nh\u1ed1t 4 ch\u00fa th\u1ecf, nh\u01b0ng 1 ch\u00fa l\u1ea1i ch\u01b0a c\u00f3 chu\u1ed3ng. B\u1ea1n h\u00e3y x\u1ebfp l\u1ea1i c\u00e1c t\u1ea5m g\u1ed7 \u0111\u1ec3 c\u00f3 \u0111\u1ee7 5 chu\u1ed3ng cho m\u1ed7i ch\u00fa th\u1ecf c\u00f3 m\u1ed9t chu\u1ed3ng ri\u00eang. 52","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i gi\u1ea3i : B\u00e0i to\u00e1n c\u00f3 nhi\u1ec1u c\u00e1ch x\u1ebfp. Xin n\u00eau ra ba c\u00e1ch x\u1ebfp nh\u01b0 sau: B\u00e0i 93: M\u1ed9t ph\u00e2n x\u01b0\u1edfng c\u00f3 25 ng\u01b0\u1eddi. H\u1ecfi r\u1eb1ng trong ph\u00e2n x\u01b0\u1edfng \u0111\u00f3 c\u00f3 th\u1ec3 c\u00f3 20 ng\u01b0\u1eddi \u00edt h\u01a1n 30 tu\u1ed5i v\u00e0 15 ng\u01b0\u1eddi nhi\u1ec1u h\u01a1n 20 tu\u1ed5i \u0111\u01b0\u1ee3c kh\u00f4ng? B\u00e0i gi\u1ea3i: V\u00ec ch\u1ec9 c\u00f3 25 ng\u01b0\u1eddi, m\u00e0 trong \u0111\u00f3 c\u00f3 20 \u00edt h\u01a1n 30 tu\u1ed5i v\u00e0 15 ng\u01b0\u1eddi nhi\u1ec1u h\u01a1n 25 tu\u1ed5i, n\u00ean s\u1ed1 ng\u01b0\u1eddi \u0111\u01b0\u1ee3c \u0111i\u1ec3m 2 l\u1ea7n l\u00e0: (20 + 15) - 25 = 10 (ng\u01b0\u1eddi) \u0110\u00e2y ch\u00ednh l\u00e0 s\u1ed1 ng\u01b0\u1eddi c\u00f3 \u0111\u1ed9 tu\u1ed5i \u00edt h\u01a1n 30 tu\u1ed5i v\u00e0 nhi\u1ec1u h\u01a1n 20 tu\u1ed5i (t\u1eeb 21 tu\u1ed5i \u0111\u1ebfn 29 tu\u1ed5i). S\u1ed1 ng\u01b0\u1eddi t\u1eeb 30 tu\u1ed5i tr\u1edf l\u00ean l\u00e0: 25 - 20 = 5 (ng\u01b0\u1eddi) S\u1ed1 ng\u01b0\u1eddi t\u1eeb 20 tu\u1ed5i tr\u1edf xu\u1ed1ng l\u00e0: 25 - 15 = 10 (ng\u01b0\u1eddi) S\u1ed1 ng\u01b0\u1eddi \u00edt h\u01a1n 30 tu\u1ed5i l\u00e0: 10 + 10 = 20 (ng\u01b0\u1eddi) S\u1ed1 ng\u01b0\u1eddi nhi\u1ec1u h\u01a1n 20 tu\u1ed5i l\u00e0: 10 + 5 = 15 (ng\u01b0\u1eddi) V\u1eady c\u00f3 th\u1ec3 c\u00f3 20 ng\u01b0\u1eddi d\u01b0\u1edbi 30 tu\u1ed5i v\u00e0 15 ng\u01b0\u1eddi tr\u00ean 20 tu\u1ed5i; trong \u0111\u00f3 t\u1eeb 21 \u0111\u1ebfn 29 tu\u1ed5i \u00edt nh\u1ea5t c\u00f3 hai ng\u01b0\u1eddi c\u00f9ng \u0111\u1ed9 tu\u1ed5i. B\u00e0i 94: T\u00ecm 4 s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp c\u00f3 t\u00edch l\u00e0 3024 B\u00e0i gi\u1ea3i: 53","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Gi\u1ea3 s\u1eed c\u1ea3 4 s\u1ed1 \u0111\u1ec1u l\u00e0 10 th\u00ec t\u00edch l\u00e0 10 x 10 x 10 x 10 = 10000 m\u00e0 10000 > 3024 n\u00ean c\u1ea3 4 s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp \u0111\u00f3 ph\u1ea3i b\u00e9 h\u01a1n 10. V\u00ec 3024 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 4 n\u00ean c\u1ea3 4 s\u1ed1 ph\u1ea3i t\u00ecm kh\u00f4ng th\u1ec3 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 5. Do \u0111\u00f3 c\u1ea3 4 s\u1ed1 ph\u1ea3i ho\u1eb7c c\u00f9ng b\u00e9 h\u01a1n 5, ho\u1eb7c c\u00f9ng l\u1edbn h\u01a1n 5. N\u1ebfu 4 s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 1; 2; 3; 4 th\u00ec: 1 x 2 x 3 x 4 = 24 < 3024 (lo\u1ea1i) N\u1ebfu 4 s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 6; 7; 8; 9 th\u00ec: 6 x 7 x 8 x 9 = 3024 (\u0111\u00fang) V\u1eady 4 s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 6; 7; 8; 9. B\u00e0i 95: C\u00f3 3 lo\u1ea1i que v\u1edbi s\u1ed1 l\u01b0\u1ee3ng v\u00e0 c\u00e1c \u0111\u1ed9 d\u00e0i nh\u01b0 sau: - 16 que c\u00f3 \u0111\u1ed9 d\u00e0i 1 cm - 20 que c\u00f3 \u0111\u1ed9 d\u00e0i 2 cm - 25 que c\u00f3 \u0111\u1ed9 d\u00e0i 3 cm H\u1ecfi c\u00f3 th\u1ec3 x\u1ebfp t\u1ea5t c\u1ea3 c\u00e1c que \u0111\u00f3 th\u00e0nh m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c kh\u00f4ng? B\u00e0i gi\u1ea3i: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i (a) v\u00e0 chi\u1ec1u r\u1ed9ng (b) \u0111\u1ec1u l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean (c\u00f9ng m\u1ed9t \u0111\u01a1n v\u1ecb \u0111o) th\u00ec chu vi (P) c\u1ee7a h\u00ecnh \u0111\u00f3 ph\u1ea3i l\u00e0 s\u1ed1 ch\u1eb5n: P = (a + b) x 2 T\u1ed5ng \u0111\u1ed9 d\u00e0i c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c que l\u00e0: 1 x 16 + 2 x 20 + 3 x 25 = 131 (cm) V\u00ec 131 l\u00e0 s\u1ed1 l\u1ebb n\u00ean kh\u00f4ng th\u1ec3 x\u1ebfp t\u1ea5t c\u1ea3 c\u00e1c que \u0111\u00f3 th\u00e0nh m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c. B\u00e0i 96: H\u00e3y ph\u00e1t hi\u1ec7n ra m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa c\u00e1c s\u1ed1 r\u1ed3i s\u1eed d\u1ee5ng m\u1ed1i li\u00ean h\u1ec7 \u0111\u00f3 \u0111\u1ec3 \u0111i\u1ec1n s\u1ed1 h\u1ee3p l\u00fd v\u00e0o (?) B\u00e0i gi\u1ea3i: \u0110\u1ec3 cho g\u1ecdn, ta k\u00fd hi\u1ec7u c\u00e1c s\u1ed1 tr\u00ean nh\u1eefng \u00f4 tr\u00f2n theo b\u1ea3ng sau: 54","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i L\u1ea5y A chia cho K: 72 : 9 = L\u1ea5y G chia cho C: 8 : 1 = L\u1ea5y B chia cho H: 16 : 2 = L\u1ea5y E chia cho D: 24 : 3 = \u0111\u1ec1u cho c\u00f9ng m\u1ed9t k\u1ebft qu\u1ea3 \u1edf \u00f4 \u0110. V\u1eady (?) l\u00e0 8. B\u00e0i 97: C\u00f4 gi\u00e1o y\u00eau c\u1ea7u: \u201cC\u00e1c con l\u1ea5y 6 \u0111i\u1ec3m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n, n\u1ed1i c\u00e1c \u0111i\u1ec3m \u0111\u00f3 b\u1edfi c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u00f4 b\u1edfi m\u1ef1c xanh ho\u1eb7c m\u1ef1c \u0111\u1ecf\u201d. B\u1ea1n l\u1edbp tr\u01b0\u1edfng t\u1eadp h\u1ee3p c\u00e1c h\u00ecnh v\u1ebd l\u1ea1i v\u00e0 xem, b\u1ea1n th\u1ed1t l\u00ean: \u201cB\u1ea1n n\u00e0o c\u0169ng v\u1ebd \u0111\u01b0\u1ee3c 1 tam gi\u00e1c m\u00e0 3 c\u1ea1nh c\u00f9ng m\u00e0u m\u1ef1c\u201d! B\u1ea1n h\u00e3y th\u1eed l\u00e0m l\u1ea1i xem. Ai c\u00f3 th\u1ec3 l\u1eadp lu\u1eadn \u0111\u1ec3 l\u00e0m r\u00f5 t\u00ednh ch\u1ea5t n\u00e0y? B\u00e0i gi\u1ea3i: Ta g\u1ecdi 6 \u0111i\u1ec3m n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 A1, A2, A3, A4, A5, A6. B\u1eb1ng b\u00fat xanh v\u00e0 \u0111\u1ecf ta n\u1ed1i A1 v\u1edbi 5 \u0111i\u1ec3m c\u00f2n l\u1ea1i ta \u0111\u01b0\u1ee3c 5 \u0111o\u1ea1n th\u1eb3ng c\u00f3 hai m\u00e0u xanh ho\u1eb7c \u0111\u1ecf. Theo nguy\u00ean l\u00fd \u0110ir\u00edchl\u00ea c\u00f3 \u00edt nh\u1ea5t 3 \u0111o\u1ea1n th\u1eb3ng c\u00f9ng m\u00e0u. Kh\u00f4ng l\u00e0m m\u1ea5t t\u00ednh t\u1ed5ng qu\u00e1t, ta n\u1ed1i 3 \u0111o\u1ea1n A1A2, A1A3, A1A4 b\u1eb1ng b\u00fat m\u00e0u \u0111\u1ecf. Ta n\u1ed1i ti\u1ebfp A2A4 v\u00e0 A2A3. \u0110\u1ec3 tam gi\u00e1c A1A2A3 v\u00e0 tam gi\u00e1c A1A2A4 c\u00f3 3 c\u1ea1nh kh\u00f4ng c\u00f9ng m\u00e0u th\u00ec A2A4 v\u00e0 A2A3 ph\u1ea3i t\u00f4 m\u00e0u xanh. B\u00e2y gi\u1edd ta ti\u1ebfp t\u1ee5c n\u1ed1i A3A4, ta th\u1ea5y A3A4 \u0111\u01b0\u1ee3c t\u00f4 b\u1eb1ng b\u1ea5t k\u1ef3 m\u00e0u xanh ho\u1eb7c \u0111\u1ecf th\u00ec ta c\u0169ng \u0111\u01b0\u1ee3c \u00edt nh\u1ea5t m\u1ed9t tam gi\u00e1c c\u00f3 3 c\u1ea1nh c\u00f9ng m\u00e0u (ho\u1eb7c A1A3A4 c\u00f3 3 c\u1ea1nh \u0111\u1ecf ho\u1eb7c A2A3A4 c\u00f3 3 c\u1ea1nh m\u00e0u xanh). B\u00e0i 98: Thi b\u1eafn s\u00fang H\u00f4m nay D\u0169ng \u0111i thi b\u1eafn s\u00fang. D\u0169ng b\u1eafn gi\u1ecfi l\u1eafm, D\u0169ng \u0111\u00e3 b\u1eafn h\u01a1n 11 vi\u00ean, vi\u00ean n\u00e0o c\u0169ng tr\u00fang bia v\u00e0 \u0111\u1ec1u tr\u00fang c\u00e1c v\u00f2ng 8;9;10 \u0111i\u1ec3m. K\u1ebft th\u00fac cu\u1ed9c thi, D\u0169ng \u0111\u01b0\u1ee3c 100 \u0111i\u1ec3m. D\u0169ng vui l\u1eafm. C\u00f2n c\u00e1c b\u1ea1n c\u00f3 bi\u1ebft D\u0169ng \u0111\u00e3 b\u1eafn bao nhi\u00eau vi\u00ean v\u00e0 k\u1ebft qu\u1ea3 b\u1eafn v\u00e0o c\u00e1c v\u00f2ng ra sao kh\u00f4ng? 55","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i gi\u1ea3i: S\u1ed1 vi\u00ean \u0111\u1ea1n D\u0169ng \u0111\u00e3 b\u1eafn ph\u1ea3i \u00edt h\u01a1n 13 vi\u00ean (v\u00ec n\u1ebfu D\u0169ng b\u1eafn 13 vi\u00ean th\u00ec D\u0169ng \u0111\u01b0\u1ee3c s\u1ed1 \u0111i\u1ec3m \u00edt nh\u1ea5t l\u00e0: 8 x 11 + 9 x 1 + 10 x 1 = 107 (\u0111i\u1ec3m) > 100 \u0111i\u1ec3m, \u0111i\u1ec1u n\u00e0y v\u00f4 l\u00fd). Theo \u0111\u1ec1 b\u00e0i D\u0169ng \u0111\u00e3 b\u1eafn h\u01a1n 11 vi\u00ean n\u00ean s\u1ed1 vi\u00ean \u0111\u1ea1n D\u0169ng \u0111\u00e3 b\u1eafn l\u00e0 12 vi\u00ean. M\u1eb7t kh\u00e1c 12 vi\u00ean \u0111\u1ec1u tr\u00fang v\u00e0o c\u00e1c v\u00f2ng 8, 9, 10 \u0111i\u1ec3m n\u00ean \u00edt nh\u1ea5t c\u00f3 10 vi\u00ean v\u00e0o v\u00f2ng 8 \u0111i\u1ec3m, 1 vi\u00ean v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m, 1 vi\u00ean v\u00e0o v\u00f2ng 10 \u0111i\u1ec3m. Do \u0111\u00f3 s\u1ed1 \u0111i\u1ec3m D\u0169ng b\u1eafn \u0111\u01b0\u1ee3c \u00edt nh\u1ea5t l\u00e0: 8 x 10 + 9 x 1 + 10 x 1 = 99 (\u0111i\u1ec3m) S\u1ed1 \u0111i\u1ec3m h\u1ee5t \u0111i so v\u1edbi th\u1ef1c t\u1ebf l\u00e0: 100 - 99 = 1 (\u0111i\u1ec3m) Nh\u01b0 v\u1eady s\u1ebd c\u00f3 1 vi\u00ean kh\u00f4ng b\u1eafn v\u00e0o v\u00f2ng 8 \u0111i\u1ec3m m\u00e0 b\u1eafn v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m; ho\u1eb7c c\u00f3 1 vi\u00ean kh\u00f4ng b\u1eafn v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m m\u00e0 b\u1eafn v\u00e0o v\u00f2ng 10 \u0111i\u1ec3m. N\u1ebfu c\u00f3 1 vi\u00ean D\u0169ng kh\u00f4ng b\u1eafn v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m m\u00e0 b\u1eafn v\u00e0o v\u00f2ng 10 \u0111i\u1ec3m th\u00ec t\u1ed5ng c\u1ed9ng s\u1ebd c\u00f3 10 vi\u00ean v\u00e0o v\u00f2ng 8 \u0111i\u1ec3m v\u00e0 2 vi\u00ean v\u00e0o v\u00f2ng 10 \u0111i\u1ec3m (lo\u1ea1i v\u00ec kh\u00f4ng c\u00f3 vi\u00ean n\u00e0o b\u1eafn v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m). V\u1eady s\u1ebd c\u00f3 1 vi\u00ean kh\u00f4ng b\u1eafn v\u00e0o v\u00f2ng 8 \u0111i\u1ec3m m\u00e0 b\u1eafn v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m, t\u1ee9c l\u00e0 c\u00f3 9 vi\u00ean v\u00e0o v\u00f2ng 8 \u0111i\u1ec3m, 2 vi\u00ean v\u00e0o v\u00f2ng 9 \u0111i\u1ec3m v\u00e0 1 vi\u00ean v\u00e0o v\u00f2ng 10 \u0111i\u1ec3m. B\u00e0i 99: Ai xem ca nh\u1ea1c? M\u1ed9t gia \u0111\u00ecnh c\u00f3 n\u0103m ng\u01b0\u1eddi: b\u00e0 n\u1ed9i, b\u1ed1, m\u1eb9 v\u00e0 hai b\u1ea1n Chi, B\u1ea3o. M\u1ed9t h\u00f4m gia \u0111\u00ecnh \u0111\u01b0\u1ee3c t\u1eb7ng 2 v\u00e9 m\u1eddi xem ca nh\u1ea1c. N\u0103m \u00fd ki\u1ebfn c\u1ee7a n\u0103m ng\u01b0\u1eddi nh\u01b0 sau: a) \u201cB\u00e0 n\u1ed9i v\u00e0 m\u1eb9 \u0111i\u201d b) \u201cB\u1ed1 v\u00e0 m\u1eb9 \u0111i\u201d c) \u201cB\u1ed1 v\u00e0 b\u00e0 n\u1ed9i \u0111i\u201d d) \u201cB\u00e0 n\u1ed9i v\u00e0 Chi \u0111i\u201d e) \u201cB\u1ed1 v\u00e0 B\u1ea3o \u0111i\u201d Sau c\u00f9ng, m\u1ecdi ng\u01b0\u1eddi theo \u00fd ki\u1ebfn c\u1ee7a b\u00e0 n\u1ed9i v\u00e0 nh\u01b0 v\u1eady trong \u00fd ki\u1ebfn c\u1ee7a m\u1ecdi ng\u01b0\u1eddi kh\u00e1c \u0111\u1ec1u c\u00f3 m\u1ed9t ph\u1ea7n \u0111\u00fang. B\u00e0 n\u1ed9i \u0111\u00e3 n\u00f3i c\u00e2u n\u00e0o? B\u00e0i gi\u1ea3i: M\u1ed9t b\u00e0i to\u00e1n l\u00f4g\u00edc c\u01a1 b\u1ea3n v\u00e0 kh\u00f3, sau \u0111\u00e2y l\u00e0 l\u1eddi gi\u1ea3i. 56","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Ta k\u00fd hi\u1ec7u theo th\u1ee9 t\u1ef1 \u201c\u0111i xem\u201d ca nh\u1ea1c: n (B\u00e0 n\u1ed9i), m (m\u1eb9), b (B\u1ed1), C (Chi) v\u00e0 B (B\u1ea3o) v\u00e0 n\u0103m ng\u01b0\u1eddi tr\u00ean khi h\u1ecd \u201ckh\u00f4ng \u0111i\u201d l\u00e0 n, m, b, C v\u00e0 B. Nh\u01b0 v\u1eady theo \u00fd ki\u1ebfn c\u1ee7a n\u0103m ng\u01b0\u1eddi l\u00e0: a) n v\u00e0 m b) b v\u00e0 m c) b v\u00e0 n d) n v\u00e0 C e) b v\u00e0 B. M\u1ed7i trong n\u0103m \u00fd tr\u00ean \u0111\u1ec1u c\u00f3 m\u1ed9t ph\u1ea7n \u0111\u00fang v\u00e0 m\u1ed9t ph\u1ea7n sai (tr\u1eeb \u00fd c\u1ee7a b\u00e0!). C\u00e2u m\u00e0 b\u00e0 n\u1ed9i n\u00f3i l\u00e0 \u0111\u00fang v\u1edbi c\u1ea3 n\u0103m \u00fd tr\u00ean. - N\u1ebfu ch\u1ecdn c\u00e2u a) th\u00ec kh\u00f4ng c\u00f3 e t\u1ee9c b v\u00e0 B. - N\u1ebfu ch\u1ecdn c\u00e2u b) th\u00ec kh\u00f4ng c\u00f3 d t\u1ee9c n v\u00e0 C. - N\u1ebfu ch\u1ecdn c\u00e2u c) th\u00ec c\u00e1c \u00fd ki\u1ebfn kh\u00e1c c\u00f3 m\u1ed9t ph\u1ea7n \u0111\u00fang. B\u00e0 n\u1ed9i \u0111\u00e3 n\u00f3i c\u00e2u c) B\u00e0i 100: Ch\u01a1i b\u1ed1c di\u00eam Tr\u00ean m\u1eb7t b\u00e0n c\u00f3 18 que di\u00eam. Hai ng\u01b0\u1eddi tham gia cu\u1ed9c ch\u01a1i: M\u1ed7i ng\u01b0\u1eddi l\u1ea7n l\u01b0\u1ee3t \u0111\u1ebfn phi\u00ean m\u00ecnh l\u1ea5y ra m\u1ed9t s\u1ed1 que di\u00eam. M\u1ed7i l\u1ea7n, m\u1ed7i ng\u01b0\u1eddi l\u1ea5y ra kh\u00f4ng qu\u00e1 4 que. Ng\u01b0\u1eddi n\u00e0o l\u1ea5y \u0111\u01b0\u1ee3c s\u1ed1 que cu\u1ed1i c\u00f9ng th\u00ec ng\u01b0\u1eddi \u0111\u00f3 th\u1eafng. N\u1ebfu b\u1ea1n \u0111\u01b0\u1ee3c b\u1ed1c tr\u01b0\u1edbc, b\u1ea1n c\u00f3 ch\u1eafc ch\u1eafn th\u1eafng \u0111\u01b0\u1ee3c kh\u00f4ng? B\u00e0i gi\u1ea3i: Gi\u1ea3 s\u1eed r\u1eb1ng A v\u00e0 B tham gia cu\u1ed9c ch\u01a1i m\u00e0 A l\u1ea5y di\u00eam tr\u01b0\u1edbc. \u0110\u1ec3 ch\u1eafc th\u1eafng th\u00ec tr\u01b0\u1edbc l\u1ea7n cu\u1ed1i c\u00f9ng A ph\u1ea3i \u0111\u1ec3 l\u1ea1i 5 que di\u00eam, tr\u01b0\u1edbc \u0111\u00f3 A ph\u1ea3i \u0111\u1ec3 l\u1ea1i 10 que di\u00eam v\u00e0 l\u1ea7n b\u1ed1c \u0111\u1ea7u ti\u00ean A \u0111\u1ec3 l\u1ea1i 15 que di\u00eam, khi \u0111\u00f3 d\u00f9 B c\u00f3 b\u1ed1c bao nhi\u00eau que th\u00ec v\u1eabn c\u00f2n l\u1ea1i s\u1ed1 que \u0111\u1ec3 A ch\u1ec9 c\u1ea7n b\u1ed1c m\u1ed9t l\u1ea7n l\u00e0 h\u1ebft.Mu\u1ed1n v\u1eady th\u00ec l\u1ea7n tr\u01b0\u1edbc \u0111\u00f3 A ph\u1ea3i \u0111\u1ec3 l\u1ea1i 10 que di\u00eam , khi \u0111\u00f3 d\u00f9 B b\u1ed1c bao nhi\u00eau que v\u1eabn c\u00f2n l\u1ea1i s\u1ed1 que m\u00e0 A c\u00f3 th\u1ec3 b\u1ed1c \u0111\u1ec3 c\u00f2n l\u1ea1i 5 que . T\u01b0\u01a1ng t\u1ef1 nh\u01b0 th\u1ebf th\u00ec l\u1ea7n b\u1ed1c \u0111\u1ea7u ti\u00ean A ph\u1ea3i \u0111\u1ec3 l\u1ea1i 15 que di\u00eam . V\u1edbi \\\" chi\u1ebfn l\u01b0\u1ee3c\\\" n\u00e0y