100-de-thi-hoc-sinh-gioi-mon-toan-lop-5

["C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i gi\u1ea3i : V\u1eady ta s\u1eafp x\u1ebfp \u0111\u01b0\u1ee3c c\u00e1c ph\u00e2n s\u1ed1 nh\u01b0 sau : T\u1ed5ng hai ph\u00e2n s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t l\u00e0 : T\u1ed5ng hai ph\u00e2n s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t l\u00e0 : Do \u0111\u00f3 t\u1ed5ng b\u1ed1n ph\u00e2n s\u1ed1 m\u00e0 Th\u0103ng v\u00e0 Long \u0111\u00e3 ch\u1ecdn l\u00e0 : B\u00e0i 6 : T\u00ecm c\u00e1c ch\u1eef s\u1ed1 a v\u00e0 b th\u1ecfa m\u00e3n : B\u00e0i gi\u1ea3i : V\u00ec 1\/3 l\u00e0 ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n n\u00ean a chia h\u1ebft cho 3 ho\u1eb7c b chia h\u1ebft cho 3. 51","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 Gi\u1ea3 s\u1eed a chia h\u1ebft cho 3, v\u00ec 1\/a < 1\/3 n\u00ean a > 3 m\u00e0 a < 10 do \u0111\u00f3 a = 6 ; 9. V\u1eady a = b = 6. B\u00e0i 7 : Vi\u1ebft li\u00ean ti\u1ebfp c\u00e1c s\u1ed1 t\u1eeb tr\u00e1i sang ph\u1ea3i theo c\u00e1ch sau : S\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 1, s\u1ed1 th\u1ee9 hai l\u00e0 2, s\u1ed1 th\u1ee9 ba l\u00e0 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng c\u1ee7a t\u1ed5ng s\u1ed1 th\u1ee9 nh\u1ea5t v\u00e0 s\u1ed1 th\u1ee9 hai, s\u1ed1 th\u1ee9 t\u01b0 l\u00e0 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng c\u1ee7a t\u1ed5ng s\u1ed1 th\u1ee9 hai v\u00e0 s\u1ed1 th\u1ee9 ba. C\u1ee9 ti\u1ebfp t\u1ee5c nh\u01b0 th\u1ebf ta \u0111\u01b0\u1ee3c d\u00e3y c\u00e1c s\u1ed1 nh\u01b0 sau : 1235831459437...... Trong d\u00e3y tr\u00ean c\u00f3 xu\u1ea5t hi\u1ec7n s\u1ed1 2005 hay kh\u00f4ng ? B\u00e0i gi\u1ea3i : Gi\u1ea3 s\u1eed trong s\u1ed1 t\u1ea1o b\u1edfi c\u00e1ch vi\u1ebft nh\u01b0 tr\u00ean c\u00f3 xu\u1ea5t hi\u1ec7n nh\u00f3m ch\u1eef 2005 th\u00ec ta c\u00f3 : 2 + 0 l\u00e0 s\u1ed1 c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 0 (v\u00f4 l\u00ed). V\u1eady trong d\u00e3y tr\u00ean kh\u00f4ng th\u1ec3 xu\u1ea5t hi\u1ec7n s\u1ed1 2005. B\u00e0i 8 : C\u00f3 5 \u0111\u1ed9i tham gia d\u1ef1 thi to\u00e1n \u0111\u1ed3ng \u0111\u1ed9i. T\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u1ea3 5 \u0111\u1ed9i l\u00e0 144 \u0111i\u1ec3m v\u00e0 th\u1eadt th\u00fa v\u1ecb l\u00e0 c\u1ea3 5 \u0111\u1ed9i \u0111\u1ec1u \u0111\u1ea1t m\u1ed9t trong ba gi\u1ea3i : nh\u1ea5t (30 \u0111i\u1ec3m) ; nh\u00ec (29 \u0111i\u1ec3m) ; ba (28 \u0111i\u1ec3m). Ch\u1ee9ng minh s\u1ed1 \u0111\u1ed9i \u0111\u1ea1t gi\u1ea3i ba h\u01a1n s\u1ed1 \u0111\u1ed9i \u0111\u1ea1t gi\u1ea3i nh\u1ea5t \u0111\u00fang m\u1ed9t \u0111\u1ed9i. B\u00e0i gi\u1ea3i : Ta th\u1ea5y trung b\u00ecnh c\u1ed9ng \u0111i\u1ec3m c\u1ee7a m\u1ed9t \u0111\u1ed9i gi\u1ea3i nh\u1ea5t v\u00e0 m\u1ed9t \u0111\u1ed9i gi\u1ea3i ba ch\u00ednh l\u00e0 s\u1ed1 \u0111i\u1ec3m c\u1ee7a m\u1ed9t \u0111\u1ed9i gi\u1ea3i nh\u00ec. N\u1ebfu s\u1ed1 \u0111\u1ed9i \u0111\u1ea1t gi\u1ea3i nh\u1ea5t b\u1eb1ng s\u1ed1 \u0111\u1ed9i \u0111\u1ea1t gi\u1ea3i ba th\u00ec t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u1ea3 5 \u0111\u1ed9i l\u00e0 : 29 x 5 = 145 (\u0111i\u1ec3m) > 144 \u0111i\u1ec3m, kh\u00f4ng th\u1ecfa m\u00e3n. N\u1ebfu s\u1ed1 \u0111\u1ed9i gi\u1ea3i nh\u1ea5t nhi\u1ec1u h\u01a1n s\u1ed1 \u0111\u1ed9i gi\u1ea3i ba th\u00ec t\u1ed5ng \u0111i\u1ec3m 5 \u0111\u1ed9i l\u1edbn h\u01a1n 145, c\u0169ng kh\u00f4ng th\u1ecfa m\u00e3n. Do \u0111\u00f3 s\u1ed1 \u0111\u1ed9i gi\u1ea3i nh\u1ea5t ph\u1ea3i \u00edt h\u01a1n s\u1ed1 \u0111\u1ed9i gi\u1ea3i ba. Khi \u0111\u00f3 ta x\u1ebfp m\u1ed9t \u0111\u1ed9i gi\u1ea3i nh\u1ea5t v\u00e0 m\u1ed9t \u0111\u1ed9i gi\u1ea3i ba l\u00e0m th\u00e0nh m\u1ed9t c\u1eb7p th\u00ec c\u1eb7p n\u00e0y s\u1ebd c\u00f3 t\u1ed5ng s\u1ed1 \u0111i\u1ec3m b\u1eb1ng hai \u0111\u1ed9i gi\u1ea3i nh\u00ec. S\u1ed1 \u0111\u1ed9i gi\u1ea3i ba th\u1eeba ra (kh\u00f4ng \u0111\u01b0\u1ee3c x\u1ebfp c\u1eb7p v\u1edbi m\u1ed9t \u0111\u1ed9i gi\u1ea3i nh\u1ea5t) ch\u00ednh l\u00e0 s\u1ed1 \u0111i\u1ec3m m\u00e0 t\u1ed5ng \u0111i\u1ec3m c\u1ee7a 5 \u0111\u1ed9i nh\u1ecf h\u01a1n 145. V\u00ec v\u1eady s\u1ed1 \u0111\u1ed9i gi\u1ea3i ba nhi\u1ec1u h\u01a1n s\u1ed1 \u0111\u1ed9i gi\u1ea3i nh\u1ea5t bao nhi\u00eau th\u00ec t\u1ed5ng \u0111i\u1ec3m c\u1ee7a 5 \u0111\u1ed9i s\u1ebd nh\u1ecf h\u01a1n 52","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 145 b\u1ea5y nhi\u00eau. V\u00ec t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u1ea3 5 \u0111\u1ed9i l\u00e0 144 \u0111i\u1ec3m n\u00ean s\u1ed1 \u0111\u1ed9i gi\u1ea3i ba nhi\u1ec1u h\u01a1n s\u1ed1 \u0111\u1ed9i gi\u1ea3i nh\u1ea5t l\u00e0 145 - 144 = 1. B\u00e0i 9 : 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 10 : T\u00edch sau \u0111\u00e2y c\u00f3 t\u1eadn c\u00f9ng b\u1eb1ng ch\u1eef s\u1ed1 n\u00e0o ? B\u00e0i gi\u1ea3i : T\u00edch c\u1ee7a b\u1ed1n th\u1eeba s\u1ed1 2 l\u00e0 2 x 2 x 2 x 2 = 