bao gi\u1edd A c\u0169ng l\u00e0 ng\u01b0\u1eddi th\u1eafng cu\u1ed9c. B\u00e0i 101: T\u00f4 m\u00e0u H\u00ecnh b\u00ean g\u1ed3m 6 \u0111\u1ec9nh A, B, C, D, E, F v\u00e0 c\u00e1c c\u1ea1nh n\u1ed1i m\u1ed9t s\u1ed1 \u0111\u1ec9nh v\u1edbi nhau. Ta t\u00f4 m\u00e0u c\u00e1c \u0111\u1ec9nh sao cho hai \u0111\u1ec9nh \u0111\u01b0\u1ee3c n\u1ed1i b\u1edfi m\u1ed9t c\u1ea1nh 57","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i ph\u1ea3i \u0111\u01b0\u1ee3c t\u00f4 b\u1edfi hai m\u00e0u kh\u00e1c nhau. H\u1ecfi ph\u1ea3i c\u1ea7n \u00edt nh\u1ea5t l\u00e0 bao nhi\u00eau m\u00e0u \u0111\u1ec3 l\u00e0m vi\u1ec7c \u0111\u00f3? B\u00e0i gi\u1ea3i: T\u1ea5t c\u1ea3 c\u00e1c \u0111\u1ec9nh A, B, C, D, E \u0111\u1ec1u n\u1ed1i v\u1edbi \u0111\u1ec9nh F n\u00ean \u0111\u1ec9nh F ph\u1ea3i t\u00f4 m\u00e0u kh\u00e1c v\u1edbi c\u00e1c \u0111\u1ec9nh c\u00f2n l\u1ea1i. V\u1edbi 5 \u0111\u1ec9nh c\u00f2n l\u1ea1i th\u00ec A v\u00e0 C t\u00f4 c\u00f9ng m\u1ed9t m\u00e0u. B v\u00e0 D t\u00f4 c\u00f9ng m\u1ed9t m\u00e0u, E t\u00f4 ri\u00eang m\u1ed9t m\u00e0u, nh\u01b0 v\u1eady c\u1ea7n \u00edt nh\u1ea5t 3 m\u00e0u \u0111\u1ec3 t\u00f4 5 \u0111\u1ec9nh sao cho 2 \u0111\u1ec9nh \u0111\u01b0\u1ee3c n\u1ed1i b\u1edfi m\u1ed9t c\u1ea1nh \u0111\u01b0\u1ee3c t\u00f4 b\u1edfi 2 m\u00e0u kh\u00e1c nhau. V\u1eady c\u1ea7n \u00edt nh\u1ea5t 4 m\u00e0u \u0111\u1ec3 t\u00f4 6 \u0111\u1ec9nh c\u1ee7a h\u00ecnh theo y\u00eau c\u1ea7u c\u1ee7a \u0111\u1ec1 b\u00e0i. B\u00e0i 102: \u0110i\u1ec1n s\u1ed1 tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n \u0110i\u1ec1n 6 s\u1ed1 ch\u1eb5n t\u1eeb 2 \u0111\u1ebfn 12 v\u00e0o c\u00e1c ch\u1ea5m tr\u00ean 3 v\u00f2ng tr\u00f2n sao cho t\u1ed5ng 3 s\u1ed1 n\u1eb1m tr\u00ean m\u1ed7i v\u00f2ng tr\u00f2n \u0111\u1ec1u b\u1eb1ng 18. B\u00e0i gi\u1ea3i: S\u00e1u s\u1ed1 ch\u1eb5n \u0111\u00f3 l\u00e0: 2, 4, 6, 8, 10, 12. Ta c\u00f3: 18 = 2 + 4 + 12 18 = 2 + 6 + 10 18 = 4 + 6 + 8 Tr\u00ean h\u00ecnh v\u1ebd ta th\u1ea5y c\u1ee9 hai \u0111\u01b0\u1eddng tr\u00f2n l\u1ea1i c\u00f3 m\u1ed9t \u0111i\u1ec3m chung. Nh\u01b0 v\u1eady s\u1ed1 n\u00e0o \u0111i\u1ec1n v\u00e0o \u0111i\u1ec3m chung \u0111\u00f3 s\u1ebd thu\u1ed9c hai t\u1ed5ng \u0111\u00e3 cho. Ta th\u1ea5y s\u1ed1 2, s\u1ed1 4, s\u1ed1 6 \u0111\u1ec1u l\u1eb7p l\u1ea1i hai l\u1ea7n n\u00ean ba s\u1ed1 \u0111\u00f3 \u0111\u01b0\u1ee3c \u0111i\u1ec1n v\u00e0o ba \u0111i\u1ec3m chung. C\u00e1c s\u1ed1 \u0111\u00e3 cho \u0111\u01b0\u1ee3c \u0111i\u1ec1n v\u00e0o h\u00ecnh v\u1ebd nh\u01b0 sau: 58","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 103 : T\u00ecm hai s\u1ed1 bi\u1ebft r\u1eb1ng t\u1ed5ng c\u1ee7a ch\u00fang g\u1ea5p 5 l\u1ea7n hi\u1ec7u c\u1ee7a ch\u00fang v\u00e0 t\u00edch c\u1ee7a ch\u00fang g\u1ea5p 4008 l\u1ea7n hi\u1ec7u c\u1ee7a ch\u00fang. B\u00e0i gi\u1ea3i : Coi hi\u1ec7u c\u1ee7a hai s\u1ed1 l\u00e0 1 ph\u1ea7n th\u00ec t\u1ed5ng c\u1ee7a ch\u00fang l\u00e0 5 ph\u1ea7n. Do \u0111\u00f3 s\u1ed1 l\u1edbn l\u00e0 (5 + 1) : 2 = 3 (ph\u1ea7n). S\u1ed1 b\u00e9 l\u00e0 : 3 - 1 = 2 (ph\u1ea7n). T\u00edch c\u1ee7a hai s\u1ed1 l\u00e0 : 2 x 3 = 6 (ph\u1ea7n), m\u00e0 t\u00edch hai s\u1ed1 l\u00e0 4008 n\u00ean gi\u00e1 tr\u1ecb m\u1ed9t ph\u1ea7n l\u00e0 : 4008 : 6 = 668. S\u1ed1 b\u00e9 l\u00e0 : 668 x 2 = 1336 ; s\u1ed1 l\u1edbn l\u00e0 : 668 x 3 = 2004. B\u00e0i 104 : Trong kho c\u1ee7a m\u1ed9t \u0111\u01a1n v\u1ecb d\u00e2n c\u00f4ng c\u00f2n l\u1ea1i \u0111\u00fang m\u1ed9t bao g\u1ea1o ch\u1ee9a 39 kg g\u1ea1o. B\u00e1c c\u1ea5p d\u01b0\u1ee1ng c\u1ea7n l\u1ea5y ra 11\/13 s\u1ed1 g\u1ea1o \u0111\u00f3. H\u1ecfi ch\u1ec9 v\u1edbi m\u1ed9t chi\u1ebfc c\u00e2n lo\u1ea1i c\u00e2n \u0111\u0129a v\u00e0 m\u1ed9t qu\u1ea3 c\u00e2n 1 kg, b\u00e1c c\u1ea5p d\u01b0\u1ee1ng ph\u1ea3i l\u00e0m th\u1ebf n\u00e0o \u0111\u1ec3 ch\u1ec9 sau 3 l\u1ea7n c\u00e2n l\u1ea5y ra \u0111\u1ee7 s\u1ed1 g\u1ea1o c\u1ea7n d\u00f9ng. B\u00e0i gi\u1ea3i : S\u1ed1 g\u1ea1o b\u00e1c c\u1ea5p d\u01b0\u1ee1ng c\u1ea7n l\u1ea5y ra l\u00e0 : 39 x 11\/13 = 33 (kg) S\u1ed1 g\u1ea1o c\u00f2n l\u1ea1i sau khi b\u00e1c c\u1ea5p d\u01b0\u1ee1ng l\u1ea5y l\u00e0 : 39 - 33 = 6 (kg) C\u00e1ch th\u1ef1c hi\u1ec7n c\u00e2n nh\u01b0 sau : L\u1ea7n 1 : \u0110\u1eb7t qu\u1ea3 c\u00e2n l\u00ean m\u1ed9t \u0111\u0129a c\u00e2n, \u0111\u1ed5 g\u1ea1o v\u00e0o \u0111\u0129a c\u00e2n b\u00ean kia \u0111\u1ebfn khi c\u00e2n th\u0103ng b\u1eb1ng, \u0111\u01b0\u1ee3c 1 kg g\u1ea1o. L\u1ea7n 2 : \u0110\u1eb7t qu\u1ea3 c\u00e2n sang \u0111\u0129a c\u00f3 1 kg g\u1ea1o v\u1eeba c\u00e2n \u0111\u01b0\u1ee3c r\u1ed3i \u0111\u1ed5 g\u1ea1o v\u00e0o \u0111\u0129a c\u00e2n tr\u1ed1ng \u0111\u1ebfn khi c\u00e2n th\u0103ng b\u1eb1ng, \u0111\u01b0\u1ee3c 2 kg g\u1ea1o. L\u1ea7n 3 : \u0110\u1eb7t c\u1ea3 3 kg g\u1ea1o c\u00e2n \u0111\u01b0\u1ee3c \u1edf hai l\u1ea7n tr\u00ean v\u00e0o m\u1ed9t \u0111\u0129a c\u00e2n, \u0111\u0129a c\u00e2n kia \u0111\u1ed5 g\u1ea1o v\u00e0o cho \u0111\u1ebfn khi c\u00e2n th\u0103ng b\u1eb1ng, \u0111\u01b0\u1ee3c m\u1ed7i b\u00ean 3 kg g\u1ea1o. Nh\u01b0 v\u1eady s\u1ed1 g\u1ea1o c\u00f3 \u0111\u01b0\u1ee3c sau ba l\u1ea7n c\u00e2n l\u00e0 6 kg. S\u1ed1 g\u1ea1o c\u00f2n l\u1ea1i trong bao ch\u00ednh l\u00e0 s\u1ed1 g\u1ea1o m\u00e0 b\u00e1c c\u1ea5p d\u01b0\u1ee1ng c\u1ea7n d\u00f9ng. 59","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 105 : Lan n\u00f3i m\u1ed9t s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 b\u1ea5t k\u00ec s\u1ebd b\u1eb1ng 1\/5 s\u1ed1 vi\u1ebft theo th\u1ee9 t\u1ef1 ng\u01b0\u1ee3c l\u1ea1i. \u0110\u1ed1 b\u1ea1n bi\u1ebft Lan n\u00f3i \u0111\u00fang hay sai ? B\u00e0i gi\u1ea3i : G\u1ecdi s\u1ed1 \u0111\u00f3 l\u00e0 (a > 0 ; a, b, c, d < 10). S\u1ed1 vi\u1ebft theo th\u1ee9 t\u1ef1 ng\u01b0\u1ee3c l\u1ea1i l\u00e0 Theo \u0111\u1ea7u b\u00e0i ta c\u00f3 : Nh\u01b0ng d x 5 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 0 ho\u1eb7c 5 (kh\u00e1c 1) n\u00ean kh\u00f4ng t\u00ecm \u0111\u01b0\u1ee3c gi\u00e1 tr\u1ecb c\u1ee7a a ho\u1eb7c d. V\u1eady b\u1ea1n Lan n\u00f3i sai. B\u00e0i 106 : B\u00e1c Phong c\u00f3 m\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt, chi\u1ec1u r\u1ed9ng m\u1ea3nh \u0111\u1ea5t d\u00e0i 8 m. B\u00e1c ng\u0103n m\u1ea3nh \u0111\u00f3 th\u00e0nh hai ph\u1ea7n, m\u1ed9t ph\u1ea7n \u0111\u1ec3 l\u00e0m nh\u00e0, ph\u1ea7n c\u00f2n l\u1ea1i \u0111\u1ec3 l\u00e0m v\u01b0\u1eddn. Di\u1ec7n t\u00edch ph\u1ea7n \u0111\u1ea5t l\u00e0m nh\u00e0 b\u1eb1ng 1\/2 di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t c\u00f2n chu vi ph\u1ea7n \u0111\u1ea5t l\u00e0m nh\u00e0 b\u1eb1ng 2\/3 chu vi m\u1ea3nh \u0111\u1ea5t. T\u00ednh di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t c\u1ee7a b\u00e1c. B\u00e0i gi\u1ea3i : C\u00f3 hai c\u00e1ch chia m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh hai ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. C\u00e1ch chia 1 : nh\u01b0 h\u00ecnh 1. H\u00ecnh 1 G\u1ecdi m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 ABCD v\u00e0 ph\u1ea7n \u0111\u1ea5t l\u00e0m nh\u00e0 l\u00e0 AMND. V\u00ec di\u1ec7n t\u00edch ph\u1ea7n \u0111\u1ea5t l\u00e0m nh\u00e0 b\u1eb1ng n\u1eeda di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t n\u00ean M, N l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a AB v\u00e0 CD. Do \u0111\u00f3 AM = MB = CN = ND. Chu vi c\u1ee7a ph\u1ea7n \u0111\u1ea5t l\u00e0m nh\u00e0 l\u00e0 : (AM + AD) x 2 = (AM + 8) x 2 = = AM x 2 + 8 x 2 = AB + 16. Chu vi c\u1ee7a m\u1ea3nh \u0111\u1ea5t l\u00e0 : (AB + AD) 2 = (AB + 8) x 2 = = AB x 2 + 8 x 2 = AB x 2 + 16. Hi\u1ec7u chu vi m\u1ea3nh \u0111\u1ea5t v\u00e0 chu vi ph\u1ea7n \u0111\u1ea5t l\u00e0m nh\u00e0 l\u00e0 : 60","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i (AB x 2 + 16) - (AB + 16) = AB. Hi\u1ec7u n\u00e0y so v\u1edbi chu vi m\u1ea3nh \u0111\u1ea5t th\u00ec chi\u1ebfm : 1 - 2\/3 = 1\/3 (chu vi m\u1ea3nh \u0111\u1ea5t) Do \u0111\u00f3 ta c\u00f3 : AB x 3 = AB x 2 + 16 AB x 3 - AB x 2 = 16 AB x (3 - 2) = 16 AB = 16 (m). V\u1eady di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t l\u00e0 : 16 x 8 = 128 (m2) C\u00e1ch chia 2 : nh\u01b0 h\u00ecnh 2. H\u00ecnh 2 L\u1eadp lu\u1eadn t\u01b0\u01a1ng t\u1ef1 tr\u01b0\u1eddng h\u1ee3p tr\u00ean, ta t\u00ecm \u0111\u01b0\u1ee3c AB = 4 m. \u0110i\u1ec1u n\u00e0y v\u00f4 l\u00ed v\u00ec AB l\u00e0 chi\u1ec1u d\u00e0i c\u1ee7a m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt, \u0111\u01b0\u01a1ng nhi\u00ean ph\u1ea3i l\u1edbn h\u01a1n 8 m. Do \u0111\u00f3 tr\u01b0\u1eddng h\u1ee3p n\u00e0y b\u1ecb lo\u1ea1i. B\u00e0i 107 : Cho m\u1ed9t ph\u00e9p chia hai s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 d\u01b0. T\u1ed5ng c\u00e1c s\u1ed1 : s\u1ed1 b\u1ecb chia, s\u1ed1 chia, s\u1ed1 th\u01b0\u01a1ng v\u00e0 s\u1ed1 d\u01b0 l\u00e0 769. S\u1ed1 th\u01b0\u01a1ng l\u00e0 15 v\u00e0 s\u1ed1 d\u01b0 l\u00e0 s\u1ed1 d\u01b0 l\u1edbn nh\u1ea5t c\u00f3 th\u1ec3 c\u00f3 trong ph\u00e9p chia \u0111\u00f3. H\u00e3y t\u00ecm s\u1ed1 b\u1ecb chia v\u00e0 s\u1ed1 chia trong ph\u00e9p chia. B\u00e0i gi\u1ea3i : S\u1ed1 d\u01b0 trong ph\u00e9p chia l\u00e0 s\u1ed1 d\u01b0 l\u1edbn nh\u1ea5t n\u00ean k\u00e9m s\u1ed1 chia 1 \u0111\u01a1n v\u1ecb. Ta c\u00f3 s\u01a1 \u0111\u1ed3 sau: Theo s\u01a1 \u0111\u1ed3, n\u1ebfu g\u1ecdi s\u1ed1 chia l\u00e0 1 ph\u1ea7n, th\u00eam 1 \u0111\u01a1n v\u1ecb v\u00e0o s\u1ed1 d\u01b0 v\u00e0 s\u1ed1 b\u1ecb chia th\u00ec t\u1ed5ng s\u1ed1 ph\u1ea7n c\u1ee7a s\u1ed1 chia, s\u1ed1 b\u1ecb chia v\u00e0 s\u1ed1 d\u01b0 (m\u1edbi) g\u1ed3m : 15 + 1 + 1 + 1 = 18 (ph\u1ea7n) nh\u01b0 v\u1eady. Khi \u0111\u00f3 t\u1ed5ng c\u1ee7a s\u1ed1 chia, s\u1ed1 b\u1ecb chia v\u00e0 s\u1ed1 d\u01b0 (m\u1edbi) l\u00e0 : 769 - 15 + 1 + 1 = 756. S\u1ed1 chia l\u00e0 : 756 : 18 = 42 S\u1ed1 d\u01b0 l\u00e0 : 42 - 1 = 41 S\u1ed1 b\u1ecb chia l\u00e0 : 42 x 15 + 41 = 671 61","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 108 : S\u1ed1 t\u00e1o c\u1ee7a An, B\u00ecnh v\u00e0 Chi l\u00e0 nh\u01b0 nhau. An cho \u0111i 17 qu\u1ea3, B\u00ecnh cho \u0111i 19 qu\u1ea3 th\u00ec l\u00fac n\u00e0y s\u1ed1 t\u00e1o c\u1ee7a Chi g\u1ea5p 5 l\u1ea7n t\u1ed5ng s\u1ed1 t\u00e1o c\u00f2n l\u1ea1i c\u1ee7a An v\u00e0 B\u00ecnh. H\u1ecfi l\u00fac \u0111\u1ea7u m\u1ed7i b\u1ea1n c\u00f3 bao nhi\u00eau qu\u1ea3 t\u00e1o ? B\u00e0i gi\u1ea3i : N\u1ebfu coi s\u1ed1 t\u00e1o c\u1ee7a Chi g\u1ed3m 5 ph\u1ea7n th\u00ec t\u1ed5ng s\u1ed1 t\u00e1o c\u1ee7a An v\u00e0 B\u00ecnh l\u00e0 10 ph\u1ea7n. S\u1ed1 t\u00e1o m\u00e0 An v\u00e0 B\u00ecnh \u0111\u00e3 cho \u0111i l\u00e0 : 17 + 19 = 36 (qu\u1ea3) V\u00ec s\u1ed1 t\u00e1o c\u1ee7a Chi g\u1ea5p 5 l\u1ea7n t\u1ed5ng s\u1ed1 t\u00e1o c\u00f2n l\u1ea1i c\u1ee7a An v\u00e0 B\u00ecnh n\u00ean s\u1ed1 t\u00e1o c\u00f2n l\u1ea1i c\u1ee7a hai b\u1ea1n g\u1ed3m 1 ph\u1ea7n. Nh\u01b0 v\u1eady An v\u00e0 B\u00ecnh \u0111\u00e3 cho \u0111i s\u1ed1 ph\u1ea7n l\u00e0 : 10 - 1 = 9 (ph\u1ea7n) V\u1eady s\u1ed1 t\u00e1o c\u1ee7a Chi l\u00e0 : (36 : 9) x 5 = 20 (qu\u1ea3) V\u00ec ba b\u1ea1n c\u00f3 s\u1ed1 t\u00e1o b\u1eb1ng nhau n\u00ean m\u1ed7i b\u1ea1n l\u00fac \u0111\u1ea7u c\u00f3 20 qu\u1ea3. B\u00e0i 109 : Con s\u1ed1 n\u00e0o trong c\u00e1c s\u1ed1 2, 3, 4, 5 c\u1ea7n thay v\u00e0o d\u1ea5u ch\u1ea5m h\u1ecfi (?) \u0111\u1ec3 h\u1ee3p l\u00f4gic ? B\u00e0i gi\u1ea3i : G\u1ecdi s\u1ed1 thay v\u00e0o h\u00ecnh tr\u00f2n l\u00e0 a, s\u1ed1 thay v\u00e0o tam gi\u00e1c l\u00e0 b v\u00e0 s\u1ed1 thay v\u00e0o h\u00ecnh vu\u00f4ng l\u00e0 c, ta c\u00f3 : a + 3 x b = 22. V\u00ec 3 x b chia h\u1ebft cho 3 ; 22 chia cho 3 d\u01b0 1 n\u00ean a chia cho 3 d\u01b0 1 (*). Ta l\u1ea1i c\u00f3 2 x a + 2 x c = 10, c nh\u1ecf nh\u1ea5t l\u00e0 2 n\u00ean a l\u1edbn nh\u1ea5t l\u00e0 (10 - 2 x 2) : 2 = 3 (**). T\u1eeb (*) v\u00e0 (**) ta c\u00f3 a = 1. Do \u0111\u00f3 1 + 3 x b = 22 ; b = (22 - 1) : 3 = 7 ; c = (10 - 2 x 1) : 2 = 4. V\u1eady s\u1ed1 c\u1ea7n thay v\u00e0o d\u1ea5u ch\u1ea5m h\u1ecfi \u0111\u1ec3 h\u1ee3p l\u00f4gic l\u00e0 s\u1ed1 4. B\u00e0i 110 : H\u00e3y d\u00f9ng t\u1ea5t c\u1ea3 c\u00e1c ch\u1eef s\u1ed1, m\u1ed7i ch\u1eef s\u1ed1 m\u1ed9t l\u1ea7n \u0111\u1ec3 vi\u1ebft n\u0103m s\u1ed1 t\u1ef1 nhi\u00ean, trong \u0111\u00f3 c\u00f3 m\u1ed9t s\u1ed1 l\u1ea7n l\u01b0\u1ee3t b\u1eb1ng 1\/2 ; 1\/3 ; 1\/4 v\u00e0 1\/5 c\u00e1c s\u1ed1 c\u00f2n l\u1ea1i. B\u00e0i gi\u1ea3i : G\u1ecdi 5 s\u1ed1 t\u1ef1 nhi\u00ean x\u1ebfp theo th\u1ee9 t\u1ef1 t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn l\u00e0 A ; B ; C ; D ; E. 62","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i N\u1ebfu A c\u00f3 1 ch\u1eef s\u1ed1 th\u00ec E kh\u00f4ng v\u01b0\u1ee3t qu\u00e1 9 x 5 = 45. Nh\u01b0 th\u1ebf c\u00f3 4 s\u1ed1 c\u00f3 kh\u00f4ng qu\u00e1 2 ch\u1eef s\u1ed1 n\u00ean m\u1edbi ch\u1ec9 d\u00f9ng kh\u00f4ng qu\u00e1 9 ch\u1eef s\u1ed1 (2 x 4 + 1 = 9). V\u1eady A c\u00f3 nhi\u1ec1u h\u01a1n 1 ch\u1eef s\u1ed1. N\u1ebfu E c\u00f3 3 ch\u1eef s\u1ed1 th\u00ec A c\u00f3 \u00edt nh\u1ea5t 2 ch\u1eef s\u1ed1 (v\u00ec 100 : 5 = 20). Nh\u01b0 v\u1eady c\u00f3 4 s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 v\u00e0 1 s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 n\u00ean ph\u1ea3i d\u00f9ng nhi\u1ec1u h\u01a1n 10 ch\u1eef s\u1ed1 (2 x 4 + 3 = 11). V\u1eady c\u1ea3 5 s\u1ed1 ph\u1ea3i l\u00e0 c\u00e1c s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 v\u00e0 E l\u1edbn h\u01a1n 45 chia h\u1ebft cho 5. V\u1eady E c\u00f3 th\u1ec3 l\u00e0 : 95 ; 90 ; 85 ; 80 ; 75 ; 70 ; 65 ; 60 ; 55 ; 50. Ta c\u00f3 b\u1ea3ng l\u1ef1a ch\u1ecdn sau : S\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 18, s\u1ed1 th\u1ee9 hai l\u00e0 36, s\u1ed1 th\u1ee9 ba l\u00e0 54, s\u1ed1 th\u1ee9 t\u01b0 l\u00e0 72 v\u00e0 s\u1ed1 th\u1ee9 5 l\u00e0 90. B\u00e0i 111 : B\u1ea1n h\u00e3y x\u00f3a nh\u1eefng ch\u1eef s\u1ed1 n\u00e0o \u0111\u00f3 \u0111\u1ec3 \u0111\u01b0\u1ee3c ph\u00e9p t\u00ednh \u0111\u00fang : 151 x 375 = 450. B\u00e0i gi\u1ea3i : Hai th\u1eeba s\u1ed1 \u1edf v\u1ebf tr\u00e1i \u0111\u1eb3ng th\u1ee9c ch\u1ec9 c\u00f3 c\u00e1c ch\u1eef s\u1ed1 l\u1ebb n\u00ean d\u00f9 x\u00f3a c\u00e1c ch\u1eef s\u1ed1 nh\u01b0 th\u1ebf n\u00e0o th\u00ec k\u1ebft qu\u1ea3 ph\u00e9p nh\u00e2n c\u0169ng l\u00e0 m\u1ed9t s\u1ed1 l\u1ebb. V\u1eady v\u1ebf ph\u1ea3i ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 45 ho\u1eb7c 5. Tr\u01b0\u1eddng h\u1ee3p 1 : K\u1ebft qu\u1ea3 ph\u00e9p nh\u00e2n l\u00e0 45 ta c\u00f3 m\u1ed9t c\u00e1ch x\u00f3a : Tr\u01b0\u1eddng h\u1ee3p 2 : K\u1ebft qu\u1ea3 ph\u00e9p nh\u00e2n l\u00e0 5 ta c\u00f3 hai c\u00e1ch x\u00f3a : 63","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 112 : C\u00f3 hai t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng m\u00e0 s\u1ed1 \u0111o c\u00e1c c\u1ea1nh l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean chia h\u1ebft cho 3. \u0110\u1eb7t t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng nh\u1ecf l\u00ean t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng l\u1edbn th\u00ec di\u1ec7n t\u00edch ph\u1ea7n t\u1ea5m b\u00eca kh\u00f4ng b\u1ecb ch\u1ed3ng l\u00ean l\u00e0 63 cm2. T\u00ecm c\u1ea1nh c\u1ee7a m\u1ed7i t\u1ea5m b\u00eca \u0111\u00f3. B\u00e0i gi\u1ea3i : Ta \u0111\u1eb7t t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng nh\u1ecf l\u00ean t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng l\u1edbn sao cho c\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf tr\u00f9ng kh\u00edt v\u1edbi c\u1ea1nh h\u00ecnh vu\u00f4ng l\u1edbn. G\u1ecdi hai h\u00ecnh vu\u00f4ng l\u00e0 ABCD v\u00e0 AEGH. Di\u1ec7n t\u00edch ph\u1ea7n t\u1ea5m b\u00eca kh\u00f4ng b\u1ecb ch\u1ed3ng l\u00ean bao g\u1ed3m hai h\u00ecnh ch\u1eef nh\u1eadt BCKE v\u00e0 DKGH. Hai h\u00ecnh ch\u1eef nh\u1eadt n\u00e0y c\u00f3 BE = DH (ch\u00ednh l\u00e0 hi\u1ec7u s\u1ed1 \u0111o c\u00e1c c\u1ea1nh c\u1ee7a hai h\u00ecnh vu\u00f4ng). Chuy\u1ec3n h\u00ecnh ch\u1eef nh\u1eadt BCKE xu\u1ed1ng b\u00ean c\u1ea1nh h\u00ecnh ch\u1eef nh\u1eadt DKGH ta \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt GKMN. Khi \u0111\u00f3 ta c\u00f3 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt HDMN l\u00e0 63 cm2. Ta th\u1ea5y h\u00ecnh ch\u1eef nh\u1eadt HDMN c\u00f3 chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng ch\u00ednh l\u00e0 t\u1ed5ng v\u00e0 hi\u1ec7u s\u1ed1 \u0111o hai c\u1ea1nh h\u00ecnh vu\u00f4ng. V\u00ec hai h\u00ecnh vu\u00f4ng \u0111\u1ec1u c\u00f3 s\u1ed1 \u0111o c\u00e1c c\u1ea1nh l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean chia h\u1ebft cho 3, n\u00ean t\u1ed5ng v\u00e0 hi\u1ec7u s\u1ed1 \u0111o hai c\u1ea1nh h\u00ecnh vu\u00f4ng c\u0169ng ph\u1ea3i l\u00e0 s\u1ed1 chia h\u1ebft cho 3. Do \u0111\u00f3 chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt HDMN \u0111\u1ec1u l\u00e0 s\u1ed1 chia h\u1ebft cho 3. V\u00ec 63 = 1 x 63 = 3 x 21 = 7 x 9 n\u00ean chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt HDMN ph\u1ea3i l\u00e0 21 cm v\u00e0 3 cm. V\u1eady \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u1ee7a t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng nh\u1ecf l\u00e0 : (21 - 3) : 2 = 9 (cm) \u0110\u1ed9 d\u00e0i c\u1ea1nh c\u1ee7a t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng l\u1edbn l\u00e0 : 9 + 3 = 12 (cm) B\u00e0i 113 : So s\u00e1nh M v\u00e0 N bi\u1ebft : B\u00e0i gi\u1ea3i : 64","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 114 : M\u1ed9t b\u1ea3ng \u00f4 vu\u00f4ng g\u1ed3m 3 d\u00f2ng v\u00e0 8 c\u1ed9t nh\u01b0 h\u00ecnh v\u1ebd. Tr\u00ean m\u1ed7i d\u00f2ng ta \u0111i\u1ec1n c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 8 v\u00e0o m\u1ed7i \u00f4 theo th\u1ee9 t\u1ef1 t\u00f9y \u00fd (m\u1ed7i \u00f4 m\u1ed9t s\u1ed1 v\u00e0 m\u1ed7i s\u1ed1 ch\u1ec9 \u0111i\u1ec1n m\u1ed9t l\u1ea7n) sao cho t\u1ed5ng c\u00e1c s\u1ed1 \u1edf 8 c\u1ed9t \u0111\u1ec1u b\u1eb1ng nhau. B\u1ea1n Nhi cho r\u1eb1ng c\u00f3 th\u1ec3 l\u00e0m \u0111\u01b0\u1ee3c c\u00f2n b\u1ea1n T\u00edn kh\u1eb3ng \u0111\u1ecbnh kh\u00f4ng \u0111i\u1ec1n \u0111\u01b0\u1ee3c. H\u1ecfi ai \u0111\u00fang, ai sai ? B\u00e0i gi\u1ea3i : Gi\u1ea3 s\u1eed c\u00f3 th\u1ec3 \u0111i\u1ec1n \u0111\u01b0\u1ee3c theo y\u00eau c\u1ea7u b\u00e0i to\u00e1n (B\u1ea1n Nhi n\u00f3i \u0111\u00fang). T\u1ed5ng c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 8 l\u00e0 : 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36. M\u1ed7i d\u00f2ng \u0111i\u1ec1n c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 8 n\u00ean t\u1ed5ng c\u00e1c s\u1ed1 tr\u00ean 3 d\u00f2ng trong b\u1ea3ng \u00f4 vu\u00f4ng \u0111\u00f3 l\u00e0 : 36 x 3 = 108. V\u00ec t\u1ed5ng c\u00e1c s\u1ed1 \u1edf 8 c\u1ed9t \u0111\u1ec1u b\u1eb1ng nhau n\u00ean t\u1ed5ng t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 trong b\u1ea3ng \u00f4 vu\u00f4ng ph\u1ea3i l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho 8. Nh\u01b0ng 108 kh\u00f4ng chia h\u1ebft cho 8 n\u00ean \u0111i\u1ec1u gi\u1ea3 s\u1eed \u1edf tr\u00ean l\u00e0 sai t\u1ee9c l\u00e0 b\u1ea1n Nhi n\u00f3i sai v\u00e0 b\u1ea1n T\u00edn n\u00f3i \u0111\u00fang. B\u00e0i 115 : N\u1ebfu \u0111\u1ebfm c\u00e1c ch\u1eef s\u1ed1 ghi t\u1ea5t c\u1ea3 c\u00e1c ng\u00e0y trong n\u0103m 2004 tr\u00ean t\u1edd l\u1ecbch treo t\u01b0\u1eddng th\u00ec s\u1ebd \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 bao nhi\u00eau ? B\u00e0i gi\u1ea3i : N\u0103m 2004 l\u00e0 n\u0103m nhu\u1eadn c\u00f3 366 ng\u00e0y. M\u1ed9t n\u0103m c\u00f3 12 th\u00e1ng, m\u1ed7i th\u00e1ng c\u00f3 9 ng\u00e0y t\u1eeb m\u00f9ng 1 \u0111\u1ebfn m\u00f9ng 9 l\u00e0 nh\u1eefng ng\u00e0y \u0111\u01b0\u1ee3c vi\u1ebft b\u1eb1ng c\u00e1c s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1. Nh\u01b0 v\u1eady s\u1ed1 ng\u00e0y \u0111\u01b0\u1ee3c vi\u1ebft b\u1eb1ng s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1 l\u00e0 : 9 x 12 = 108 (ng\u00e0y). S\u1ed1 ng\u00e0y c\u00f2n l\u1ea1i trong n\u0103m \u0111\u01b0\u1ee3c vi\u1ebft b\u1eb1ng s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 l\u00e0 : 65","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i 366 - 108 = 258 (ng\u00e0y). V\u1eady \u0111\u1ebfm c\u00e1c ch\u1eef s\u1ed1 ghi t\u1ea5t c\u1ea3 c\u00e1c ng\u00e0y c\u1ee7a n\u0103m 2004 tr\u00ean t\u1edd l\u1ecbch th\u00ec ta \u0111\u01b0\u1ee3c : 1 x 108 + 2 x 258 = 624 (ch\u1eef s\u1ed1). B\u00e0i 116 : Cho : H\u00e3y so s\u00e1nh S v\u00e0 1\/2. B\u00e0i gi\u1ea3i : B\u00e0i 117 : Cho m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean, n\u1ebfu vi\u1ebft th\u00eam m\u1ed9t ch\u1eef s\u1ed1 v\u00e0o b\u00ean ph\u1ea3i s\u1ed1 \u0111\u00f3 ta \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi h\u01a1n s\u1ed1 \u0111\u00e3 cho \u0111\u00fang 2004 \u0111\u01a1n v\u1ecb. T\u00ecm s\u1ed1 \u0111\u00e3 cho v\u00e0 ch\u1eef s\u1ed1 vi\u1ebft th\u00eam. B\u00e0i gi\u1ea3i : C\u00e1ch 1 : Khi vi\u1ebft th\u00eam m\u1ed9t ch\u1eef s\u1ed1 n\u00e0o \u0111\u00f3 v\u00e0o b\u00ean ph\u1ea3i m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00e3 cho ta \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi b\u1eb1ng 10 l\u1ea7n s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00f3 c\u1ed9ng th\u00eam ch\u00ednh ch\u1eef s\u1ed1 vi\u1ebft th\u00eam. G\u1ecdi ch\u1eef s\u1ed1 vi\u1ebft th\u00eam l\u00e0 a, ta c\u00f3 s\u01a1 \u0111\u1ed3 : 9 l\u1ea7n s\u1ed1 \u0111\u00e3 cho l\u00e0 : 2004 - a. S\u1ed1 \u0111\u00e3 cho l\u00e0 : (2004 - a) : 9. V\u00ec s\u1ed1 \u0111\u00e3 cho l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean n\u00ean 2004 - a ph\u1ea3i chia h\u1ebft cho 9, s\u1ed1 2004 chia 9 d\u01b0 6 n\u00ean a chia cho 9 ph\u1ea3i d\u01b0 6, m\u00e0 a l\u00e0 ch\u1eef s\u1ed1 n\u00ean a = 6. S\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00e3 cho l\u00e0 (2004 - 6) : 9 = 222. C\u00e1ch 2 : G\u1ecdi s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00e3 cho l\u00e0 A ch\u1eef s\u1ed1 vi\u1ebft th\u00eam l\u00e0 x th\u00ec s\u1ed1 m\u1edbi l\u00e0 . Ta c\u00f3 - A = 2004 66","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i A x 10 + x - A = 2004 (ph\u00e2n t\u00edch s\u1ed1) A x 10 - A + x = 2004 A x (10 - 1) + x = 2004 (m\u1ed9t s\u1ed1 nh\u00e2n v\u1edbi m\u1ed9t t\u1ed5ng) A x 9 + x = 2004 V\u00ec A x 9 chia h\u1ebft cho 9 ; 2004 chia 9 d\u01b0 6 n\u00ean x chia cho 9 ph\u1ea3i d\u01b0 6. V\u00ec x l\u00e0 ch\u1eef s\u1ed1 n\u00ean x = 6. Ta c\u00f3 : A x 9 + 6 = 2004 A x 9 = 2004 - 6 A x 9 = 1998 A = 1998 : 9 A = 222. V\u1eady s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00e3 cho l\u00e0 222 ; ch\u1eef s\u1ed1 vi\u1ebft th\u00eam l\u00e0 6. B\u00e0i 118 : M\u1ed9t t\u1edd gi\u1ea5y h\u00ecnh vu\u00f4ng c\u00f3 di\u1ec7n t\u00edch l\u00e0 72 cm2 th\u00ec \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a t\u1edd gi\u1ea5y \u0111\u00f3 d\u00e0i bao nhi\u00eau ? B\u00e0i gi\u1ea3i : G\u1ecdi t\u1edd gi\u1ea5y h\u00ecnh vu\u00f4ng l\u00e0 ABCD. N\u1ed1i hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i O (h\u00ecnh v\u1ebd). H\u00ecnh vu\u00f4ng \u0111\u01b0\u1ee3c chia th\u00e0nh 4 tam gi\u00e1c vu\u00f4ng nh\u1ecf c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Di\u1ec7n t\u00edch tam gi\u00e1c AOB l\u00e0 : 72 : 4 = 18 (cm2). V\u00ec di\u1ec7n t\u00edch tam gi\u00e1c AOB b\u1eb1ng (OA x OB) : 2, do \u0111\u00f3 (OA x OB) : 2 = 18 (cm2). Suy ra OA x OB = 36 (cm2). V\u00ec OA = OB m\u00e0 36 = 6 x 6 n\u00ean OA = 6 (cm). V\u00ec AC = 2 x OA n\u00ean \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a t\u1edd gi\u1ea5y \u0111\u00f3 l\u00e0 : 6 x 2 = 12 (cm). 67","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 119 : Trong \u0111\u1ee3t tr\u1ed3ng c\u00e2y \u0111\u1ea7u n\u0103m, l\u1edbp 5A c\u1eed m\u1ed9t s\u1ed1 b\u1ea1n \u0111i tr\u1ed3ng c\u00e2y v\u00e0 tr\u1ed3ng \u0111\u01b0\u1ee3c 180 c\u00e2y, m\u1ed7i h\u1ecdc sinh tr\u1ed3ng \u0111\u01b0\u1ee3c 8 ho\u1eb7c 9 c\u00e2y. T\u00ednh s\u1ed1 h\u1ecdc sinh tham gia tr\u1ed3ng c\u00e2y, bi\u1ebft s\u1ed1 h\u1ecdc sinh tham gia l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho 3. B\u00e0i gi\u1ea3i : N\u1ebfu m\u1ed7i b\u1ea1n tr\u1ed3ng 9 c\u00e2y th\u00ec s\u1ed1 ng\u01b0\u1eddi tham gia s\u1ebd \u00edt nh\u1ea5t v\u00e0 ch\u00ednh l\u00e0 : 180 : 9 = 20 (ng\u01b0\u1eddi). V\u00ec 180 : 8 = 22 (d\u01b0 4) n\u00ean s\u1ed1 ng\u01b0\u1eddi tham gia nhi\u1ec1u nh\u1ea5t l\u00e0 22 ng\u01b0\u1eddi v\u00e0 khi \u0111\u00f3 c\u00f3 4 ng\u01b0\u1eddi tr\u1ed3ng 9 c\u00e2y, c\u00f2n l\u1ea1i m\u1ed7i ng\u01b0\u1eddi tr\u1ed3ng 8 c\u00e2y. Theo \u0111\u1ea7u b\u00e0i s\u1ed1 ng\u01b0\u1eddi tham gia l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho 3 n\u00ean c\u00f3 21 b\u1ea1n tham gia. B\u00e0i 120 : Ch\u1ee9ng minh r\u1eb1ng kh\u00f4ng th\u1ec3 thay c\u00e1c ch\u1eef b\u1eb1ng c\u00e1c ch\u1eef s\u1ed1 \u0111\u1ec3 c\u00f3 ph\u00e9p t\u00ednh \u0111\u00fang : - = 2004 B\u00e0i gi\u1ea3i : C\u00e1ch 1 : \u0110\u1eb7t t\u00ednh : X\u00e9t ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb : C\u00f3 2 tr\u01b0\u1eddng h\u1ee3p x\u1ea3y ra : Tr\u01b0\u1eddng h\u1ee3p 1 : I > C. Khi \u0111\u00f3 ph\u00e9p tr\u1eeb \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb kh\u00f4ng c\u00f3 nh\u1edb sang h\u00e0ng ch\u1ee5c. \u1ede ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c : U - O = 0 hay U = O. \u1ede ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m : V - H = 0 hay V = H. Do \u0111\u00f3 (v\u00ec \u1edf ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn C < I). Tr\u01b0\u1eddng h\u1ee3p 2 : I < C. Khi \u0111\u00f3 ph\u00e9p tr\u1eeb \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb c\u00f3 nh\u1edb 1 sang h\u00e0ng ch\u1ee5c. Do \u0111\u00f3 \u1edf h\u00e0ng ch\u1ee5c : U - O - 1 = 0 hay U - O = 1 n\u00ean O < U. Ph\u00e9p tr\u1eeb kh\u00f4ng c\u00f3 nh\u1edb sang h\u00e0ng tr\u0103m. \u1edf h\u00e0ng tr\u0103m : V - H = 0 hay V = H. V\u00ec th\u1ebf (v\u00ec \u1edf ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c ngh\u00ecn O < U). V\u1eady ta kh\u00f4ng th\u1ec3 thay th\u1ebf c\u00e1c ch\u1eef b\u1eb1ng c\u00e1c ch\u1eef s\u1ed1 \u0111\u1ec3 c\u00f3 ph\u00e9p t\u00ednh nh\u01b0 \u0111\u00e3 cho. C\u00e1ch 2 : D\u00f9ng t\u00ednh ch\u1ea5t chia h\u1ebft c\u1ee7a m\u1ed9t hi\u1ec7u : Ta th\u1ea5y 2 s\u1ed1 v\u00e0 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 b\u1eb1ng nhau n\u00ean c\u1ea3 2 s\u1ed1 s\u1ebd c\u00f3 c\u00f9ng s\u1ed1 d\u01b0 khi chia cho 9, do \u0111\u00f3 hi\u1ec7u c\u1ee7a hai s\u1ed1 ch\u1eafc ch\u1eafn s\u1ebd chia h\u1ebft cho 9. M\u00e0 2004 kh\u00f4ng chia h\u1ebft cho 9, do \u0111\u00f3 hi\u1ec7u c\u1ee7a hai s\u1ed1 kh\u00f4ng th\u1ec3 b\u1eb1ng 2004. 