16 v\u00e0 2003 : 4 = 500 (d\u01b0 3) n\u00ean ta c\u00f3 th\u1ec3 vi\u1ebft t\u00edch c\u1ee7a 2003 th\u1eeba s\u1ed1 2 d\u01b0\u1edbi d\u1ea1ng t\u00edch c\u1ee7a 500 nh\u00f3m (m\u1ed7i nh\u00f3m l\u00e0 t\u00edch c\u1ee7a b\u1ed1n th\u1eeba s\u1ed1 2) v\u00e0 t\u00edch c\u1ee7a ba th\u1eeba s\u1ed1 2 c\u00f2n l\u1ea1i. V\u00ec t\u00edch c\u1ee7a c\u00e1c th\u1eeba s\u1ed1 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 6 c\u0169ng l\u00e0 s\u1ed1 c\u00f3 t\u1eadn c\u00f9ng b\u1eb1ng 6 n\u00ean t\u00edch c\u1ee7a 500 nh\u00f3m tr\u00ean c\u00f3 t\u1eadn c\u00f9ng l\u00e0 6. Do 2 x 2 x 2 = 8 n\u00ean khi nh\u00e2n s\u1ed1 c\u00f3 t\u1eadn c\u00f9ng b\u1eb1ng 6 v\u1edbi 8 th\u00ec ta \u0111\u01b0\u1ee3c s\u1ed1 c\u00f3 t\u1eadn c\u00f9ng b\u1eb1ng 8 (v\u00ec 6 x 8 = 48). V\u1eady t\u00edch c\u1ee7a 2003 th\u1eeba s\u1ed1 2 s\u1ebd l\u00e0 s\u1ed1 c\u00f3 t\u1eadn c\u00f9ng b\u1eb1ng 8. 53","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i 11 : M\u1ed9t ng\u01b0\u1eddi mang cam \u0111i \u0111\u1ed5i l\u1ea5y t\u00e1o v\u00e0 l\u00ea. C\u1ee9 9 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 t\u00e1o v\u00e0 1 qu\u1ea3 l\u00ea, 5 qu\u1ea3 t\u00e1o th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea. N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ed5i h\u1ebft s\u1ed1 cam mang \u0111i th\u00ec \u0111\u01b0\u1ee3c 17 qu\u1ea3 t\u00e1o v\u00e0 13 qu\u1ea3 l\u00ea. H\u1ecfi ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i bao nhi\u00eau qu\u1ea3 cam ? B\u00e0i gi\u1ea3i : 9 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 t\u00e1o v\u00e0 1 qu\u1ea3 l\u00ea n\u00ean 18 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 4 qu\u1ea3 t\u00e1o v\u00e0 2 qu\u1ea3 l\u00ea. V\u00ec 5 qu\u1ea3 t\u00e1o \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea n\u00ean 18 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c : 4 + 5 = 9 (qu\u1ea3 t\u00e1o). Do \u0111\u00f3 2 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 qu\u1ea3 t\u00e1o. C\u1ee9 5 qu\u1ea3 t\u00e1o \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea n\u00ean 10 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea. V\u1eady 5 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 qu\u1ea3 l\u00ea. S\u1ed1 cam ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i \u0111\u1ec3 \u0111\u1ed5i \u0111\u01b0\u1ee3c 17 qu\u1ea3 t\u00e1o v\u00e0 13 qu\u1ea3 l\u00ea l\u00e0 : 2 x 17 + 5 x 13 = 99 (qu\u1ea3). B\u00e0i 12 : T\u00ecm m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean sao cho khi l\u1ea5y 1\/3 s\u1ed1 \u0111\u00f3 chia cho 1\/17 s\u1ed1 \u0111\u00f3 th\u00ec c\u00f3 d\u01b0 l\u00e0 100. B\u00e0i gi\u1ea3i : V\u00ec 17 x 3 = 51 n\u00ean \u0111\u1ec3 d\u1ec5 l\u00ed lu\u1eadn, ta gi\u1ea3 s\u1eed s\u1ed1 t\u1ef1 nhi\u00ean c\u1ea7n t\u00ecm \u0111\u01b0\u1ee3c chia ra th\u00e0nh 51 ph\u1ea7n b\u1eb1ng nhau. Khi \u1ea5y 1\/3 s\u1ed1 \u0111\u00f3 l\u00e0 51 : 3 = 17 (ph\u1ea7n) ; 1\/17 s\u1ed1 \u0111\u00f3 l\u00e0 51 : 17 = 3 (ph\u1ea7n). V\u00ec 17 : 3 = 5 (d\u01b0 2) n\u00ean 2 ph\u1ea7n c\u1ee7a s\u1ed1 \u0111\u00f3 c\u00f3 gi\u00e1 tr\u1ecb l\u00e0 100 suy ra s\u1ed1 \u0111\u00f3 l\u00e0 : 100 : 2 x 51 = 2550. B\u00e0i 13 : Tu\u1ed5i c\u1ee7a con hi\u1ec7n nay b\u1eb1ng 1\/2 hi\u1ec7u tu\u1ed5i c\u1ee7a b\u1ed1 v\u00e0 tu\u1ed5i con. B\u1ed1n n\u0103m tr\u01b0\u1edbc, tu\u1ed5i con b\u1eb1ng 1\/3 hi\u1ec7u tu\u1ed5i c\u1ee7a b\u1ed1 v\u00e0 tu\u1ed5i con. H\u1ecfi khi tu\u1ed5i con b\u1eb1ng 1\/4 hi\u1ec7u tu\u1ed5i c\u1ee7a b\u1ed1 v\u00e0 tu\u1ed5i c\u1ee7a con th\u00ec tu\u1ed5i c\u1ee7a m\u1ed7i ng\u01b0\u1eddi l\u00e0 bao nhi\u00eau ? B\u00e0i gi\u1ea3i : Hi\u1ec7u s\u1ed1 tu\u1ed5i c\u1ee7a b\u1ed1 v\u00e0 con kh\u00f4ng \u0111\u1ed5i. Tr\u01b0\u1edbc \u0111\u00e2y 4 n\u0103m tu\u1ed5i con b\u1eb1ng 1\/3 hi\u1ec7u n\u00e0y, do \u0111\u00f3 4 n\u0103m ch\u00ednh l\u00e0 : 1\/2 - 1\/3 = 1\/6 (hi\u1ec7u s\u1ed1 tu\u1ed5i c\u1ee7a b\u1ed1 v\u00e0 con). S\u1ed1 tu\u1ed5i b\u1ed1 h\u01a1n con l\u00e0 : 4 : 1\/6 = 24 (tu\u1ed5i). Khi tu\u1ed5i con b\u1eb1ng 1\/4 hi\u1ec7u s\u1ed1 tu\u1ed5i c\u1ee7a b\u1ed1 v\u00e0 con th\u00ec tu\u1ed5i con l\u00e0 : 24 x 1\/4 = 6 (tu\u1ed5i). L\u00fac \u0111\u00f3 tu\u1ed5i b\u1ed1 l\u00e0 : 6 + 24 = 30 (tu\u1ed5i). B\u00e0i 14 : Hoa c\u00f3 m\u1ed9t s\u1ee3i d\u00e2y d\u00e0i 16 m\u00e9t. B\u00e2y gi\u1edd Hoa c\u1ea7n c\u1eaft \u0111o\u1ea1n d\u00e2y \u0111\u00f3 \u0111\u1ec3 c\u00f3 \u0111o\u1ea1n d\u00e2y d\u00e0i 10 m\u00e9t m\u00e0 trong tay Hoa ch\u1ec9 c\u00f3 m\u1ed9t c\u00e1i k\u00e9o. C\u00e1c b\u1ea1n c\u00f3 bi\u1ebft Hoa c\u1eaft th\u1ebf n\u00e0o kh\u00f4ng ? B\u00e0i gi\u1ea3i : Xin n\u00eau 2 c\u00e1ch c\u1eaft nh\u01b0 sau : C\u00e1ch 1 : G\u1eadp \u0111\u00f4i s\u1ee3i d\u00e2y li\u00ean ti\u1ebfp 3 l\u1ea7n, khi \u0111\u00f3 s\u1ee3i d\u00e2y s\u1ebd \u0111\u01b0\u1ee3c chia th\u00e0nh 8 ph\u1ea7n b\u1eb1ng nhau. 