68","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i N\u00f3i c\u00e1ch kh\u00e1c ta kh\u00f4ng th\u1ec3 thay c\u00e1c ch\u1eef b\u1eb1ng c\u00e1c ch\u1eef s\u1ed1 \u0111\u1ec3 c\u00f3 ph\u00e9p t\u00ednh \u0111\u00fang. B\u00e0i 121 : S\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u1ee7a m\u1ed9t quy\u1ec3n s\u00e1ch l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3. Bi\u1ebft r\u1eb1ng cu\u1ed1n s\u00e1ch \u0111\u00f3 tr\u00ean 100 trang v\u00e0 \u00edt h\u01a1n 500 trang. H\u1ecfi cu\u1ed1n s\u00e1ch \u0111\u00f3 c\u00f3 bao nhi\u00eau trang ? B\u00e0i gi\u1ea3i : V\u00ec cu\u1ed1n s\u00e1ch \u0111\u00f3 tr\u00ean 100 trang v\u00e0 \u00edt h\u01a1n 500 trang n\u00ean s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3 l\u00e0 m\u1ed9t s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1. G\u1ecdi s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3 l\u00e0 v\u1edbi a, b, c l\u00e0 c\u00e1c ch\u1eef s\u1ed1 v\u00e0 a kh\u00e1c 0. C\u00e1c s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch l\u00e0 c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn . C\u00f3 9 trang c\u00f3 1 ch\u1eef s\u1ed1 n\u00ean c\u1ea7n 9 ch\u1eef s\u1ed1 \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang cho c\u00e1c trang n\u00e0y. C\u00f3 90 trang c\u00f3 2 ch\u1eef s\u1ed1 n\u00ean c\u1ea7n 2 x 90 = 180 (ch\u1eef s\u1ed1) \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang cho c\u00e1c trang n\u00e0y. S\u1ed1 trang c\u00f3 3 ch\u1eef s\u1ed1 l\u00e0 - 99 trang. S\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u00f3 3 ch\u1eef s\u1ed1 l\u00e0 : 3 x ( - 99) S\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3 l\u00e0 : 9 + 180 + 3 x ( - 99) = 189 + 3 x - 297 = 3 x - 180. V\u00ec s\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch l\u00e0 s\u1ed1 chia h\u1ebft cho s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3 n\u00ean chia h\u1ebft cho hay 108 chia h\u1ebft cho. Suy ra ch\u00ednh b\u1eb1ng 108. V\u1eady cu\u1ed1n s\u00e1ch \u0111\u00f3 c\u00f3 108 trang. B\u00e0i 122 : Cha hi\u1ec7n nay 43 tu\u1ed5i. N\u1ebfu t\u00ednh sang n\u0103m th\u00ec tu\u1ed5i cha v\u1eeba g\u1ea5p 4 tu\u1ed5i con hi\u1ec7n nay. H\u1ecfi l\u00fac con m\u1ea5y tu\u1ed5i th\u00ec tu\u1ed5i cha g\u1ea5p 5 l\u1ea7n tu\u1ed5i con ? C\u00f3 bao gi\u1edd tu\u1ed5i cha g\u1ea5p 4 l\u1ea7n tu\u1ed5i con kh\u00f4ng ? V\u00ec sao ? B\u00e0i gi\u1ea3i : Tu\u1ed5i c\u1ee7a cha sang n\u0103m l\u00e0 : 43 + 1 = 44 (tu\u1ed5i) Tu\u1ed5i c\u1ee7a con hi\u1ec7n nay l\u00e0 : 44 : 4 = 11 (tu\u1ed5i) Tu\u1ed5i cha h\u01a1n tu\u1ed5i con l\u00e0 : 43 - 11 = 32 (tu\u1ed5i) Khi tu\u1ed5i cha g\u1ea5p 5 l\u1ea7n tu\u1ed5i con th\u00ec cha v\u1eabn h\u01a1n con 32 tu\u1ed5i. Ta c\u00f3 s\u01a1 \u0111\u1ed3 khi tu\u1ed5i cha g\u1ea5p 5 l\u1ea7n tu\u1ed5i con nh\u01b0 sau : 69","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Nh\u00ecn v\u00e0o s\u01a1 \u0111\u1ed3 ta th\u1ea5y : Tu\u1ed5i con khi \u0111\u00f3 l\u00e0 : 32 : (5 - 1) = 8 (tu\u1ed5i) N\u1ebfu tu\u1ed5i cha g\u1ea5p 4 l\u1ea7n tu\u1ed5i con, khi \u0111\u00f3 tu\u1ed5i con l\u00e0 1 ph\u1ea7n th\u00ec tu\u1ed5i cha l\u00e0 4 ph\u1ea7n nh\u01b0 th\u1ebf. Tu\u1ed5i cha h\u01a1n tu\u1ed5i con s\u1ed1 ph\u1ea7n l\u00e0 : 4 - 1 = 3 (ph\u1ea7n), khi \u0111\u00f3 cha c\u0169ng v\u1eabn h\u01a1n con 32 tu\u1ed5i ; 32 kh\u00f4ng chia h\u1ebft cho 3 n\u00ean kh\u00f4ng bao gi\u1edd tu\u1ed5i cha g\u1ea5p 4 l\u1ea7n tu\u1ed5i con (v\u00ec ta coi tu\u1ed5i con h\u00e0ng n\u0103m l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean). B\u00e0i 123 : C\u00f3 4 b\u00ecnh (\u0111\u00e1nh s\u1ed1 l\u00e0 1, 2, 3, 4) \u0111\u1ef1ng s\u1ed1 l\u01b0\u1ee3ng c\u00e1c h\u00f2n bi b\u1eb1ng nhau. L\u1ea5y ra t\u1eeb b\u00ecnh th\u1ee9 nh\u1ea5t m\u1ed9t s\u1ed1 vi\u00ean bi, l\u1ea5y g\u1ea5p \u0111\u00f4i s\u1ed1 \u0111\u00f3 t\u1eeb b\u00ecnh th\u1ee9 hai, l\u1ea5y g\u1ea5p ba s\u1ed1 \u0111\u00f3 t\u1eeb b\u00ecnh th\u1ee9 ba v\u00e0 cu\u1ed1i c\u00f9ng l\u1ea5y g\u1ea5p b\u1ed1n s\u1ed1 \u0111\u00f3 t\u1eeb b\u00ecnh th\u1ee9 t\u01b0. Khi \u0111\u00f3 t\u1ed5ng s\u1ed1 bi c\u00f2n l\u1ea1i trong c\u1ea3 b\u1ed1n b\u00ecnh l\u00e0 40 vi\u00ean v\u00e0 b\u00ecnh th\u1ee9 t\u01b0 c\u00f2n l\u1ea1i \u0111\u00fang 1 vi\u00ean bi. H\u1ecfi ban \u0111\u1ea7u s\u1ed1 l\u01b0\u1ee3ng bi trong b\u1ed1n b\u00ecnh l\u00e0 bao nhi\u00eau ? B\u00e0i gi\u1ea3i : S\u1ed1 bi l\u1ea5y ra t\u1eeb b\u00ecnh 1 l\u00e0 : (40 - 1 x 4) : (3 + 2 + 1) = 6 (vi\u00ean). L\u00fac \u0111\u1ea7u s\u1ed1 l\u01b0\u1ee3ng bi trong b\u1ed1n b\u00ecnh l\u00e0 : (6 x 4 + 1) x 4 = 100 (vi\u00ean). B\u00e0i 124 : T\u1eeb m\u1ed9t t\u1edd gi\u1ea5y k\u1ebb \u00f4 vu\u00f4ng, b\u1ea1n Khang c\u1eaft ra m\u1ed9t h\u00ecnh sao b\u1ed1n c\u00e1nh nh\u01b0 h\u00ecnh b\u00ean. H\u00ecnh sao n\u00e0y c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng m\u1ea5y \u00f4 vu\u00f4ng ? B\u00e0i gi\u1ea3i : C\u00e1ch 1 : Di\u1ec7n t\u00edch h\u00ecnh sao \u0111\u00fang b\u1eb1ng di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng g\u1ed3m 16 \u00f4 vu\u00f4ng tr\u1eeb \u0111i di\u1ec7n t\u00edch b\u1ed1n h\u00ecnh tam gi\u00e1c b\u1eb1ng nhau. M\u1ed7i tam gi\u00e1c n\u00e0y c\u00f3 di\u1ec7n t\u00edch l\u00e0 2 \u00f4 vu\u00f4ng. Do \u0111\u00f3 di\u1ec7n t\u00edch h\u00ecnh sao l\u00e0 : 16 - 2 x 4 = 8 (\u00f4 vu\u00f4ng). 70","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i C\u00e1ch 2 : C\u1eaft gh\u00e9p \u0111\u1ec3 t\u1eeb h\u00ecnh sao ta c\u00f3 h\u00ecnh m\u1edbi m\u00e0 h\u00ecnh n\u00e0y di\u1ec7n t\u00edch \u0111\u00fang b\u1eb1ng 8 \u00f4 vu\u00f4ng. B\u00e0i 125 : M\u1ed9t \u0111o\u00e0n t\u00e0u h\u1ecfa d\u00e0i 200 m l\u01b0\u1edbt qua m\u1ed9t ng\u01b0\u1eddi \u0111i xe \u0111\u1ea1p ng\u01b0\u1ee3c chi\u1ec1u v\u1edbi t\u00e0u h\u1ebft 12 gi\u00e2y. T\u00ednh v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u, bi\u1ebft v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi \u0111i xe \u0111\u1ea1p l\u00e0 18 km\/gi\u1edd. B\u00e0i gi\u1ea3i : \u0110o\u00e0n t\u00e0u h\u1ecfa d\u00e0i 200 m l\u01b0\u1edbt qua ng\u01b0\u1eddi \u0111i xe \u0111\u1ea1p h\u1ebft 12 gi\u00e2y, c\u00f3 ngh\u0129a l\u00e0 sau 12 gi\u00e2y t\u1ed5ng qu\u00e3ng \u0111\u01b0\u1eddng t\u00e0u h\u1ecfa v\u00e0 xe \u0111\u1ea1p \u0111i l\u00e0 200 m. Nh\u01b0 v\u1eady t\u1ed5ng v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u h\u1ecfa v\u00e0 xe \u0111\u1ea1p l\u00e0 : 200 : 12 = 50\/3(m\/gi\u00e2y), 50\/3 m\/gi\u00e2y = 60 km\/gi\u1edd. V\u1eadn t\u1ed1c c\u1ee7a xe \u0111\u1ea1p l\u00e0 18 km\/gi\u1edd, th\u00ec v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u h\u1ecfa l\u00e0 : 60 - 18 = 42 (km\/gi\u1edd). B\u00e0i 126 : Cho s\u1ed1 g\u1ed3m b\u1ed1n ch\u1eef s\u1ed1 c\u00f3 ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m l\u00e0 9 v\u00e0 ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c l\u00e0 7. T\u00ecm s\u1ed1 \u0111\u00e3 cho bi\u1ebft s\u1ed1 \u0111\u00f3 chia h\u1ebft cho 5 v\u00e0 27. B\u00e0i gi\u1ea3i : G\u1ecdi s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 (a kh\u00e1c 0 ; a ; b <10) V\u00ec chia h\u1ebft cho 5 n\u00ean b = 0 ho\u1eb7c b = 5. V\u00ec chia h\u1ebft cho 27 n\u00ean chia h\u1ebft cho 9. Thay b = 0 ta c\u00f3 chia h\u1ebft cho 9 n\u00ean a = 2. Th\u1eed 2970 : 27 = 110 (\u0111\u00fang). Thay b = 5 ta c\u00f3 chia h\u1ebft cho 9 n\u00ean a = 6. Th\u1eed 6975 : 27 = 258 (d\u01b0 9) tr\u00e1i v\u1edbi \u0111i\u1ec1u ki\u1ec7n b\u00e0i to\u00e1n. V\u1eady s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 2970. B\u00e0i 127 : Ba l\u1edbp 5A, 5B v\u00e0 5C tr\u1ed3ng c\u00e2y nh\u00e2n d\u1ecbp \u0111\u1ea7u xu\u00e2n. Trong \u0111\u00f3 s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A v\u00e0 l\u1edbp 5B tr\u1ed3ng \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n s\u1ed1 c\u00e2y c\u1ee7a 5B v\u00e0 5C l\u00e0 3 c\u00e2y. S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B v\u00e0 5C tr\u1ed3ng \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n s\u1ed1 c\u00e2y c\u1ee7a 5A v\u00e0 5C l\u00e0 1 c\u00e2y. T\u00ednh s\u1ed1 c\u00e2y tr\u1ed3ng \u0111\u01b0\u1ee3c c\u1ee7a m\u1ed7i l\u1edbp. Bi\u1ebft r\u1eb1ng t\u1ed5ng s\u1ed1 c\u00e2y tr\u1ed3ng \u0111\u01b0\u1ee3c c\u1ee7a ba l\u1edbp l\u00e0 43 c\u00e2y. 71","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i gi\u1ea3i : C\u00e1ch 1 : V\u00ec s\u1ed1 c\u00e2y l\u1edbp 5A v\u00e0 l\u1edbp 5B tr\u1ed3ng \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B v\u00e0 5C l\u00e0 3 c\u00e2y n\u00ean s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A h\u01a1n s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5C l\u00e0 3 c\u00e2y. S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B v\u00e0 5C tr\u1ed3ng \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A v\u00e0 5C l\u00e0 1 c\u00e2y n\u00ean s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B tr\u1ed3ng \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A l\u00e0 1 c\u00e2y. Ta c\u00f3 s\u01a1 \u0111\u1ed3 : Ba l\u1ea7n s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5C l\u00e0 : 43 - (3 + 3 + 1) = 36 (c\u00e2y) S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5C l\u00e0 : 36 : 3 = 12 (c\u00e2y). S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A l\u00e0 : 12 + 3 = 15 (c\u00e2y). S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B l\u00e0 : 15 + 1 = 16 (c\u00e2y). C\u00e1ch 2 : Hai l\u1ea7n t\u1ed5ng s\u1ed1 c\u00e2y c\u1ee7a 3 l\u1edbp l\u00e0 : 43 x 2 = 86 (c\u00e2y). Ta c\u00f3 s\u01a1 \u0111\u1ed3 : S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A v\u00e0 5C tr\u1ed3ng \u0111\u01b0\u1ee3c l\u00e0 : (86 - 3 - 1 - 1) : 3 = 27 (c\u00e2y). S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B l\u00e0 : 43 - 27 = 16 (c\u00e2y). S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5B v\u00e0 5C l\u00e0 : 27 + 1 = 28 (c\u00e2y). S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5C l\u00e0 : 28 - 16 = 12 (c\u00e2y). S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 5A l\u00e0 : 43 - 28 = 15 (c\u00e2y). B\u00e0i 128 : M\u1ed9t d\u00e3y c\u00f3 7 \u00f4 vu\u00f4ng g\u1ed3m 3 \u00f4 \u0111en v\u00e0 4 \u00f4 tr\u1eafng \u0111\u01b0\u1ee3c s\u1eafp x\u1ebfp nh\u01b0 h\u00ecnh v\u1ebd. Cho ph\u00e9p m\u1ed7i l\u1ea7n ch\u1ecdn hai \u00f4 t\u00f9y \u00fd v\u00e0 \u0111\u1ed5i m\u00e0u ch\u00fang (t\u1eeb \u0111en sang tr\u1eafng v\u00e0 t\u1eeb tr\u1eafng sang \u0111en). H\u1ecfi r\u1eb1ng n\u1ebfu l\u00e0m nh\u01b0 tr\u00ean nhi\u1ec1u l\u1ea7n th\u00ec c\u00f3 th\u1ec3 nh\u1eadn \u0111\u01b0\u1ee3c d\u00e3y \u00f4 vu\u00f4ng c\u00f3 m\u00e0u xen k\u1ebd nhau nh\u01b0 sau hay kh\u00f4ng ? 72","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i gi\u1ea3i : Nh\u00ecn v\u00e0o h\u00ecnh v\u1ebd ta th\u1ea5y \u1edf h\u00ecnh ban \u0111\u1ea7u c\u00f3 3 \u00f4 \u0111en v\u00e0 4 \u00f4 tr\u1eafng, c\u00f2n h\u00ecnh l\u00fac sau c\u00f3 4 \u00f4 \u0111en v\u00e0 3 \u00f4 tr\u1eafng. Khi ch\u1ecdn hai \u00f4 t\u00f9y \u00fd \u0111\u1ec3 \u0111\u1ed5i m\u00e0u c\u1ee7a ch\u00fang (t\u1eeb \u0111en sang tr\u1eafng v\u00e0 t\u1eeb tr\u1eafng sang \u0111en) th\u00ec c\u00f3 ba kh\u1ea3 n\u0103ng x\u1ea3y ra : - Ch\u1ecdn hai \u00f4 tr\u1eafng : Khi \u0111\u00f3 hai \u00f4 tr\u1eafng \u0111\u01b0\u1ee3c ch\u1ecdn s\u1ebd \u0111\u1ed5i th\u00e0nh hai \u00f4 \u0111en, do \u0111\u00f3 s\u1ed1 \u00f4 \u0111en t\u0103ng l\u00ean 2 \u00f4. - Ch\u1ecdn hai \u00f4 \u0111en : Khi \u0111\u00f3 hai \u00f4 \u0111en \u0111\u01b0\u1ee3c ch\u1ecdn s\u1ebd \u0111\u1ed5i th\u00e0nh hai \u00f4 tr\u1eafng, do \u0111\u00f3 s\u1ed1 \u00f4 \u0111en gi\u1ea3m \u0111i 2 \u00f4. - Ch\u1ecdn m\u1ed9t \u00f4 \u0111en v\u00e0 m\u1ed9t \u00f4 tr\u1eafng : Khi \u0111\u00f3 \u00f4 tr\u1eafng \u0111\u1ed5i th\u00e0nh \u00f4 \u0111en v\u00e0 \u00f4 \u0111en \u0111\u1ed5i th\u00e0nh \u00f4 tr\u1eafng, do \u0111\u00f3 s\u1ed1 \u00f4 \u0111en gi\u1eef nguy\u00ean. Do v\u1eady khi th\u1ef1c hi\u1ec7n vi\u1ec7c ch\u1ecdn hai \u00f4 \u0111\u1ec3 \u0111\u1ed5i m\u00e0u c\u1ee7a ch\u00fang th\u00ec s\u1ed1 l\u01b0\u1ee3ng \u00f4 \u0111en ho\u1eb7c t\u0103ng l\u00ean 2 \u00f4, ho\u1eb7c gi\u1ea3m \u0111i 2 \u00f4, ho\u1eb7c gi\u1eef nguy\u00ean. \u0110i\u1ec1u \u0111\u00f3 c\u00f3 ngh\u0129a l\u00e0 n\u1ebfu ch\u1ecdn hai \u00f4 t\u00f9y \u00fd v\u00e0 \u0111\u1ed5i m\u00e0u ch\u00fang nhi\u1ec1u l\u1ea7n th\u00ec s\u1ed1 \u00f4 \u0111en v\u1eabn lu\u00f4n lu\u00f4n l\u00e0 m\u1ed9t s\u1ed1 l\u1ebb. V\u00ec h\u00ecnh sau c\u00f3 4 \u00f4 \u0111en n\u00ean kh\u00f4ng th\u1ec3 th\u1ef1c hi\u1ec7n \u0111\u01b0\u1ee3c. B\u00e0i 129 : M\u1ed9t t\u1edd gi\u1ea5y h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c g\u1ea5p theo \u0111\u01b0\u1eddng ch\u00e9o nh\u01b0 h\u00ecnh v\u1ebd. Di\u1ec7n t\u00edch h\u00ecnh nh\u1eadn \u0111\u01b0\u1ee3c b\u1eb1ng 5\/8 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u. Bi\u1ebft di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 m\u00e0u l\u00e0 18 cm2. T\u00ednh di\u1ec7n t\u00edch t\u1edd gi\u1ea5y ban \u0111\u1ea7u. B\u00e0i gi\u1ea3i : 73","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Khi g\u1ea5p t\u1edd gi\u1ea5y h\u00ecnh ch\u1eef nh\u1eadt theo \u0111\u01b0\u1eddng ch\u00e9o (\u0111\u01b0\u1eddng n\u00e9t \u0111\u1ee9t) th\u00ec ph\u1ea7n h\u00ecnh tam gi\u00e1c \u0111\u01b0\u1ee3c t\u00f4 m\u00e0u b\u1ecb x\u1ebfp ch\u1ed3ng l\u00ean nhau. Do \u0111\u00f3 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u1edbn h\u01a1n di\u1ec7n t\u00edch h\u00ecnh nh\u1eadn \u0111\u01b0\u1ee3c ch\u00ednh l\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u01b0\u1ee3c t\u00f4 m\u00e0u. Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u gi\u1ea3m \u0111i b\u1eb1ng 1 - 5\/8 = 3\/8 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u. Do v\u1eady di\u1ec7n t\u00edch tam gi\u00e1c t\u00f4 m\u00e0u b\u1eb1ng 3\/8 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u, hay 3\/8 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u b\u1eb1ng 18 cm2. V\u1eady di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u00e0 : 18 : 3\/8 = 48 (cm2) B\u00e0i 130. Ch\u1ee9ng t\u1ecf r\u1eb1ng k\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p nh\u00e2n sau : 3 x 3 x 3 x ... x 3 (2000 th\u1eeba s\u1ed1 3) l\u00e0 s\u1ed1 c\u00f3 \u00edt h\u01a1n 1001 ch\u1eef s\u1ed1. L\u1eddi gi\u1ea3i. Trong t\u00edch s\u1ed1 A = 3 x 3 x 3 x ... x 3 g\u1ed3m 2000 th\u1eeba s\u1ed1 3, k\u1ebft h\u1ee3p t\u1eebng c\u1eb7p s\u1ed1 3 \u0111\u01b0\u1ee3c A = (3 x 3) (3 x 3) ... (3 x 3) = 9 x 9 x ... x 9 g\u1ed3m 1000 th\u1eeba s\u1ed1 9. X\u00e9t s\u1ed1 B = 9 x 10 x ...x 10 th\u1eeba s\u1ed1 10 n\u00ean s\u1ed1 B = 90...0 c\u00f3 999 ch\u1eef s\u1ed1 0 v\u00e0 1 ch\u1eef s\u1ed1 9, ngh\u0129a l\u00e0 c\u00f3 1000 ch\u1eef s\u1ed1. V\u00ec 9 < 10 n\u00ean A = 9 x 9 x ... x 9 < B = 9 x10 x ... x 10 V\u1eady s\u1ed1 A c\u00f3 \u00edt h\u01a1n 1001 ch\u1eef s\u1ed1. B\u00e0i 131. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD. Bi\u1ebft r\u1eb1ng di\u1ec7n t\u00edch ph\u1ea7n m\u00e0u v\u00e0ng l\u00e0 20cm2 v\u00e0 I l\u00e0 \u0111i\u1ec3m chia AB th\u00e0nh 2 ph\u1ea7n b\u1eb1ng nhau. L\u1eddi gi\u1ea3i. K\u00ed hi\u1ec7u S l\u00e0 di\u1ec7n t\u00edch c\u1ee7a m\u1ed9t h\u00ecnh. N\u1ed1i D v\u1edbi I. Qua I v\u00e0 C v\u1ebd c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng IP v\u00e0 CQ vu\u00f4ng g\u00f3c v\u1edbi BD, IH vu\u00f4ng g\u00f3c v\u1edbi DC. Ta c\u00f3 SADB = SCDB = 1\/2 SABCD SDIB = 1\/2 SADB (v\u00ec c\u00f3 chung \u0111\u01b0\u1eddng cao DA, IB = 1\/2 AB), SDIB = 1\/2 SDBC. M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y DB 74","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i N\u00ean IP = 1\/2 CQ. SIDK = 1\/2 SCDK (v\u00ec c\u00f3 chung \u0111\u00e1y DK v\u00e0 IP = 1\/2 CQ) SCDI = SIDK + SDKC = 3SDIK. Ta c\u00f3 : SADI = 1\/2 AD x AI, SDIC = 1\/2 IH x DC M\u00e0 IH = AD, AI = 1\/2 DC, SDIC = 2SADI n\u00ean SADI = 3\/2 SDIK V\u00ec AIKD l\u00e0 ph\u1ea7n \u0111\u01b0\u1ee3c t\u00f4 m\u00e0u v\u00e0ng n\u00ean SAIKD = 20(cm2) SDAI + SIDK = 20(cm2) SDAI + 2\/3 SADI = 20(cm2) SDAI = (3 x 20)\/5 = 12 (cm2) M\u1eb7t kh\u00e1c SDAI = 1\/2 SDAB (c\u00f9ng chung chi\u1ec1u cao DA, AI = 1\/2 AB) = 1\/4 SABCD suy ra SABCD = 4 x SDAI = 4 x 12 = 48 (cm2). B\u00e0i 132. N\u1ebfu trong m\u1ed9t th\u00e1ng n\u00e0o \u0111\u00f3 m\u00e0 c\u00f3 3 ng\u00e0y th\u1ee9 b\u1ea3y \u0111\u1ec1u l\u00e0 c\u00e1c ng\u00e0y ch\u1eb5n th\u00ec ng\u00e0y 25 c\u1ee7a th\u00e1ng \u0111\u00f3 s\u1ebd l\u00e0 ng\u00e0y th\u1ee9 m\u1ea5y ? L\u1eddi gi\u1ea3i. C\u00e1ch 1. Trong m\u1ed9t th\u00e1ng n\u00e0o \u0111\u00f3 c\u00f3 ba ng\u00e0y th\u1ee9 b\u1ea3y l\u00e0 ng\u00e0y ch\u1eb5n th\u00ec ch\u1eafc ch\u1eafn c\u00f2n c\u00f3 hai ng\u00e0y th\u1ee9 B\u1ea3y l\u00e0 ng\u00e0y l\u1ebb. N\u0103m ng\u00e0y th\u1ee9 B\u1ea3y \u0111\u00f3 s\u1eafp x\u1ebfp nh\u01b0 sau : Th\u1ee9 B\u1ea3y (1) Th\u1ee9 B\u1ea3y (2) ch\u1eb5n l\u1ebb Th\u1ee9 B\u1ea3y (3) Th\u1ee9 B\u1ea3y (4) Th\u1ee9 B\u1ea3y (5) ch\u1eafn l\u1ebb ch\u1eb5n S\u1ed1 ng\u00e0y nhi\u1ec1u nh\u1ea5t trong m\u1ed9t th\u00e1ng l\u00e0 31 ng\u00e0y. Th\u00e1ng n\u00e0y c\u00f3 4 tu\u1ea7n v\u00e0 3 ng\u00e0y. N\u1ebfu th\u1ee9 b\u1ea3y \u0111\u1ea7u ti\u00ean l\u00e0 ng\u00e0y m\u00f9ng 4 th\u00ec th\u00e1ng \u0111\u00f3 s\u1ebd c\u00f3 s\u1ed1 ng\u00e0y l\u00e0: 4 + 7 x 4 = 32 (ng\u00e0y) ; tr\u00e1i v\u1edbi l\u1ecbch th\u00f4ng th\u01b0\u1eddng. V\u00ec th\u1ebf th\u1ee9 b\u1ea3y \u0111\u1ea7u ti\u00ean (1) ph\u1ea3i l\u00e0 ng\u00e0y m\u00f9ng 2; th\u1ee9 7 th\u1ee9 t\u01b0 s\u1ebd l\u00e0 ng\u00e0y: 2 + 7 x 3 = 23 V\u1eady ng\u00e0y 25 c\u1ee7a th\u00e1ng \u0111\u00f3 l\u00e0 ng\u00e0y th\u1ee9 hai. C\u00e1ch 2. L\u1eadp b\u1ea3ng theo tu\u1ea7n l\u1ec5 : 123456 7 75","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Trong 3 c\u1ed9t \u0111\u1ea7u ti\u00ean ch\u1ec9 c\u00f3 c\u1ed9t 2 th\u00edch h\u1ee3p v\u1edbi \u0111\u1ea7u b\u00e0i to\u00e1n. C\u1ed9t n\u00e0y c\u00f3 5 ng\u00e0y th\u1ee9 b\u1ea3y. V\u00ec ng\u00e0y 23 l\u00e0 th\u1ee9 b\u1ea3y, n\u00ean ng\u00e0y 25 l\u00e0 th\u1ee9 hai. B\u00e0i 133. B\u1ed1n b\u1ea1n Xu\u00e2n, H\u1ea1, Thu, \u0110\u00f4ng c\u00f3 t\u1ea5t c\u1ea3 61 vi\u00ean bi. Xu\u00e2n c\u00f3 s\u1ed1 bi \u00edt nh\u1ea5t, \u0110\u00f4ng c\u00f3 s\u1ed1 bi nhi\u1ec1u nh\u1ea5t v\u00e0 l\u00e0 s\u1ed1 l\u1ebb, Thu c\u00f3 s\u1ed1 bi g\u1ea5p 9 l\u1ea7n s\u1ed1 bi c\u1ee7a H\u1ea1. H\u00e3y cho bi\u1ebft m\u1ed7i b\u1ea1n c\u00f3 bao nhi\u00eau vi\u00ean bi ? L\u1eddi gi\u1ea3i. + S\u1ed1 bi c\u1ee7a Thu g\u1ea5p 9 l\u1ea7n s\u1ed1 bi c\u1ee7a H\u1ea1 n\u00ean t\u1ed5ng s\u1ed1 bi c\u1ee7a Thu v\u00e0 H\u1ea1 l\u00e0 m\u1ed9t s\u1ed1 ch\u1eb5n. T\u1ed1ng s\u1ed1 bi c\u1ee7a b\u1ed1n b\u1ea1n l\u00e0 s\u1ed1 l\u1ebb, s\u1ed1 bi c\u1ee7a \u0110\u00f4ng l\u00e0 s\u1ed1 l\u1ebb, t\u1ed5ng s\u1ed1 bi c\u1ee7a H\u1ea1 v\u00e0 Thu l\u00e0 s\u1ed1 l\u1ebb ; do \u0111\u00f3 s\u1ed1 bi c\u1ee7a Xu\u00e2n ph\u1ea3i l\u00e0 s\u1ed1 ch\u1eb5n. + S\u1ed1 bi c\u1ee7a H\u1ea1 ph\u1ea3i l\u00e0 s\u1ed1 b\u00e9 h\u01a1n 4 v\u00ec n\u1ebfu s\u1ed1 \u0111\u00f3 l\u00e0 4 th\u00ec s\u1ed1 bi c\u1ee7a Thu l\u00e0 4 x 9 = 36. Khi \u0111\u00f3 \u00edt nh\u1ea5t \u0110\u00f4ng c\u00f3 s\u1ed1 bi l\u00e0 37 th\u00ec ch\u1ec9 ri\u00eang t\u1ed5ng s\u1ed1 bi c\u1ee7a Thu v\u00e0 \u0110\u00f4ng \u0111\u00e3 v\u01b0\u1ee3t qu\u00e1 t\u1ed5ng s\u1ed1 bi c\u1ee7a b\u1ed1n b\u1ea1n (36 + 37 = 73 > 61). + N\u1ebfu s\u1ed1 bi c\u1ee7a Xu\u00e2n l\u00e0 2 th\u00ec s\u1ed1 bi c\u1ee7a H\u1ea1 l\u00e0 3, s\u1ed1 bi c\u1ee7a Thu l\u00e0 27 (3 x 9 = 27) S\u1ed1 bi c\u1ee7a \u0110\u00f4ng l\u00e0 : 61 - (2 + 3 + 27) = 29 (vi\u00ean). B\u00e0i 134. Thay c\u00e1c ch\u1eef c\u00e1i d\u01b0\u1edbi \u0111\u00e2y b\u1edfi c\u00e1c ch\u1eef s\u1ed1 (ch\u1eef c\u00e1i kh\u00e1c nhau th\u00ec thay b\u1edfi c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau) sao cho k\u1ebft qu\u1ea3 c\u00e1c ph\u00e9p t\u00ednh d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t. CHUC + MUNG + THAY + CO + NHAN + NGAY - 20 - 11 L\u1eddi gi\u1ea3i. V\u00ec N xu\u1ea5t hi\u1ec7n \u1edf nh\u1eefng h\u00e0ng cao nh\u1ea5t v\u00e0 nhi\u1ec1u l\u1ea7n nh\u1ea5t n\u00ean N ph\u1ea3i b\u1eb1ng 9 \u0111\u1ec3 k\u1ebft qu\u1ea3 l\u1edbn nh\u1ea5t. Ti\u1ebfp \u0111\u00f3 C xu\u1ea5t hi\u1ec7n \u1edf h\u00e0ng cao nh\u1ea5t c\u00f2n l\u1ea1i gi\u1ed1ng M v\u00e0 T nh\u01b0ng C c\u00f2n \u1edf hai h\u00e0ng kh\u00e1c n\u1eefa n\u00ean C b\u1eb1ng 8. N\u1ebfu M l\u00e0 7 th\u00ec T l\u00e0 6 v\u00e0 ng\u01b0\u1ee3c l\u1ea1i, k\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p to\u00e1n kh\u00f4ng thay \u0111\u1ed5i. V\u1edbi l\u1eadp lu\u1eadn nh\u01b0 tr\u00ean th\u00ec H b\u1eb1ng 5, U b\u1eb1ng 4 v\u00e0 G l\u00e0 3. T\u1eeb \u0111\u00f3 A b\u1eb1ng 2, Y b\u1eb1ng 1 v\u00e0 O l\u00e0 0. V\u1eady ta c\u00f3 2 \u0111\u00e1p s\u1ed1 : 8548 + 6493 + 7521 + 80 + 9529 + 9321 - 20 - 11 = 41461 76","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i v\u00e0 8548 + 7493 + 6521 + 80 + 9529 + 9321 - 20 - 11 = 41461 B\u00e0i 135 : Th\u0103ng \u0111\u1ed1 Long bi\u1ebft \u0111\u01b0\u1ee3c s\u1ed1 h\u1ecdc sinh c\u1ee7a tr\u01b0\u1eddng Th\u0103ng cu\u1ed1i n\u0103m h\u1ecdc v\u1eeba r\u1ed3i c\u00f3 bao nhi\u00eau h\u1ecdc sinh \u0111\u01b0\u1ee3c nh\u1eadn th\u01b0\u1edfng ? Bi\u1ebft r\u1eb1ng s\u1ed1 h\u1ecdc sinh \u0111\u01b0\u1ee3c nh\u1eadn th\u01b0\u1edfng l\u00e0 s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 v\u00e0 r\u1ea5t th\u00fa v\u1ecb l\u00e0 ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m, ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb gi\u1ed1ng nhau. N\u1ebfu nh\u00e2n s\u1ed1 n\u00e0y v\u1edbi 6 th\u00ec \u0111\u01b0\u1ee3c t\u00edch l\u00e0 s\u1ed1 c\u0169ng c\u00f3 ba ch\u1eef s\u1ed1 v\u00e0 trong t\u00edch \u0111\u00f3 c\u00f3 m\u1ed9t ch\u1eef s\u1ed1 2. B\u00e0i gi\u1ea3i : G\u1ecdi s\u1ed1 phi t\u00ecm l\u00e0 aba(a kh\u00e1c b;a ; b nh\u1ecf ho\u1eb7c b\u1eb1ng 9). Theo \u0111\u1ea7u b\u00e0i ta c\u00f3: aba x 6 = deg (d kh\u00e1c 0 ; d; e; g nh\u1ecf h\u01a1n ho\u1eb7c b\u1eb1ng 9).N\u1ebfu a l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng 2 th\u00ec t\u00edch nhi\u1ec1u h\u01a1n 3 ch\u1eef s\u1ed1.V\u1eady a = 1. Ta c\u00f3 1b1x 6 = deg ( deg c\u00f3 m\u1ed9t ch\u1eef s\u1ed1 2). Do \u0111\u00f3 : g = 1 x 6 = 6 v\u00e0 d l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng 6. V\u00ec th\u1ebf : e = 2 V\u00ec b x 6 = n\u00ean b = 2 ho\u1eb7c b = 7. N\u1ebfu b = 2 th\u00ec 121 x 6 = 726 (\u0110\u00fang) N\u1ebfu b = 7 th\u00ec 171 x 6 = 1026 (Lo\u1ea1i) V\u1eady s\u1ed1 h\u1ecdc s\u1ecbnh nh\u1eadn th\u01b0\u1edfng l\u00e0 121 b\u1ea1n. B\u00e0i 136 : Em h\u00e3y di chuy\u1ec3n hai que di\u00eam l\u1ea1i \u0111\u00fang v\u1ecb tr\u00ed \u0111\u1ec3 k\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh l\u00e0 \u0111\u00fang : B\u00e0i gi\u1ea3i : C\u00e1ch 1 : Ta chuy\u1ec3n que di\u00eam \u1edf gi\u1eefa ch\u1eef s\u1ed1 8 \u0111\u1ec3 c\u00f3 ch\u1eef s\u1ed1 0. L\u1ea5y que di\u00eam \u0111\u00f3 gh\u00e9p v\u00e0o ch\u1eef s\u1ed1 5 c\u1ee7a s\u1ed1 502 \u0111\u1ec3 \u0111\u01b0\u1ee3c s\u1ed1 602. L\u1ea5y 1 que di\u00eam \u1edf ch\u1eef s\u1ed1 3 c\u1ee7a s\u1ed1 2003 v\u00e0 \u0111\u1eb7t v\u00e0o v\u1ecb tr\u00ed kh\u00e1c c\u1ee7a ch\u1eef s\u1ed1 3 \u0111\u00f3 \u0111\u1ec3 chuy\u1ec3n s\u1ed1 2003 th\u00e0nh s\u1ed1 2002, ta c\u00f3 ph\u00e9p t\u00ednh \u0111\u00fang : C\u00e1ch 2 : Ta chuy\u1ec3n que di\u00eam \u1edf gi\u1eefa s\u1ed1 8 \u0111\u1ec3 c\u00f3 ch\u1eef s\u1ed1 0. l\u1ea5y que di\u00eam \u0111\u00f3 gh\u00e9p v\u00e0o ch\u1eef s\u1ed1 5 c\u1ee7a s\u1ed1 502 \u0111\u1ec3 \u0111\u01b0\u1ee3c s\u1ed1 602. L\u1ea5y 1 que di\u00eam \u1edf ch\u1eef s\u1ed1 2 c\u1ee7a s\u1ed1 602 v\u00e0 \u0111\u1eb7t v\u00e0o v\u1ecb tr\u00ed kh\u00e1c c\u1ee7a ch\u1eef s\u1ed1 2 \u0111\u00f3 \u0111\u1ec3 chuy\u1ec3n s\u1ed1 602 th\u00e0nh s\u1ed1 603, ta c\u00f3 ph\u00e9p t\u00ednh \u0111\u00fang : B\u00e0i 137 : M\u1ed9t b\u1ea1n ch\u1ecdn hai s\u1ed1 t\u1ef1 nhi\u00ean tu\u1ef3 \u00fd, t\u00ednh t\u1ed5ng c\u1ee7a ch\u00fang r\u1ed3i l\u1ea5y t\u1ed5ng \u0111\u00f3 nh\u00e2n v\u1edbi ch\u00ednh n\u00f3. B\u1ea1n \u1ea5y c\u0169ng l\u00e0m t\u01b0ng t\u1ef1 \u0111\u1ed1i v\u1edbi hi\u1ec7u c\u1ee7a hai s\u1ed1 m\u00e0 77","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i m\u00ecnh \u0111\u00e3 ch\u1ecdn \u0111\u00f3. Cu\u1ed1i c\u00f9ng c\u1ed9ng hai t\u00edch t\u00ecm \u0111\u01b0\u1ee3c v\u1edbi nhau. H\u1ecfi r\u1eb1ng t\u1ed5ng c\u1ee7a hai t\u00edch \u0111\u00f3 l\u00e0 s\u1ed1 ch\u1eb5n hay s\u1ed1 l\u1ebb ? V\u00ec sao ? B\u00e0i gi\u1ea3i : S\u1ebd x\u1ea3y ra m\u1ed9t trong hai tr\u01b0\u1eddng h\u1ee3p : C hai s\u1ed1 \u0111\u1ec1u ch\u1eb5n (ho\u1eb7c \u0111\u1ec1u l\u1ebb) ; m\u1ed9t s\u1ed1 ch\u1eb5n v\u00e0 m\u1ed9t s\u1ed1 l\u1ebb. a) Hai s\u1ed1 ch\u1eb5n (ho\u1eb7c hai s\u1ed1 l\u1ebb). T\u1ed5ng, hi\u1ec7u c\u1ee7a hai s\u1ed1 \u0111\u00f3 l\u00e0 s\u1ed1 ch\u1eb5n. S\u1ed1 ch\u1eb5n nh\u00e2n v\u1edbi ch\u00ednh n\u00f3 \u0111\u01b0\u1ee3c s\u1ed1 ch\u1eb5n. Do \u0111\u00f3 c\u1ed9ng hai t\u00edch (l\u00e0 hai s\u1ed1 ch\u1eb5n) ph\u1ea3i \u0111\u01b0\u1ee3c s\u1ed1 ch\u1eb5n. b) M\u1ed9t s\u1ed1 ch\u1eb5n v\u00e0 m\u1ed9t s\u1ed1 l\u1ebb. T\u1ed5ng, hi\u1ec7u c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 s\u1ed1 l\u1ebb. S\u1ed1 l\u1ebb nh\u00e2n v\u1edbi ch\u00ednh n\u00f3 \u0111\u01b0\u1ee3c s\u1ed1 l\u1ebb. Do \u0111\u00f3 c\u1ed9ng hai t\u00edch (l\u00e0 hai s\u1ed1 l\u1ebb) ph\u1ea3i \u0111\u01b0\u1ee3c s\u1ed1 ch\u1eb5n. V\u1eady theo \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n th\u00ec k\u1ebft qu\u1ea3 c\u1ee7a b\u00e0i to\u00e1n ph\u1ea3i l\u00e0 s\u1ed1 ch\u1eb5n. B\u00e0i 138 : a) H\u00e3y ph\u00e2n t\u00edch 20 th\u00e0nh t\u1ed5ng c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean sao cho t\u00edch c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean \u1ea5y c\u0169ng b\u1eb1ng 20. b) B\u1ea1n c\u00f3 th\u1ec3 l\u00e0m nh\u01b0 th\u1ebf v\u1edbi b\u1ea5t k\u00ec s\u1ed1 t\u1ef1 nhi\u00ean n\u00e0o \u0111\u01b0\u1ee3c kh\u00f4ng ? B\u00e0i gi\u1ea3i : Ph\u00e2n t\u00edch 20 th\u00e0nh t\u00edch c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean kh\u00e1c 1. 20 = 2 x 2 x 5 = 4 x 5 = 10 x 2 Tr\u01b0\u1eddng h\u1ee3p : 2 x 2 x 5 = 20 th\u00ec t\u1ed5ng c\u1ee7a ch\u00fang l\u00e0 : 2+ 2 + 5 = 9. V\u1eady \u0111\u1ec3 t\u1ed5ng b\u1eb1ng 20 th\u00ec ph\u1ea3i th\u00eam v\u00e0o : 20 - 9 = 11, ta thay 11 b\u1eb1ng t\u1ed5ng c\u1ee7a 11 s\u1ed1 1 khi \u0111\u00f3 t\u00edch s\u1ebd kh\u00f4ng thay \u0111\u1ed5i. L\u00ed lu\u1eadn t\u01b0\u01a1ng t\u1ef1 v\u1edbi c\u00e1c tr\u01b0\u1eddng h\u1ee3p : 20 = 4 x 5 v\u00e0 20 = 10 x 2. Ta c\u00f3 3 c\u00e1ch ph\u00e2n t\u00edch nh\u01b0 sau : C\u00e1ch 1 : 20 = 2 x 2 x 5 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1. 20 = 2 + 2 + 5 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1. C\u00e1ch 2 : 20 = 4 x 5 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1. 20 = 4 + 5 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1. 78","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i C\u00e1ch 3 : 20 = 10 x 2 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1. 