54","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 \u0110\u1ed9 d\u00e0i m\u1ed7i ph\u1ea7n chia l\u00e0 : 16 : 8 = 2 (m) C\u1eaft \u0111i 3 ph\u1ea7n b\u1eb1ng nhau th\u00ec c\u00f2n l\u1ea1i 5 ph\u1ea7n. Khi \u0111\u00f3 \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n d\u00e2y c\u00f2n l\u1ea1i l\u00e0 : 2 x 5 = 10 (m) C\u00e1ch 2 : G\u1eadp \u0111\u00f4i s\u1ee3i d\u00e2y li\u00ean ti\u1ebfp 2 l\u1ea7n, khi \u0111\u00f3 s\u1ee3i d\u00e2y s\u1ebd \u0111\u01b0\u1ee3c chia th\u00e0nh 4 ph\u1ea7n b\u1eb1ng nhau. \u0110\u1ed9 d\u00e0i m\u1ed7i ph\u1ea7n chia l\u00e0 : 16 : 4 = 4 (m) \u0110\u00e1nh d\u1ea5u m\u1ed9t ph\u1ea7n chia \u1edf m\u1ed9t \u0111\u1ea7u d\u00e2y, ph\u1ea7n \u0111o\u1ea1n d\u00e2y c\u00f2n l\u1ea1i \u0111\u01b0\u1ee3c g\u1eadp \u0111\u00f4i l\u1ea1i, c\u1eaft \u0111i m\u1ed9t ph\u1ea7n \u1edf \u0111\u1ea7u b\u00ean kia th\u00ec \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n d\u00e2y c\u1eaft \u0111i l\u00e0 : (16 - 4) : 2 = 6 (m) Do \u0111\u00f3 \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n d\u00e2y c\u00f2n l\u1ea1i l\u00e0 : 16 - 6 = 10 (m) B\u00e0i 15 : M\u1ed9t th\u1eeda ru\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c chia th\u00e0nh 2 m\u1ea3nh, m\u1ed9t m\u1ea3nh nh\u1ecf tr\u1ed3ng rau v\u00e0 m\u1ea3nh c\u00f2n l\u1ea1i tr\u1ed3ng ng\u00f4 (h\u00ecnh v\u1ebd). Di\u1ec7n t\u00edch c\u1ee7a m\u1ea3nh tr\u1ed3ng ng\u00f4 g\u1ea5p 6 l\u1ea7n di\u1ec7n t\u00edch c\u1ee7a m\u1ea3nh tr\u1ed3ng rau. Chu vi m\u1ea3nh tr\u1ed3ng ng\u00f4 g\u1ea5p 4 l\u1ea7n chu vi m\u1ea3nh tr\u1ed3ng rau. T\u00ednh di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng ban \u0111\u1ea7u, bi\u1ebft chi\u1ec1u r\u1ed9ng c\u1ee7a n\u00f3 l\u00e0 5 m\u00e9t. B\u00e0i gi\u1ea3i : Di\u1ec7n t\u00edch m\u1ea3nh tr\u1ed3ng ng\u00f4 g\u1ea5p 6 l\u1ea7n di\u1ec7n t\u00edch m\u1ea3nh tr\u1ed3ng rau m\u00e0 hai m\u1ea3nh c\u00f3 chung m\u1ed9t c\u1ea1nh n\u00ean c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a m\u1ea3nh tr\u1ed3ng ng\u00f4 g\u1ea5p 6 l\u1ea7n c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a m\u1ea3nh tr\u1ed3ng rau. G\u1ecdi c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a m\u1ea3nh tr\u1ed3ng rau l\u00e0 a th\u00ec c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a m\u1ea3nh tr\u1ed3ng ng\u00f4 l\u00e0 a x 6. V\u00ec chu vi m\u1ea3nh tr\u1ed3ng ng\u00f4 (P1) g\u1ea5p 4 l\u1ea7n chu vi m\u1ea3nh tr\u1ed3ng rau (P2) n\u00ean