20 = 10 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1. b) M\u1ed9t s\u1ed1 chia h\u1ebft cho 1 v\u00e0 ch\u00ednh n\u00f3 s\u1ebd kh\u00f4ng l\u00e0m \u0111\u01b0\u1ee3c nh\u01b0 tr\u00ean v\u00ec t\u00edch c\u1ee7a 1v\u1edbi ch\u00ednh n\u00f3 lu\u00f4n nh\u1ecf h\u01a1n t\u1ed5ng c\u1ee7a 1 v\u1edbi ch\u00ednh n\u00f3. B\u00e0i 139 : T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean a nh\u1ecf nh\u1ea5t sao cho a chia cho 2 d\u01b0 1, chia cho 5 d\u01b0 1, chia cho 7 d\u01b0 3 v\u00e0 chia h\u1ebft cho 9. B\u00e0i gi\u1ea3i : V\u00ec a chia cho 2 d\u01b0 1 n\u00ean a l\u00e0 s\u1ed1 l\u1ebb. V\u00ec a chia cho 5 d\u01b0 1 n\u00ean a c\u00f3 t\u1eadn c\u00f9ng l\u00e0 1 ho\u1eb7c 6. Do \u0111\u00f3 a ph\u1ea3i c\u00f3 t\u1eadn c\u00f9ng l\u00e0 1. - N\u1ebfu a l\u00e0 s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 th\u00ec do a chia h\u1ebft cho 9 n\u00ean a = 81, lo\u1ea1i v\u00ec 81 : 7 = 11 d\u01b0 4 (tr\u00e1i v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u0111\u1ec1 b\u00e0i). - N\u1ebfu a l\u00e0 s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 th\u00ec \u0111\u1ec3 a nh\u1ecf nh\u1ea5t th\u00ec ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m ph\u1ea3i l\u00e0 1. Khi \u0111\u00f3 \u0111\u1ec3 a chia h\u1ebft cho 9 th\u00ec theo d\u1ea5u hi\u1ec7u chia h\u1ebft cho 9 ta c\u00f3 ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c phi l\u00e0 7 (\u0111\u1ec3 1 + 7 + 1 = 9 9). V\u00ec 171 : 7 = 24 d\u01b0 3 n\u00ean a = 171. V\u1eady s\u1ed1 ph\u1ea3i t\u00ecm nh\u1ecf nh\u1ea5t th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u0111\u1ec1 b\u00e0i l\u00e0 171. B\u00e0i 140 : S\u1ed1 n\u00e0y n\u1eb1m trong ph\u1ea1m vi c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 58. Khi vi\u1ebft \\\"n\u00f3\\\" kh\u00f4ng s\u1eed d\u1ee5ng c\u00e1c ch\u1eef s\u1ed1 1 ; 2 ; 3. Ngo\u00e0i ra \\\"n\u00f3\\\" l\u00e0 s\u1ed1 l\u1ebb v\u00e0 kh\u00f4ng chia h\u1ebft cho c\u00e1c s\u1ed1 3 ; 5 ; 7. V\u1eady \\\"n\u00f3\\\" l\u00e0 s\u1ed1 n\u00e0o ? B\u00e0i gi\u1ea3i : N\u00f3 l\u00e0 s\u1ed1 l\u1ebb n\u1eb1m trong ph\u1ea1m vi c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 58, khi vi\u1ebft n\u00f3 kh\u00f4ng s\u1eed d\u1ee5ng c\u00e1c ch\u1eef s\u1ed1 1 ; 2 ; 3 n\u00ean n\u00f3 c\u00f3 th\u1ec3 l\u00e0 : 5 ; 7 ; 9 ; 45 ; 47 ; 49 ; 55 ; 57 ; 59. Nh\u01b0ng n\u00f3 kh\u00f4ng chia h\u1ebft cho 3 ; 5 ; 7 n\u00ean trong c\u00e1c s\u1ed1 tr\u00ean ch\u1ec9 c\u00f3 s\u1ed1 47 l\u00e0 th\u1ecfa m\u00e3n. V\u1eady n\u00f3 l\u00e0 s\u1ed1 47. B\u00e0i 141 : B\u1ea1n T\u00e2n th\u1ef1c hi\u1ec7n ph\u00e9p chia m\u1ed9t s\u1ed1 cho 12 th\u00ec d\u01b0 1 v\u00e0 chia s\u1ed1 \u0111\u00f3 cho 14 th\u00ec d\u01b0 2. B\u1ea1n h\u00e3y ch\u1ee9ng t\u1ecf T\u00e2n \u0111\u00e3 l\u00e0m sai \u00edt nh\u1ea5t m\u1ed9t ph\u00e9p t\u00ednh. 79","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i gi\u1ea3i : A = 12 x p + 1 = 14 x q + 2 (v\u1edbi p ; q l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean) Ta th\u1ea5y : 12 x p l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean A = 12 x p + 1 l\u00e0 s\u1ed1 l\u1ebb. 14 x q l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean A = 14 x q + 2 l\u00e0 s\u1ed1 ch\u1eb5n. A kh\u00f4ng th\u1ec3 v\u1eeba l\u1ebb v\u1eeba ch\u1eb5n n\u00ean ch\u1eafc ch\u1eafn c\u00f3 \u00edt nh\u1ea5t m\u1ed9t ph\u00e9p t\u00ednh sai. B\u00e0i 142 : V\u01b0\u1eddn c\u00e2y b\u00e0 Th\u01b0\u1ee3c c\u00f3 s\u1ed1 c\u00e2y ch\u01b0a \u0111\u1ebfn 100 v\u00e0 c\u00f3 4 lo\u1ea1i c\u00e2y : xo\u00e0i, cam, m\u00edt, b\u01b0\u1edfi. Trong \u0111\u00f3 s\u1ed1 c\u00e2y xo\u00e0i chi\u1ebfm 1\/5 s\u1ed1 c\u00e2y, s\u1ed1 c\u00e2y cam chi\u1ebfm 1\/6 s\u1ed1 c\u00e2y, s\u1ed1 c\u00e2y b\u01b0\u1edfi chi\u1ebfm1\/4 s\u1ed1 c\u00e2y v\u00e0 c\u00f2n l\u1ea1i l\u00e0 m\u00edt. H\u00e3y t\u00ednh xem m\u1ed7i lo\u1ea1i c\u00f3 bao nhi\u00eau c\u00e2y? B\u00e0i gi\u1ea3i : S\u1ed1 c\u00e2y xo\u00e0i chi\u1ebfm 1\/5 s\u1ed1 c\u00e2y, s\u1ed1 c\u00e2y cam chi\u1ebfm 1\/6 s\u1ed1 c\u00e2y, s\u1ed1 c\u00e2y b\u01b0\u1edfi chi\u1ebfm 1\/4 s\u1ed1 c\u00e2y n\u00ean s\u1ed1 c\u00e2y trong v\u01b0\u1eddn ph\u1ea3i chia h\u1ebft cho 4, 5, 6. M\u00e0 6 = 2 x 3 n\u00ean s\u1ed1 c\u00e2y trong v\u01b0\u1eddn ph\u1ea3i chia h\u1ebft cho 3, 4, 5. S\u1ed1 nh\u1ecf h\u01a1n 100 chia h\u1ebft cho 3, 4, 5 l\u00e0 60. V\u1eady s\u1ed1 c\u00e2y trong v\u01b0\u1eddn l\u00e0 60 c\u00e2y. S\u1ed1 c\u00e2y xo\u00e0i trong v\u01b0\u1eddn l\u00e0 : 60 : 5 = 12 (c\u00e2y) S\u1ed1 c\u00e2y cam trong v\u01b0\u1eddn l\u00e0 : 60 : 6 = 10 (c\u00e2y) S\u1ed1 c\u00e2y b\u01b0\u1edfi trong v\u01b0\u1eddn l\u00e0 : 60 : 4 = 15 (c\u00e2y) S\u1ed1 c\u00e2y m\u00edt trong V\u01b0\u1eddn l\u00e0 : 60 - (12 + 10 + 15) = 23 (c\u00e2y) \u0110\u00e1p s\u1ed1 : xo\u00e0i : 12 c\u00e2y ; cam : 10 c\u00e2y ; b\u01b0\u1edfi : 15 c\u00e2y ; m\u00edt : 23 c\u00e2y B\u00e0i 143 : B\u1ea1n h\u00e3y chia t\u1ea5m b\u00eca b\u00ean d\u01b0\u1edbi th\u00e0nh 6 ph\u1ea7n gi\u1ed1ng h\u1ec7t nhau v\u1ec1 h\u00ecnh d\u1ea1ng v\u00e0 m\u1ed7i ph\u1ea7n c\u00f3 m\u1ed9t b\u00f4ng hoa. B\u00e0i gi\u1ea3i : Ta chia t\u1ea5m b\u00eca th\u00e0nh c\u00e1c \u00f4 vu\u00f4ng nh\u1ecf b\u1eb1ng nhau nh\u01b0 trong h\u00ecnh v\u1ebd sau : Nh\u00ecn h\u00ecnh v\u1ebd ta th\u1ea5y t\u1ed5ng s\u1ed1 \u00f4 vu\u00f4ng nh\u1ecf l\u00e0 18 \u00f4. Do \u0111\u00f3 khi chia t\u1ea5m b\u00eca th\u00e0nh 6 ph\u1ea7n gi\u1ed1ng h\u1ec7t nhau v\u1ec1 h\u00ecnh d\u1ea1ng th\u00ec m\u1ed7i ph\u1ea7n s\u1ebd c\u00f3 s\u1ed1 \u00f4 l\u00e0 : 18 : 6 = 3 (\u00f4) v\u00e0 h\u00ecnh 80","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i d\u1ea1ng m\u1ed7i ph\u1ea7n ph\u1ea3i c\u00f3 d\u1ea1ng h\u00ecnh ch\u1eef L. Ta c\u00f3 c\u00e1ch chia nh\u01b0 sau : (c\u1eaft theo \u0111\u01b0\u1eddng m\u00e0u) B\u00e0i 144 : Cho d\u00e3y c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp : 2 ; 4 ; 6 ; 8 ; ... ; 998 ; 1000. Sau khi \u0111i\u1ec1n th\u00eam c\u00e1c d\u1ea5u + ho\u1eb7c d\u1ea5u - v\u00e0o gi\u1eefa c\u00e1c s\u1ed1 theo \u00fd m\u00ecnh, b\u1ea1n B\u00ecnh th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 2002 ; b\u1ea1n Minh th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 2006. Ai t\u00ednh \u0111\u00fang ? B\u00e0i gi\u1ea3i : T\u1eeb 2 \u0111\u1ebfn 1000 c\u00f3 : (1000 - 2) : 2 + 1 = 500 (s\u1ed1 ch\u1eb5n) T\u1ed5ng c\u00e1c s\u1ed1 \u0111\u00f3 : N = (1000 + 2) x 500 : 2 = 250500. S\u1ed1 n\u00e0y chia h\u1ebft cho 4. Khi thay + a th\u00e0nh - a th\u00ec N b\u1ecb gi\u1ea3m \u0111i a x 2 c\u0169ng l\u00e0 s\u1ed1 chia h\u1ebft cho 4. Do \u0111\u00f3 k\u1ebft qu\u1ea3 cu\u1ed1i c\u00f9ng ph\u1ea3i l\u00e0 s\u1ed1 chia h\u1ebft cho 4. B\u00ecnh t\u00ednh \u0111\u01b0\u1ee3c 2002, Minh t\u00ednh \u0111\u01b0\u1ee3c 2006 \u0111\u1ec1u l\u00e0 s\u1ed1 kh\u00f4ng chia h\u1ebft cho 4. V\u1eady c\u1ea3 hai b\u1ea1n \u0111\u1ec1u t\u00ednh sai. B\u00e0i 145 : Tr\u01b0\u1eddng Ti\u1ec3u h\u1ecdc Xu\u00e2n \u0110\u1ec9nh tham gia h\u1ed9i kh\u1ecfe Ph\u00f9 \u0110\u1ed5ng, c\u00f3 11 h\u1ecdc sinh \u0111o\u1ea1t gi\u1ea3i, trong \u0111\u00f3 c\u00f3 6 em gi\u00e0nh \u00edt nh\u1ea5t 2 gi\u1ea3i, c\u00f3 4 em gi\u00e0nh \u00edt nh\u1ea5t 3 gi\u1ea3i v\u00e0 c\u00f3 2 em gi\u00e0nh m\u1ed7i ng\u01b0\u1eddi 4 gi\u1ea3i. H\u1ecfi tr\u01b0\u1eddng \u0111\u00f3 \u0111\u00e3 gi\u00e0nh \u0111\u01b0\u1ee3c bao nhi\u00eau gi\u1ea3i ? B\u00e0i gi\u1ea3i : C\u00f3 11 em \u0111o\u1ea1t gi\u1ea3i, trong \u0111\u00f3 c\u00f3 6 em gi\u00e0nh \u00edt nh\u1ea5t 2 gi\u1ea3i n\u00ean s\u1ed1 h\u1ecdc sinh gi\u00e0nh m\u1ed7i em 1 gi\u1ea3i l\u00e0 : 11 - 6 = 5 (em). C\u00f3 6 em gi\u00e0nh \u00edt nh\u1ea5t 2 gi\u1ea3i, trong \u0111\u00f3 c\u00f3 4 em gi\u00e0nh \u00edt nh\u1ea5t 3 gi\u1ea3i n\u00ean s\u1ed1 em gi\u00e0nh m\u1ed7i em 2 gi\u1ea3i l\u00e0 : 6 - 4 = 2 (em). C\u00f3 4 em gi\u00e0nh \u00edt nh\u1ea5t 3 gi\u1ea3i trong \u0111\u00f3 c\u00f3 c\u00f3 2 em gi\u00e0nh m\u1ed7i em 4 gi\u1ea3i n\u00ean s\u1ed1 em gi\u00e0nh m\u1ed7i em 3 gi\u1ea3i l\u00e0 : 4 - 2 = 2 (em). S\u1ed1 em gi\u00e0nh t\u1eeb 1 \u0111\u1ebfn 4 gi\u1ea3i l\u00e0 : 5 + 2 + 2 + 2 = 11 (em). Do \u0111\u00f3 kh\u00f4ng c\u00f3 em n\u00e0o gi\u00e0nh \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n 4 gi\u1ea3i. V\u1eady s\u1ed1 gi\u1ea3i m\u00e0 tr\u01b0\u1eddng \u0111\u00f3 gi\u00e0nh \u0111\u01b0\u1ee3c l\u00e0 : 1 x 5 + 2 x 2 + 3 x 2 + 4 x 2 = 23 (gi\u1ea3i). B\u00e0i 146 : T\u00ednh nhanh t\u1ed5ng sau : 81","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i gi\u1ea3i : \u0110\u1eb7t t\u1ed5ng tr\u00ean b\u1eb1ng A ta c\u00f3 : B\u00e0i 147 : T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean a \u0111\u1ec3 bi\u1ec3u th\u1ee9c : A = 4010 - 2005 : (2006 - a) c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. B\u00e0i gi\u1ea3i : \u0110\u1ec3 A c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t th\u00ec s\u1ed1 tr\u1eeb 2005 : (2006 - a) c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t kh\u00f4ng v\u01b0\u1ee3t qu\u00e1 4010. \u0110\u1ec3 2005 : (2006 - a) c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t th\u00ec s\u1ed1 chia (2006 - a) c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t l\u1edbn h\u01a1n 0. V\u1eady 2006 - a = 1 a = 2006 - 1 a = 2005. B\u00e0i 148 : M\u1ed9t l\u1edbp c\u00f3 29 h\u1ecdc sinh. Trong m\u1ed9t l\u1ea7n ki\u1ec3m tra ch\u00ednh t\u1ea3. b\u1ea1n Xu\u00e2n m\u1eafc 9 l\u1ed7i, c\u00f2n c\u00e1c b\u1ea1n trong l\u1edbp m\u1eafc \u00edt l\u1ed7i h\u01a1n. Ch\u1ee9ng minh r\u1eb1ng : Trong l\u1edbp c\u00f3 \u00edt nh\u1ea5t 4 b\u1ea1n c\u00f3 s\u1ed1 l\u1ed7i b\u1eb1ng nhau (k\u1ec3 c\u1ea3 tr\u01b0\u1eddng h\u1ee3p s\u1ed1 l\u1ed7i b\u1eb1ng 0). B\u00e0i gi\u1ea3i : V\u00ec c\u00e1c b\u1ea1n trong l\u1edbp \u0111\u1ec1u c\u00f3 \u00edt l\u1ed7i h\u01a1n Xu\u00e2n, n\u00ean c\u00e1c b\u1ea1n ch\u1ec9 c\u00f3 s\u1ed1 l\u1ed7i t\u1eeb 0 \u0111\u1ebfn 8. Tr\u1eeb Xu\u00e2n ra th\u00ec s\u1ed1 b\u1ea1n c\u00f2n l\u1ea1i l\u00e0 : 29 - 1 = 28 (b\u1ea1n). N\u1ebfu chia c\u00e1c b\u1ea1n c\u00f2n l\u1ea1i th\u00e0nh c\u00e1c nh\u00f3m theo s\u1ed1 l\u1ed7i th\u00ec t\u1ed1i \u0111a c\u00f3 9 nh\u00f3m. N\u1ebfu m\u1ed7i nh\u00f3m c\u00f3 kh\u00f4ng qu\u00e1 3 b\u1ea1n th\u00ec 9 nh\u00f3m s\u1ebd c\u00f3 kh\u00f4ng qu\u00e1 3 x 9 = 27 (b\u1ea1n). \u0110i\u1ec1u n\u00e0y m\u00e2u thu\u1eabn v\u1edbi s\u1ed1 b\u1ea1n c\u00f2n l\u1ea1i l\u00e0 28 82","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i b\u1ea1n. Ch\u1ee9ng t\u1ecf \u00edt nh\u1ea5t ph\u1ea3i c\u00f3 m\u1ed9t nh\u00f3m c\u00f3 qu\u00e1 3 b\u1ea1n t\u1ee9c l\u00e0 trong l\u1edbp c\u00f3 \u00edt nh\u1ea5t c\u00f3 4 b\u1ea1n c\u00f3 s\u1ed1 l\u1ed7i b\u1eb1ng nhau. B\u00e0i 149 : H\u1ee3p t\u00e1c x\u00e3 H\u00f2a B\u00ecnh d\u1ef1 \u0111\u1ecbnh x\u00e2y d\u1ef1ng m\u1ed9t khu vui ch\u01a1i cho tr\u1ebb em trong x\u00e3. V\u00ec th\u1ebf h\u1ecd \u0111\u00e3 m\u1edf r\u1ed9ng m\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u1ec3 di\u1ec7n t\u00edch g\u1ea5p ba l\u1ea7n di\u1ec7n t\u00edch ban \u0111\u1ea7u. Chi\u1ec1u r\u1ed9ng m\u1ea3nh \u0111\u1ea5t ch\u1ec9 c\u00f3 th\u1ec3 t\u0103ng l\u00ean g\u1ea5p \u0111\u00f4i n\u00ean ph\u1ea3i m\u1edf r\u1ed9ng th\u00eam chi\u1ec1u d\u00e0i. Khi \u0111\u00f3 m\u1ea3nh \u0111\u1ea5t tr\u1edf th\u00e0nh h\u00ecnh vu\u00f4ng. H\u00e3y t\u00ednh di\u1ec7n t\u00edch khu vui ch\u01a1i \u0111\u00f3. Bi\u1ebft r\u1eb1ng chu vi m\u1ea3nh \u0111\u1ea5t ban \u0111\u1ea7u l\u00e0 56 m. B\u00e0i gi\u1ea3i : G\u1ecdi m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt l\u00fac \u0111\u1ea7u l\u00e0 ABCD, khi m\u1edf r\u1ed9ng m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u1ec3 \u0111\u01b0\u1ee3c m\u1ea3nh \u0111\u1ea5t h\u00ecnh vu\u00f4ng APMN c\u00f3 c\u1ea1nh h\u00ecnh vu\u00f4ng g\u1ea5p 2 l\u1ea7n chi\u1ec1u r\u1ed9ng m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt ABCD v\u00e0 di\u1ec7n t\u00edch g\u1ea5p 3 l\u1ea7n di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt \u1ea5y. Khi \u0111\u00f3 di\u1ec7n t\u00edch c\u1ee7a c\u00e1c m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt ABCD, DCHN, BPMH b\u1eb1ng nhau. M\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt BPMH c\u00f3 \u0111\u1ed9 d\u00e0i c\u1ea1nh BH g\u1ea5p 2 l\u1ea7n \u0111\u1ed9 d\u00e0i c\u1ea1nh AD n\u00ean N\u1eeda chu vi m\u1ea3nh \u0111\u1ea5t ban \u0111\u1ea7u l\u00e0 56 m n\u00ean AD + AB = 56 : 2 = 28 (m). Ta c\u00f3 : Chi\u1ec1u r\u1ed9ng m\u1ea3nh \u0111\u1ea5t ban \u0111\u1ea7u (AD) l\u00e0 : 28 : (3 + 4) x 3 = 12 (m). C\u1ea1nh h\u00ecnh vu\u00f4ng APMN l\u00e0 : 12 x 2 = 24 (m). 83","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Di\u1ec7n t\u00edch khu vui ch\u01a1i l\u00e0 : 24 x 24 = 576 (m2). B\u00e0i 150 : Cho (1), (2), (3), (4) l\u00e0 c\u00e1c h\u00ecnh thang vu\u00f4ng c\u00f3 k\u00edch th\u01b0\u1edbc b\u1eb1ng nhau. Bi\u1ebft r\u1eb1ng PQ = 4 cm. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD. B\u00e0i gi\u1ea3i : V\u00ec c\u00e1c h\u00ecnh thang vu\u00f4ng PQMA, QMBC, QPNC, PNDA b\u1eb1ng nhau n\u00ean : MQ = NP = QP = 4 cm v\u00e0 CN = AD. M\u1eb7t kh\u00e1c AD = NP + QM = 4 + 4 = 8 (cm) Do \u0111\u00f3 : CN = AD = 8 cm. Di\u1ec7n t\u00edch h\u00ecnh thang vu\u00f4ng PQCN l\u00e0 : (CN + PQ) x NP : 2 = (8 + 4) x 4 : 2 = 24 (cm2) Suy ra : Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD l\u00e0 : 24 x 4 = 96 (cm2) B\u00e0i 151 : M\u1ed9t \u00f4 t\u00f4 d\u1ef1 \u0111\u1ecbnh \u0111i t\u1eeb C \u0111\u1ebfn D trong 3 gi\u1edd. Do th\u1eddi ti\u1ebft x\u1ea5u n\u00ean v\u1eadn t\u1ed1c c\u1ee7a \u00f4 t\u00f4 gi\u1ea3m 14 km\/gi\u1edd v\u00e0 v\u00ec v\u1eady \u0111\u1ebfn D mu\u1ed9n 1 gi\u1edd so v\u1edbi th\u1eddi gian d\u1ef1 \u0111\u1ecbnh. T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng CD. Gi\u1ea3i : Th\u1eddi gian \u00f4 t\u00f4 th\u1ef1c \u0111i qu\u00e3ng \u0111\u01b0\u1eddng CD l\u00e0 : 3 + 1 = 4 (gi\u1edd) T\u1ec9 s\u1ed1 gi\u1eefa th\u1eddi gian d\u1ef1 \u0111\u1ecbnh v\u00e0 th\u1eddi gian th\u1ef1c \u0111i l\u00e0 3 : 4 = 3\/4. V\u00ec qu\u00e3ng \u0111\u01b0\u1eddng CD kh\u00f4ng \u0111\u1ed5i n\u00ean v\u1eadn t\u1ed1c v\u00e0 th\u1eddi gian l\u00e0 hai \u0111\u1ea1i l\u01b0\u1ee3ng t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi nhau. Do \u0111\u00f3 t\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh (vd\u1ef1 \u0111\u1ecbnh) v\u00e0 v\u1eadn t\u1ed1c th\u1ef1c \u0111i (vth\u1ef1c \u0111i) l\u00e0 4\/3. N\u1ebfu vd\u1ef1 \u0111\u1ecbnh v\u00e0 vth\u1ef1c \u0111i t\u00ednh theo \u0111\u01a1n v\u1ecb km\/gi\u1edd th\u00ec ta c\u00f3 s\u01a1 \u0111\u1ed3 sau : 84","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i V\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh \u0111i qu\u00e3ng \u0111\u01b0\u1eddng CD l\u00e0 : 14 x 4 = 56 (km\/gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng CD d\u00e0i l\u00e0 : 56 x 3 = 168 (km). B\u00e0i 152 : M\u1ed9t ca n\u00f4 xu\u00f4i d\u00f2ng t\u1eeb A \u0111\u1ebfn B h\u1ebft 5 gi\u1edd v\u00e0 ng\u01b0\u1ee3c d\u00f2ng t\u1eeb B v\u1ec1 A h\u1ebft 6 gi\u1edd. T\u00ednh kho\u1ea3ng c\u00e1ch AB bi\u1ebft v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 3 km\/gi\u1edd. Ph\u00e2n t\u00edch : \u0110\u00e2y l\u00e0 b\u00e0i to\u00e1n chuy\u1ec3n \u0111\u1ed9ng tr\u00ean d\u00f2ng n\u01b0\u1edbc. Ngo\u00e0i gi\u1ea3 thi\u1ebft m\u00e0 b\u00e0i to\u00e1n \u0111\u00e3 cho, ch\u00fang ta c\u1ea7n bi\u1ebft th\u00eam ki\u1ebfn th\u1ee9c v\u1ec1 chuy\u1ec3n \u0111\u1ed9ng tr\u00ean d\u00f2ng n\u01b0\u1edbc nh\u01b0 sau : V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng = V\u1eadn t\u1ed1c th\u1ef1c + V\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc. V\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng = V\u1eadn t\u1ed1c th\u1ef1c - V\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc. T\u1eeb \u0111\u00f3 ta c\u00f3 : V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng - V\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng = 2 x V\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc. B\u00e0i to\u00e1n n\u00e0y cho bi\u1ebft v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc n\u00ean ta t\u00ednh \u0111\u01b0\u1ee3c hi\u1ec7u v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng. Bi\u1ebft th\u1eddi gian xu\u00f4i d\u00f2ng v\u00e0 th\u1eddi gian ng\u01b0\u1ee3c d\u00f2ng ta d\u1ef1a v\u00e0o \u0111\u00f3 t\u00ecm t\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c v\u00e0 \u0111\u01b0a v\u1ec1 d\u1ea1ng to\u00e1n t\u00ecm 2 s\u1ed1 bi\u1ebft hi\u1ec7u v\u00e0 t\u1ec9. Gi\u1ea3i : Hi\u1ec7u v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng v\u00e0 v\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng ch\u00ednh l\u00e0 2 l\u1ea7n v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc n\u00ean hi\u1ec7u \u0111\u00f3 l\u00e0 : 3 x 2 = 6 (km\/gi\u1edd) T\u1ec9 s\u1ed1 th\u1eddi gian xu\u00f4i d\u00f2ng v\u00e0 th\u1eddi gian ng\u01b0\u1ee3c d\u00f2ng l\u00e0 5 : 6 = 5\/6. V\u00ec qu\u00e3ng \u0111\u01b0\u1eddng kh\u00f4ng \u0111\u1ed5i n\u00ean v\u1eadn t\u1ed1c v\u00e0 th\u1eddi gian l\u00e0 hai \u0111\u1ea1i l\u01b0\u1ee3ng t\u1ec9 l\u1ec7 ngh\u1ecbch. Do \u0111\u00f3 t\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng l\u00e0 6\/5. Ta c\u00f3 s\u01a1 \u0111\u1ed3 : V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng l\u00e0 : 6 x 6 = 36 (km\/gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0 : 36 x 5 = 180 (km). B\u00e0i 153 : Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD, g\u1ecdi M v\u00e0 N l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a AB v\u00e0 CD. N\u1ed1i DM, BN c\u1eaft AC t\u1ea1i I v\u00e0 K. Ch\u1ee9ng t\u1ecf r\u1eb1ng AI = IK = KC. 85","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Gi\u1ea3i : (\u1edf b\u00e0i n\u00e0y ta c\u1ea7n v\u1eadn d\u1ee5ng m\u1ed1i quan h\u1ec7 gi\u1eefa di\u1ec7n t\u00edch, c.\u0111\u00e1y v\u00e0 c.cao c\u1ee7a tam gi\u00e1c) Ta c\u00f3 : dt (ABC) = 2 x dt (AMD) (v\u00ec AB = 2 x AM v\u00e0 AD = BC) ; dt (DCM) = dt (ABC) (v\u00ec AB = DC v\u00e0 c.cao c\u00f9ng b\u1eb1ng BC) Suy ra dt (DCM) = 2 x dt (AMD). G\u1ecdi CH v\u00e0 AE l\u1ea7n l\u01b0\u1ee3t l\u00e0 chi\u1ec1u cao c\u1ee7a tam gi\u00e1c DCM v\u00e0 DAM xu\u1ed1ng \u0111\u00e1y DM, khi \u0111\u00f3 CH = 2 x AE. Nh\u01b0ng CH v\u00e0 AE l\u1ea7n l\u01b0\u1ee3t l\u00e0 chi\u1ec1u cao c\u1ee7a tam gi\u00e1c ICM v\u00e0 IAM c\u00f3 chung c\u1ea1nh \u0111\u00e1y IM. V\u1eady dt (ICM) = 2 x dt (IAM). M\u00e0 tam gi\u00e1c IAM v\u00e0 ICM chung chi\u1ec1u cao t\u1eeb M, do \u0111\u00f3 IC = 2 x AI, suy ra AC = 3 x AI hay AI = 1\/3 AC. L\u00e0m t\u01b0\u01a1ng t\u1ef1 v\u1edbi c\u00e1c c\u1eb7p tam gi\u00e1c ABN v\u00e0 CBN ; KCN v\u00e0 KAN ta c\u00f3 KC = 1\/3 AC. V\u1eady AI = KC = 1\/3 AC, suy ra IK = 1\/3 AC. Do \u0111\u00f3 AI = IK = KC. Ch\u00fa \u00fd : \u1edf \u0111\u00e2y \u0111\u1ec3 ch\u1ee9ng t\u1ecf c\u00e1c \u0111o\u1ea1n th\u1eb3ng b\u1eb1ng nhau ta ph\u1ea3i ch\u1ee9ng t\u1ecf c\u00e1c tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao v\u00e0 di\u1ec7n t\u00edch b\u1eb1ng nhau. B\u00e0i 154: Cho tam gi\u00e1c ABC, g\u1ecdi c\u00e1c \u0111i\u1ec3m M, N l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean c\u00e1c c\u1ea1nh AB, AC sao cho : AB = 3 x AM, AC = 3 x AN. G\u1ecdi I l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u1ea1nh BC. a) Ch\u1ee9ng t\u1ecf r\u1eb1ng t\u1ee9 gi\u00e1c BMNC l\u00e0 h\u00ecnh thang v\u00e0 BC = 3 x MN. b) Ch\u1ee9ng t\u1ecf r\u1eb1ng c\u00e1c \u0111o\u1ea1n th\u1eb3ng BN, CM, AI c\u00f9ng c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m. Gi\u1ea3i : a) V\u00ec AB = 3 x AM, AC = 3 x AN, n\u00ean MB = 2\/3 x AB, NC = 2\/3 x AC. T\u1eeb \u0111\u00f3 suy ra : dt (MBC) = 2\/3 x dt (ABC) (chung chi\u1ec1u cao t\u1eeb C dt (NCB) = 2\/3 x dt (ABC) (chung chi\u1ec1u cao t\u1eeb B) 86","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i V\u1eady dt (MBC) = dt (NCB) m\u00e0 tam gi\u00e1c MBC v\u00e0 tam gi\u00e1c NCB c\u00f3 chung \u0111\u00e1y BC, n\u00ean chi\u1ec1u cao t\u1eeb M b\u1eb1ng chi\u1ec1u cao t\u1eeb N xu\u1ed1ng \u0111\u00e1y BC hay MN song song v\u1edbi BC. Do \u0111\u00f3 BMNC l\u00e0 h\u00ecnh thang. T\u1eeb MB = 2\/3 x AB, n\u00ean dt (MBN) = 2\/3 x dt (ABN) (chung chi\u1ec1u cao t\u1eeb N) hay dt (ABN) = 2\/3 x dt (MBN). H\u01a1n n\u1eefa t\u1eeb AC = 3 x AN, n\u00ean NC = 2 x AN, do \u0111\u00f3 dt (NBC) = 2 x dt (ABN) (chung chi\u1ec1u cao t\u1eeb B) ; suy ra dt (NBC) = 3\/2 x 2 x dt (MBN) = 3 x dt (MBN). M\u00e0 tam gi\u00e1c NBC v\u00e0 tam gi\u00e1c MBN c\u00f3 chi\u1ec1u cao b\u1eb1ng nhau (c\u00f9ng l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh thang BMNC). V\u00ec v\u1eady \u0111\u00e1y BC = 3 x MN. b) G\u1ecdi BN c\u1eaft CM t\u1ea1i O. Ta s\u1ebd ch\u1ee9ng t\u1ecf AI c\u0169ng c\u1eaft BN t\u1ea1i O. Mu\u1ed1n v\u1eady, n\u1ed1i AO k\u00e9o d\u00e0i c\u1eaft BC t\u1ea1i K, ta s\u1ebd ch\u1ee9ng t\u1ecf K l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a BC (hay K tr\u00f9ng v\u1edbi I). Theo ph\u1ea7n a) ta \u0111\u00e3 c\u00f3 dt (NBC) = 2 x dt (ABN). M\u00e0 tam gi\u00e1c NBC v\u00e0 tam gi\u00e1c ABN c\u00f3 chung \u0111\u00e1y BN, n\u00ean chi\u1ec1u cao t\u1eeb C g\u1ea5p 2 l\u1ea7n chi\u1ec1u cao t\u1eeb A xu\u1ed1ng \u0111\u00e1y BN. Nh\u01b0ng \u0111\u00f3 l\u00e0 chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng c\u1ee7a hai tam gi\u00e1c BCO v\u00e0 BAO c\u00f3 chung \u0111\u00e1y BO, v\u00ec v\u1eady dt (BCO) = 2 x dt (BAO) T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 dt (BCO) = 2 x dt (CAO). Do \u0111\u00f3 dt (BAO) = dt (CAO). Hai tam gi\u00e1c BAO v\u00e0 CAO c\u00f3 chung \u0111\u00e1y AO, n\u00ean chi\u1ec1u cao t\u1eeb B b\u1eb1ng chi\u1ec1u cao t\u1eeb C xu\u1ed1ng \u0111\u00e1y AO. \u0110\u00f3 c\u0169ng l\u00e0 chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng c\u1ee7a hai tam gi\u00e1c BOK v\u00e0 COK c\u00f3 chung \u0111\u00e1y OK, v\u00ec v\u1eady dt (BOK) = dt (COK). M\u00e0 hai tam gi\u00e1c BOK v\u00e0 tam gi\u00e1c COK l\u1ea1i chung chi\u1ec1u cao t\u1eeb O, n\u00ean hai \u0111\u00e1y BK = CK hay K l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u1ea1nh BC. V\u1eady \u0111i\u1ec3m K tr\u00f9ng v\u1edbi \u0111i\u1ec3m I hay BN, CM, AI c\u00f9ng c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m O. B\u00e0i 155: M\u1ed9t vi\u00ean quan mang l\u1ec5 v\u1eadt \u0111\u1ebfn d\u00e2ng vua v\u00e0 \u0111\u01b0\u1ee3c vua ban th\u01b0\u1edfng cho m\u1ed9t qu\u1ea3 cam trong v\u01b0\u1eddn th\u01b0\u1ee3ng uy\u1ec3n, nh\u01b0ng ph\u1ea3i t\u1ef1 v\u00e0o v\u01b0\u1eddn h\u00e1i. \u0110\u01b0\u1eddng v\u00e0o v\u01b0\u1eddn th\u01b0\u1ee3ng uy\u1ec3n ph\u1ea3i qua ba c\u1ed5ng c\u00f3 l\u00ednh canh. Vi\u00ean quan \u0111\u1ebfn c\u1ed5ng th\u1ee9 nh\u1ea5t, ng\u01b0\u1eddi l\u00ednh canh giao h\u1eb9n: \u201cTa cho \u00f4ng v\u00e0o nh\u01b0ng l\u00fac ra \u00f4ng ph\u1ea3i bi\u1ebfu ta m\u1ed9t n\u1eeda s\u1ed1 cam, th\u00eam n\u1eeda qu\u1ea3\u201d. Qua c\u1ed5ng th\u1ee9 hai r\u1ed3i th\u1ee9 ba l\u00ednh canh c\u0169ng \u0111\u1ec1u giao h\u1eb9n nh\u01b0 v\u1eady. H\u1ecfi \u0111\u1ec3 c\u00f3 m\u1ed9t qu\u1ea3 cam mang v\u1ec1 th\u00ec vi\u00ean quan \u0111\u00f3 ph\u1ea3i h\u00e1i bao nhi\u00eau cam trong v\u01b0\u1eddn? 87","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Gi\u1ea3i: S\u1ed1 cam vi\u00ean quan c\u00f2n l\u1ea1i sau khi cho l\u00ednh g\u00e1c c\u1ed5ng th\u1ee9 hai (c\u1ed5ng gi\u1eefa) l\u00e0: S\u1ed1 cam vi\u00ean quan c\u00f2n l\u1ea1i sau khi cho l\u00ednh g\u00e1c c\u1ed5ng th\u1ee9 ba (c\u1ed5ng trong c\u00f9ng) l\u00e0: S\u1ed1 cam vi\u00ean quan ph\u1ea3i h\u00e1i trong v\u01b0\u1eddn l\u00e0: V\u1eady \u0111\u1ec3 c\u00f3 \u0111\u01b0\u1ee3c m\u1ed9t qu\u1ea3 cam mang v\u1ec1 th\u00ec vi\u00ean quan ph\u1ea3i h\u00e1i 15 qu\u1ea3 trong v\u01b0\u1eddn. \u0110\u00e1p s\u1ed1: 15 qu\u1ea3 cam B\u00e0i 156: C\u00f3 m\u1ed9t gi\u1ed1ng b\u00e8o c\u1ee9 m\u1ed7i ng\u00e0y l\u1ea1i n\u1edf t\u0103ng g\u1ea5p \u0111\u00f4i. N\u1ebfu ng\u00e0y \u0111\u1ea7u cho v\u00e0o m\u1eb7t h\u1ed3 m\u1ed9t c\u00e2y b\u00e8o th\u00ec 10 ng\u00e0y sau b\u00e8o lan ph\u1ee7 k\u00edn m\u1eb7t h\u1ed3. V\u1eady n\u1ebfu ban \u0111\u1ea7u cho v\u00e0o 16 c\u00e2y b\u00e8o th\u00ec m\u1ea5y ng\u00e0y sau b\u00e8o ph\u1ee7 k\u00edn m\u1eb7t h\u1ed3? Gi\u1ea3i: Ta c\u00f3 b\u1ea3ng sau bi\u1ec3u di\u1ec5n s\u1ed1 c\u00e2y b\u00e8o tr\u00ean m\u1eb7t h\u1ed3: 88","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Nh\u00ecn v\u00e0o b\u1ea3ng tr\u00ean ta th\u1ea5y: N\u1ebfu ng\u00e0y \u0111\u1ea7u cho v\u00e0o m\u1eb7t h\u1ed3 16 c\u00e2y b\u00e8o th\u00ec 6 ng\u00e0y sau b\u00e8o s\u1ebd lan ph\u1ee7 k\u00edn m\u1eb7t h\u1ed3. B\u00e0i 157 : L\u1edbp 4A tr\u1ed3ng \u0111\u01b0\u1ee3c 21 c\u00e2y ; l\u1edbp 4B tr\u1ed3ng \u0111\u01b0\u1ee3c 22 c\u00e2y ; l\u1edbp 4C tr\u1ed3ng \u0111\u01b0\u1ee3c 29 c\u00e2y ; l\u1edbp 4D tr\u1ed3ng \u0111\u01b0\u1ee3c s\u1ed1 c\u00e2y h\u01a1n trung b\u00ecnh c\u1ed9ng s\u1ed1 c\u00e2y c\u1ee7a c\u1ea3 4 l\u1edbp l\u00e0 3 c\u00e2y. H\u1ecfi l\u1edbp 4D tr\u1ed3ng \u0111\u01b0\u1ee3c bao nhi\u00eau c\u00e2y? Ph\u00e2n t\u00edch : B\u00e0i to\u00e1n n\u00e0y cho s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 4D kh\u00f4ng nh\u1eefng b\u1eb1ng trung b\u00ecnh c\u1ed9ng s\u1ed1 c\u00e2y c\u1ee7a c 4 l\u1edbp m\u00e0 c\u00f2n h\u01a1n trung b\u00ecnh c\u1ed9ng s\u1ed1 c\u00e2y c\u1ee7a b\u1ed1n l\u1edbp l\u00e0 3 c\u00e2y. D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p s\u01a1 \u0111\u1ed3 \u0111o\u1ea1n th\u1eb3ng ta c\u00f3 : T\u1ed5ng s\u1ed1 c\u00e2y c\u1ee7a 3 l\u1edbp 4A ; 4B ; 4C v\u00e0 th\u00eam 3 c\u00e2y n\u1eefa s\u1ebd l\u00e0 3 l\u1ea7n trung b\u00ecnh c\u1ed9ng s\u1ed1 c\u00e2y c\u1ee7a c\u1ea3 4 l\u1edbp. T\u1eeb \u0111\u00f3 ta t\u00ecm \u0111\u01b0\u1ee3c s\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 4D. Gi\u1ea3i : Theo b\u00e0i ra ta c\u00f3 s\u01a1 \u0111\u1ed3: Nh\u00ecn v\u00e0o s\u01a1 \u0111\u1ed3 ta c\u00f3 trung b\u00ecnh c\u1ed9ng s\u1ed1 c\u00e2y c\u1ee7a c\u1ea3 4 l\u1edbp l\u00e0 : (21 + 22 + 29 + 3) : 3 = 25 (c\u00e2y) S\u1ed1 c\u00e2y c\u1ee7a l\u1edbp 4D tr\u1ed3ng \u0111\u01b0\u1ee3c l\u00e0 : 25 + 3 = 28 (c\u00e2y) 89","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i Nh\u1eadn x\u00e9t : N\u1ebfu c\u00f3 3 s\u1ed1 a ; b ; c v\u00e0 s\u1ed1 ch\u01b0a bi\u1ebft x m\u00e0 x l\u1edbn h\u01a1n trung b\u00ecnh c\u1ed9ng c\u1ee7a c\u1ea3 4 s\u1ed1 a ; b ; c ; x l\u00e0 n \u0111\u01a1n v\u1ecb th\u00ec trung b\u00ecnh c\u1ed9ng c\u1ee7a c\u1ea3 b\u1ed1n s\u1ed1 l\u00e0: (a + b + c + n) : 3 hay (a + b + c + x) : 4 = (a + b + c + n) : 3 (V\u1eadn d\u1ee5ng gi\u1ea3i b\u00e0i t\u1eadp sau: L\u1edbp 4A tr\u1ed3ng \u0111\u01b0\u1ee3c 21 c\u00e2y ; l\u1edbp 4B tr\u1ed3ng \u0111\u01b0\u1ee3c 22 c\u00e2y ; l\u1edbp 4C tr\u1ed3ng \u0111\u01b0\u1ee3c 29 c\u00e2y. L\u1edbp 4D tr\u1ed3ng \u0111\u01b0\u1ee3c s\u1ed1 c\u00e2y k\u00e9m trung b\u00ecnh c\u1ed9ng s\u1ed1 c\u00e2y c\u1ee7a c\u1ea3 4 l\u1edbp l\u00e0 3 c\u00e2y. H\u1ecfi l\u1edbp 4D tr\u1ed3ng \u0111\u01b0\u1ee3c bao nhi\u00eau c\u00e2y?) B\u00e0i 158 : H\u01b0ng \u0111i xe \u0111\u1ea1p t\u1eeb nh\u00e0 l\u00ean huy\u1ec7n v\u1edbi v\u1eadn t\u1ed1c 12 km\/gi\u1edd. Sau \u0111\u00f3 tr\u1edf v\u1ec1 v\u1edbi v\u1eadn t\u1ed1c 10 km\/gi\u1edd. T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 l\u00ean huy\u1ec7n bi\u1ebft r\u1eb1ng th\u1eddi gian l\u00fac v\u1ec1 l\u00e2u h\u01a1n l\u00fac \u0111i l\u00e0 10 ph\u00fat. Gi\u1ea3i Nh\u1eadn x\u00e9t : Ta th\u1ea5y H\u01b0ng \u0111i v\u00e0 v\u1ec1 tr\u00ean c\u00f9ng m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 l\u00ean huy\u1ec7n. Do \u0111\u00f3 th\u1eddi gian \u0111i v\u00e0 v\u1ec1 s\u1ebd t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi v\u1eadn t\u1ed1c l\u00fac \u0111i v\u00e0 v\u1eadn t\u1ed1c l\u00fac v\u1ec1. \u1edf \u0111\u00e2y t\u1ec9 s\u1ed1 v\u1ec1 v\u1eadn t\u1ed1c gi\u1eefa l\u00fac \u0111i v\u00e0 l\u00fac v\u1ec1 l\u00e0 12\/10 = 6\/5. V\u1eady t\u1ec9 s\u1ed1 gi\u1eefa th\u1eddi gian \u0111i v\u00e0 th\u1eddi gian v\u1ec1 l\u00e0 5\/6. M\u00e0 th\u1eddi gian l\u00fac v\u1ec1 l\u00e2u h\u01a1n l\u00fac \u0111i l\u00e0 10 ph\u00fat hay nhi\u1ec1u h\u01a1n 10 ph\u00fat. T\u1eeb \u0111\u00f3 ta c\u00f3 s\u01a1 \u0111\u1ed3 : Th\u1eddi gian l\u00fac v\u1ec1 h\u1ebft l\u00e0 :10 : (6 - 5) x 6 = 60 (ph\u00fat) \u0110\u1ed5i : 60 ph\u00fat = 1 gi\u1edd Qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 l\u00ean huy\u1ec7n l\u00e0 : 10 x 1 = 10 (km) \u0110\u00e1p s\u1ed1 : 10 km. B\u00e0i 159 : Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 75 cm2. Tr\u00ean BC l\u1ea5y M sao cho BM = 2\/3 BC. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABM. Nh\u1eadn x\u00e9t : Ta th\u1ea5y tam gi\u00e1c ABM v\u00e0 tam gi\u00e1c ABC c\u00f3 c\u00f9ng chi\u1ec1u cao l\u00e0 AH ; hai \u0111\u00e1y t\u01b0\u01a1ng \u1ee9ng l\u00e0 BM v\u00e0 BC. Do \u0111\u00f3 \u0111\u00e1y v\u00e0 di\u1ec7n t\u00edch l\u00e0 hai \u0111\u1ea1i l\u01b0\u1ee3ng t\u1ec9 l\u1ec7 thu\u1eadn v\u1edbi nhau. \u1edf \u0111\u00e2y t\u1ec9 s\u1ed1 v\u1ec1 hai \u0111\u00e1y l\u00e0 : BM\/BC = 2\/3. V\u1eady t\u1ec9 s\u1ed1 v\u1ec1 di\u1ec7n t\u00edch c\u1ee7a hai tam gi\u00e1cABM v\u00e0 ABC l\u00e0 2\/3. V\u00ec di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 75 cm2, n\u00ean di\u1ec7n t\u00edch tam gi\u00e1c ABM 90","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i l\u00e0 : 75 : 3 x 2 = 50 (cm2). \u0110\u00e1p s\u1ed1 : 50 cm2 B\u00e0i 160: C\u00f4 gi\u00e1o x\u1ebfp ch\u1ed7 ng\u1ed3i cho h\u1ecdc sinh l\u1edbp 4A. N\u1ebfu x\u1ebfp m\u1ed7i b\u00e0n 4 b\u1ea1n th\u00ec thi\u1ebfu m\u1ed9t b\u00e0n. N\u1ebfu x\u1ebfp m\u1ed7i b\u00e0n 5 b\u1ea1n th\u00ec th\u1eeba m\u1ed9t b\u00e0n. H\u1ecfi l\u1edbp \u0111\u00f3 c\u00f3 bao nhi\u00eau b\u00e0n, bao nhi\u00eau h\u1ecdc sinh ? Nh\u1eadn x\u00e9t : S\u1ed1 h\u1ecdc sinh kh\u00f4ng \u0111\u1ed5i n\u00ean s\u1ed1 b\u00e0n v\u00e0 s\u1ed1 h\u1ecdc sinh x\u1ebfp \u1edf m\u1ed7i b\u00e0n l\u00e0 hai \u0111\u1ea1i l\u01b0\u1ee3ng t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi nhau. S\u1ed1 b\u00e0n c\u1ea7n c\u00f3 \u0111\u1ec3 x\u1ebfp 4 b\u1ea1n 1 b\u00e0n nhi\u1ec1u h\u01a1n s\u1ed1 b\u00e0n c\u1ea7n c\u00f3 \u0111\u1ec3 x\u1ebfp 5 b\u1ea1n 1 b\u00e0n l\u00e0 : 1 + 1 = 2 (b\u00e0n) \u1ede \u0111\u00e2y t\u1ec9 s\u1ed1 gi\u1eefa s\u1ed1 b\u1ea1n x\u1ebfp \u1edf m\u1ed9t b\u00e0n 4 b\u1ea1n v\u00e0 m\u1ed9t b\u00e0n 5 b\u1ea1n l\u00e0. Do \u0111\u00f3 t\u1ec9 s\u1ed1 gi\u1eefa s\u1ed1 b\u00e0n khi x\u1ebfp m\u1ed9t b\u00e0n 4 b\u1ea1n v\u00e0 m\u1ed9t b\u00e0n 5 b\u1ea1n l\u00e0 . V\u1eady ta c\u00f3 s\u01a1 \u0111\u1ed3 : S\u1ed1 b\u00e0n c\u1ea7n \u0111\u1ee7 \u0111\u1ec3 x\u1ebfp 4 b\u1ea1n m\u1ed9t b\u00e0n l\u00e0 : 2 : (5 - 4) x 5 = 10 (b\u00e0n) S\u1ed1 b\u00e0n l\u1edbp 4A l\u00e0 : 10 - 1 = 9 (b\u00e0n) S\u1ed1 h\u1ecdc sinh l\u1edbp 4A l\u00e0 : 4 x 9 + 4 = 40 (h\u1ecdc sinh) \u0110\u00e1p s\u1ed1 : 9 b\u00e0n ; 40 h\u1ecdc sinh. B\u00e0i 161: \u201cB\u1ea1n Y\u1ebfn c\u00f3 m\u1ed9t b\u00f3 hoa h\u1ed3ng \u0111em t\u1eb7ng c\u00e1c b\u1ea1n c\u00f9ng l\u1edbp. L\u1ea7n \u0111\u1ea7u Y\u1ebfn t\u1eb7ng m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng v\u00e0 th\u00eam 1 b\u00f4ng. L\u1ea7n th\u1ee9 hai Y\u1ebfn t\u1eb7ng m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i v\u00e0 th\u00eam 2 b\u00f4ng. L\u1ea7n th\u1ee9 ba Y\u1ebfn t\u1eb7ng m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i v\u00e0 th\u00eam 3 b\u00f4ng. Cu\u1ed1i c\u00f9ng Y\u1ebfn c\u00f2n l\u1ea1i 1 b\u00f4ng h\u1ed3ng d\u00e0nh cho m\u00ecnh. H\u1ecfi Y\u1ebfn \u0111\u00e3 t\u1eb7ng bao nhi\u00eau b\u00f4ng h\u1ed3ng ?\u201d B\u00e0i gi\u1ea3i *C\u00e1ch 1 : Ta c\u00f3 s\u01a1 \u0111\u1ed3 v\u1ec1 s\u1ed1 c\u00e1c b\u00f4ng h\u1ed3ng : S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi Y\u1ebfn t\u1eb7ng l\u1ea7n th\u1ee9 hai l\u00e0 : (1 + 3) x 2 = 8 (b\u00f4ng) S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi Y\u1ebfn t\u1eb7ng l\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0 : ( 8 + 2) x 2 = 20 (b\u00f4ng) S\u1ed1 b\u00f4ng h\u1ed3ng l\u00fac \u0111\u1ea7u Y\u1ebfn c\u00f3 l\u00e0 : (20 + 1) x 2 = 42 (b\u00f4ng) 91","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i S\u1ed1 b\u00f4ng h\u1ed3ng Y\u1ebfn \u0111\u00e3 t\u1eb7ng c\u00e1c b\u1ea1n l\u00e0 : 42 - 1 = 41 (b\u00f4ng) \u0110\u00e1p s\u1ed1 : 41 b\u00f4ng h\u1ed3ng. *C\u00e1ch 2 : G\u1ecdi s\u1ed1 b\u00f4ng h\u1ed3ng l\u00fac \u0111\u1ea7u Y\u1ebfn c\u00f3 l\u00e0 a. S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi Y\u1ebfn cho b\u1ea1n l\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0 : a : 2 - 1 (b\u00f4ng h\u1ed3ng) S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau Y\u1ebfn cho b\u1ea1n l\u1ea7n th\u1ee9 hai l\u00e0 : (a : 2 - 1) : 2 - 2 (b\u00f4ng h\u1ed3ng) S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi Y\u1ebfn cho b\u1ea1n l\u1ea7n th\u1ee9 ba l\u00e0 : ((a : 2 - 1) : 2 - 2) : 2 - 3 (b\u00f4ng h\u1ed3ng) Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 : ((a : 2 - 1) : 2 - 2) : 2 - 3 = 1 (b\u00f4ng h\u1ed3ng) ((a : 2 - 1) : 2 - 2) : 2 = 1 + 3 (b\u00f4ng h\u1ed3ng) ((a : 2 - 1) : 2 - 2) : 2 = 4 (b\u00f4ng h\u1ed3ng) (a : 2 - 1) : 2 - 2 = 4 x 2 (b\u00f4ng h\u1ed3ng) (a : 2 - 1) : 2 - 2 = 8 (b\u00f4ng h\u1ed3ng) (a : 2 - 1) : 2 = 8 + 2 (b\u00f4ng h\u1ed3ng) (a : 2 - 1) : 2 = 10 (b\u00f4ng h\u1ed3ng) a : 2 - 1 = 10 x 2 (b\u00f4ng h\u1ed3ng) a : 2 - 1 = 20 (b\u00f4ng h\u1ed3ng) a : 2 = 20 + 1 (b\u00f4ng h\u1ed3ng) a : 2 = 21 (b\u00f4ng h\u1ed3ng) a = 21 x 2 (b\u00f4ng h\u1ed3ng) a = 42 (b\u00f4ng h\u1ed3ng) S\u1ed1 b\u00f4ng h\u1ed3ng m\u00e0 Y\u1ebfn \u0111\u00e3 t\u1eb7ng c\u00e1c b\u1ea1n l\u00e0 : 42 - 1 = 41 (b\u00f4ng h\u1ed3ng) \u0110\u00e1p s\u1ed1 : 41 b\u00f4ng h\u1ed3ng. *C\u00e1ch 3 : Bi\u1ec3u th\u1ecb : A l\u00e0 s\u1ed1 b\u00f4ng h\u1ed3ng l\u00fac \u0111\u1ea7u Y\u1ebfn c\u00f3. B l\u00e0 s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi cho l\u1ea7n th\u1ee9 nh\u1ea5t. C l\u00e0 s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi cho l\u1ea7n th\u1ee9 hai. Ta c\u00f3 l\u01b0u \u0111\u1ed3 sau : 92","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi Y\u1ebfn cho l\u1ea7n th\u1ee9 2 l\u00e0 : (1 + 3) x 2 = 8 (b\u00f4ng h\u1ed3ng) S\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi Y\u1ebfn cho l\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0 : (8 + 2) x 2 = 20 (b\u00f4ng h\u1ed3ng) S\u1ed1 b\u00f4ng h\u1ed3ng l\u00fac \u0111\u1ea7u Y\u1ebfn c\u00f3 l\u00e0 : (20 + 1) x 2 = 42 (b\u00f4ng h\u1ed3ng) S\u1ed1 b\u00f4ng h\u1ed3ng Y\u1ebfn t\u1eb7ng c\u00e1c b\u1ea1n l\u00e0 : 42 - 1 = 41 (b\u00f4ng h\u1ed3ng) \u0110\u00e1p s\u1ed1 : 41 b\u00f4ng h\u1ed3ng. Nh\u1eadn x\u00e9t : C\u00e1ch gi\u1ea3i 1 l\u00e0 c\u00e1ch gi\u1ea3i th\u00f4ng th\u01b0\u1eddng m\u00e0 h\u1ecdc sinh ti\u1ec3u h\u1ecdc l\u1ef1a ch\u1ecdn \u0111\u1ec3 gi\u1ea3i. M\u1ee5c \u0111\u00edch c\u1ee7a vi\u1ec7c v\u1ebd s\u01a1 \u0111\u1ed3 nh\u1eb1m gi\u00fap h\u1ecdc sinh d\u1ec5 d\u00e0ng nh\u00ecn th\u1ea5y c\u00e1c m\u1ed1i li\u00ean h\u1ec7 trong b\u00e0i to\u00e1n. Tuy nhi\u00ean, \u0111\u1ed1i v\u1edbi c\u00e1c em h\u1ecdc sinh kh\u00e1 gi\u1ecfi th\u00ec vi\u1ec7c v\u1ebd s\u01a1 \u0111\u1ed3 l\u00e0 kh\u00f4ng c\u1ea7n thi\u1ebft khi c\u00e1c em \u0111\u00e3 th\u00e0nh th\u1ea1o. \u0110\u1ed1i v\u1edbi c\u00e1ch gi\u1ea3i 2, nhi\u1ec1u ng\u01b0\u1eddi cho r\u1eb1ng, khi gi\u1ea3i b\u1eb1ng c\u00e1ch n\u00e0y l\u00e0 kh\u00f4ng v\u1eeba s\u1ee9c \u0111\u1ed1i v\u1edbi h\u1ecdc sinh ti\u1ec3u h\u1ecdc. \u0110i\u1ec1u \u0111\u00f3 kh\u00f4ng \u0111\u00fang, v\u00ec th\u1ef1c ra h\u1ecdc sinh ch\u1ec9 c\u1ea7n v\u1eadn d\u1ee5ng c\u00e1c ki\u1ebfn th\u1ee9c c\u01a1 b\u1ea3n \u0111\u00e3 h\u1ecdc trong ch\u01b0\u01a1ng tr\u00ecnh ti\u1ec3u h\u1ecdc l\u00e0 t\u00ecm th\u00e0nh ph\u1ea7n ch\u01b0a bi\u1ebft c\u1ee7a ph\u00e9p t\u00ednh v\u00e0 c\u0103n c\u1ee9 v\u00e0o d\u1eef ki\u1ec7n \u0111\u00e3 cho \u0111\u1ec3 \u0111\u01b0a ra l\u1eddi gi\u1ea3i. V\u00ed d\u1ee5 \u1edf b\u01b0\u1edbc 1, h\u1ecdc sinh th\u1ef1c hi\u1ec7n t\u00ecm s\u1ed1 b\u1ecb tr\u1eeb khi bi\u1ebft s\u1ed1 tr\u1eeb v\u00e0 hi\u1ec7u, b\u01b0\u1edbc 2 h\u1ecdc sinh th\u1ef1c hi\u1ec7n t\u00ecm s\u1ed1 b\u1ecb chia khi bi\u1ebft th\u01b0\u01a1ng v\u00e0 s\u1ed1 chia v.v... \u1ede c\u00e1ch gi\u1ea3i 3, ch\u00fang ta th\u1ea5y khi cho \u0111i m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng Y\u1ebfn c\u00f3 th\u00ec c\u00f2n l\u1ea1i m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng. Sau \u0111\u00f3 l\u1ea1i cho th\u00eam 1 b\u00f4ng h\u1ed3ng n\u1eefa, ngh\u0129a l\u00e0 s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi cho l\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0 m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng l\u00fac \u0111\u1ea7u b\u1edbt \u0111i 1 b\u00f4ng. T\u01b0\u01a1ng t\u1ef1 nh\u01b0 v\u1eady s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi cho l\u1ea7n th\u1ee9 hai ch\u00ednh l\u00e0 m\u1ed9t n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng sau khi cho l\u1ea7n th\u1ee9 nh\u1ea5t r\u1ed3i b\u1edbt \u0111i 2 b\u00f4ng. 1 b\u00f4ng h\u1ed3ng d\u00e0nh cho Y\u1ebfn ch\u00ednh l\u00e0 1 n\u1eeda s\u1ed1 b\u00f4ng h\u1ed3ng c\u00f2n l\u1ea1i sau khi cho l\u1ea7n th\u1ee9 hai b\u1edbt \u0111i 3 b\u00f4ng. T\u1edbi \u0111\u00e2y, mu\u1ed1n t\u00ecm C ta l\u1ea5y (1 + 3) x 2. T\u01b0\u01a1ng t\u1ef1, ta t\u00ecm \u0111\u01b0\u1ee3c s\u1ed1 b\u00f4ng h\u1ed3ng l\u00fac \u0111\u1ea7u Y\u1ebfn c\u00f3 (A). B\u00e0i 162: H\u00e3y cho bi\u1ebft 2\/7 c\u1ee7a 75 l\u00e0 bao nhi\u00eau? Gi\u1ea3i :Ta c\u00f3 s\u01a1 \u0111\u1ed3: 2\/5 c\u1ee7a 75 l\u00e0 : 75 : 5 x 2 = 30 hay 75 x 2\/5 = 30. 93","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 163 : T\u00ecm 3\/4 c\u1ee7a 5\/6 Gi\u1ea3i : Ta c\u00f3 s\u01a1 \u0111\u1ed3 : 3\/4 c\u1ee7a 5\/6 l\u00e0 : 5\/6 : 4 x 3 = 5\/8 hay 5\/6 x 3\/4 = 5\/8. B\u00e0i 164 : Bi\u1ebft 2\/3 c\u1ee7a m\u1ed9t s\u1ed1 l\u00e0 20. H\u00e3y t\u00ecm s\u1ed1 \u0111\u00f3. Gi\u1ea3i : Ta c\u00f3 s\u01a1 \u0111\u1ed3 : S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 : 20 : 2 x 3 = 30 hay 20 : 2\/3 = 30. B\u00e0i 165: Bi\u1ebft 8\/9 c\u1ee7a m\u1ed9t s\u1ed1 l\u00e0 2\/3. T\u00ecm s\u1ed1 \u0111\u00f3. Gi\u1ea3i : Ta c\u00f3 s\u01a1 \u0111\u1ed3 : S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 : 2\/3 : 8 x 9 = 3\/4 hay 2\/3 : 8\/9 = 3\/4. B\u00e0i 166 : C\u00f3 t\u1ea5t c\u1ea3 720 kg g\u1ea1o g\u1ed3m 3 lo\u1ea1i : 1\/6 s\u1ed1 g\u1ea1o l\u00e0 g\u1ea1o th\u01a1m, 3\/8 s\u1ed1 g\u1ea1o l\u00e0 g\u1ea1o n\u1ebfp, c\u00f2n l\u1ea1i l\u00e0 g\u1ea1o t\u1ebb. T\u00ednh s\u1ed1 kg g\u1ea1o m\u1ed7i lo\u1ea1i. Gi\u1ea3i : 1\/6 s\u1ed1 g\u1ea1o l\u00e0 g\u1ea1o th\u01a1m, n\u00ean kh\u1ed1i l\u01b0\u1ee3ng g\u1ea1o th\u01a1m l\u00e0 :720 x 1\/6 = 120 (kg) 3\/8 s\u1ed1 g\u1ea1o l\u00e0 g\u1ea1o n\u1ebfp, n\u00ean kh\u1ed1i l\u01b0\u1ee3ng g\u1ea1o n\u1ebfp l\u00e0 : 720 x 3\/8 = 270 (kg) Kh\u1ed1i l\u01b0\u1ee3ng g\u1ea1o t\u1ebb l\u00e0 : 720 - (120 + 270) = 330 (kg). \u0110\u00e1p s\u1ed1 : 120 kg, 270 kg, 330 kg B\u00e0i 167 : M\u1ed9t ng\u01b0\u1eddi b\u00e1n cam,bu\u1ed5i s\u00e1ng b\u00e1n \u0111\u01b0\u1ee3c 3\/5 s\u1ed1 cam mang \u0111i, bu\u1ed5i chi\u1ec1u b\u00e1n th\u00eam \u0111\u01b0\u1ee3c 52 qu\u1ea3 v\u00e0 s\u1ed1 cam c\u00f2n l\u1ea1i \u0111\u00fang b\u1eb1ng 1\/8 s\u1ed1 cam \u0111\u00e3 b\u00e1n. T\u00ednh s\u1ed1 qu\u1ea3 cam m\u00e0 ng\u01b0\u1eddi \u0111\u00f3 \u0111\u00e3 mang \u0111i b\u00e1n. Gi\u1ea3i : S\u1ed1 cam c\u00f2n l\u1ea1i b\u1eb1ng 1\/8 s\u1ed1 cam \u0111\u00e3 b\u00e1n, hay \u0111\u00fang b\u1eb1ng 1\/9 s\u1ed1 cam m\u00e0 ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i b\u00e1n. S\u1ed1 cam bu\u1ed5i chi\u1ec1u ng\u01b0\u1eddi \u0111\u00f3 b\u00e1n ch\u00ednh l\u00e0 1 - (3\/5 + 1\/9) = 13\/45 s\u1ed1 cam mang \u0111i. S\u1ed1 cam bu\u1ed5i chi\u1ec1u ng\u01b0\u1eddi \u0111\u00f3 b\u00e1n l\u00e0 52 qu\u1ea3 n\u00ean s\u1ed1 cam ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i ch\u1ee3 l\u00e0 : 52 : 13\/45 = 180 (qu\u1ea3). 94","500 B\u00c0I TO\u00c1N CH\u1eccN L\u1eccC L\u1edaP 5 v\u00e0 l\u1eddi gi\u1ea3i B\u00e0i 168 : Ba ng\u01b0\u1eddi chia nhau m\u1ed9t s\u1ed1 ti\u1ec1n. Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t (NT1) l\u1ea5y 1\/4 s\u1ed1 ti\u1ec1n r\u1ed3i b\u1edbt l\u1ea1i 50000 \u0111\u1ed3ng, ng\u01b0\u1eddi th\u1ee9 hai (NT2) l\u1ea5y 3\/5 s\u1ed1 ti\u1ec1n c\u00f2n l\u1ea1i r\u1ed3i b\u1edbt l\u1ea1i 40000 \u0111\u1ed3ng. Ng\u01b0\u1eddi th\u1ee9 ba l\u1ea5y 240000 \u0111\u1ed3ng th\u00ec v\u1eeba h\u1ebft. S\u1ed1 ti\u1ec1n \u0111\u01b0\u1ee3c \u0111em chia l\u00e0 bao nhi\u00eau ? Gi\u1ea3i : Ta c\u00f3 s\u01a1 \u0111\u1ed3 sau : 2\/5 s\u1ed1 ti\u1ec1n c\u00f2n l\u1ea1i sau khi ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u1ea5y l\u00e0 : 240000 - 40000 = 200000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n c\u00f2n l\u1ea1i sau khi ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u1ea5y l\u00e0 : 200000 : 2\/5 = 500000 (\u0111\u1ed3ng). 3\/4 t\u1ed5ng s\u1ed1 ti\u1ec1n l\u00e0 : 500000 - 50000 = 450000 (\u0111\u1ed3ng) T\u1ed5ng s\u1ed1 ti\u1ec1n l\u00e0 : 450000 : 3\/4 = 600000(\u0111\u1ed3ng) \u0110\u00e1p s\u1ed1 : 600000 \u0111\u1ed3ng 95"]