n\u1eeda chu vi m\u1ea3nh tr\u1ed3ng ng\u00f4 g\u1ea5p 4 l\u1ea7n n\u1eeda chu vi m\u1ea3nh tr\u1ed3ng rau. N\u1eeda chu vi m\u1ea3nh tr\u1ed3ng ng\u00f4 h\u01a1n n\u1eeda chu vi m\u1ea3nh tr\u1ed3ng rau l\u00e0 : a x 6 + 5 - (a + 5) = 5 x a. Ta c\u00f3 s\u01a1 \u0111\u1ed3 : 55","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 \u0110\u1ed9 d\u00e0i c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a m\u1ea3nh tr\u1ed3ng rau l\u00e0 : 5 x 3 : (5 x a - 3 x a) = 7,5 (m) \u0110\u1ed9 d\u00e0i c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a m\u1ea3nh tr\u1ed3ng ng\u00f4 l\u00e0 : 7,5 x 6 = 45 (m) Di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng ban \u0111\u1ea7u l\u00e0 : (7,5 + 4,5) x 5 = 262,5 (m2) B\u00e0i 16 : T\u00f4i \u0111i b\u1ed9 t\u1eeb tr\u01b0\u1eddng v\u1ec1 nh\u00e0 v\u1edbi v\u1eadn t\u1ed1c 5 km\/gi\u1edd. V\u1ec1 \u0111\u1ebfn nh\u00e0 l\u1eadp t\u1ee9c t\u00f4i \u0111\u1ea1p xe \u0111\u1ebfn b\u01b0u \u0111i\u1ec7n v\u1edbi v\u1eadn t\u1ed1c 15 km\/gi\u1edd. Bi\u1ebft r\u1eb1ng qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 t\u1edbi tr\u01b0\u1eddng ng\u1eafn h\u01a1n qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 \u0111\u1ebfn b\u01b0u \u0111i\u1ec7n 3 km. T\u1ed5ng th\u1eddi gian t\u00f4i \u0111i t\u1eeb tr\u01b0\u1eddng v\u1ec1 nh\u00e0 v\u00e0 t\u1eeb nh\u00e0 \u0111\u1ebfn b\u01b0u \u0111i\u1ec7n l\u00e0 1 gi\u1edd 32 ph\u00fat. B\u1ea1n h\u00e3y t\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 t\u00f4i \u0111\u1ebfn tr\u01b0\u1eddng. B\u00e0i gi\u1ea3i : Th\u1eddi gian \u0111\u1ec3 \u0111i 3 km b\u1eb1ng xe \u0111\u1ea1p l\u00e0 : 3 : 15 = 0,2 (gi\u1edd) \u0110\u1ed5i : 0,2 gi\u1edd = 12 ph\u00fat. N\u1ebfu b\u1edbt 3 km qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 \u0111\u1ebfn b\u01b0u \u0111i\u1ec7n th\u00ec th\u1eddi gian \u0111i c\u1ea3 hai qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 \u0111\u1ebfn tr\u01b0\u1eddng v\u00e0 t\u1eeb nh\u00e0 \u0111\u1ebfn b\u01b0u \u0111i\u1ec7n (\u0111\u00e3 b\u1edbt 3 km) l\u00e0 : 1 gi\u1edd 32 ph\u00fat - 12 ph\u00fat = 1 gi\u1edd 20 ph\u00fat = 80 ph\u00fat. V\u1eadn t\u1ed1c \u0111i xe \u0111\u1ea1p g\u1ea5p v\u1eadn t\u1ed1c \u0111i b\u1ed9 l\u00e0 : 15 : 5 = 3 (l\u1ea7n) Khi qu\u00e3ng \u0111\u01b0\u1eddng kh\u00f4ng \u0111\u1ed5i, v\u1eadn t\u1ed1c t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi th\u1eddi gian n\u00ean th\u1eddi gian \u0111i t\u1eeb nh\u00e0 \u0111\u1ebfn tr\u01b0\u1eddng g\u1ea5p 3 l\u1ea7n th\u1eddi gian \u0111i t\u1eeb nh\u00e0 \u0111\u1ebfn th\u01b0 vi\u1ec7n (khi \u0111\u00e3 b\u1edbt \u0111i 3 km). V\u1eady : Th\u1eddi gian \u0111i t\u1eeb nh\u00e0 \u0111\u1ebfn tr\u01b0\u1eddng l\u00e0 : 80 : (1 + 3) x 3 = 60 (ph\u00fat) ; 60 ph\u00fat = 1 gi\u1edd Qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb nh\u00e0 \u0111\u1ebfn tr\u01b0\u1eddng l\u00e0 : 1 x 5 = 5 (km) B\u00e0i 17 : Cho ph\u00e2n s\u1ed1 : a) C\u00f3 th\u1ec3 x\u00f3a \u0111i trong t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 nh\u1eefng s\u1ed1 n\u00e0o m\u00e0 gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n s\u1ed1 v\u1eabn kh\u00f4ng thay \u0111\u1ed5i kh\u00f4ng ? b) N\u1ebfu ta th\u00eam s\u1ed1 2004 v\u00e0o m\u1eabu s\u1ed1 th\u00ec ph\u1ea3i th\u00eam s\u1ed1 t\u1ef1 nhi\u00ean n\u00e0o v\u00e0o t\u1eed s\u1ed1 \u0111\u1ec3 ph\u00e2n s\u1ed1 kh\u00f4ng \u0111\u1ed5i ? 56","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i gi\u1ea3i : = 45 \/ 270 = 1\/6. a) \u0110\u1ec3 gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n s\u1ed1 kh\u00f4ng \u0111\u1ed5i th\u00ec ta ph\u1ea3i x\u00f3a nh\u1eefng s\u1ed1 \u1edf m\u1eabu m\u00e0 t\u1ed5ng c\u1ee7a n\u00f3 g\u1ea5p 6 l\u1ea7n t\u1ed5ng c\u1ee7a nh\u1eefng s\u1ed1 x\u00f3a \u0111i \u1edf t\u1eed. Khi \u0111\u00f3 t\u1ed5ng c\u00e1c s\u1ed1 c\u00f2n l\u1ea1i \u1edf m\u1eabu c\u0169ng g\u1ea5p 6 l\u1ea7n t\u1ed5ng c\u00e1c s\u1ed1 c\u00f2n l\u1ea1i \u1edf t\u1eed. V\u00ec v\u1eady \u0111\u1ed5i vai tr\u00f2 c\u00e1c s\u1ed1 b\u1ecb x\u00f3a v\u1edbi c\u00e1c s\u1ed1 c\u00f2n l\u1ea1i \u1edf t\u1eed v\u00e0 m\u1eabu th\u00ec ta s\u1ebd c\u00f3 th\u00eam ph\u01b0\u01a1ng \u00e1n x\u00f3a. C\u00f3 nhi\u1ec1u c\u00e1ch x\u00f3a, xin gi\u1edbi thi\u1ec7u m\u1ed9t s\u1ed1 c\u00e1ch (s\u1ed1 c\u00e1c s\u1ed1 b\u1ecb x\u00f3a \u1edf m\u1eabu t\u0103ng d\u1ea7n v\u00e0 t\u1ed5ng chia h\u1ebft cho 6) : m\u1eabu x\u00f3a 12 th\u00ec t\u1eed x\u00f3a 2 ; m\u1eabu x\u00f3a 18 th\u00ec t\u1eed x\u00f3a 3 ho\u1eb7c x\u00f3a 1, 2 ; m\u1eabu x\u00f3a 24 ho\u1eb7c x\u00f3a 11, 13 th\u00ec t\u1eed x\u00f3a 4 ho\u1eb7c x\u00f3a 1, 3 ; m\u1eabu x\u00f3a 12, 18 ho\u1eb7c 13, 17 ho\u1eb7c 14, 16 th\u00ec t\u1eed x\u00f3a 5 ho\u1eb7c 2, 3 ho\u1eb7c 1, 4 ; m\u1eabu x\u00f3a 12, 24 ho\u1eb7c 11, 25 ho\u1eb7c 13, 23 ho\u1eb7c 14, 22 ho\u1eb7c 15, 21 ho\u1eb7c 16, 20 ho\u1eb7c 17, 19 th\u00ec t\u1eed x\u00f3a 6 ho\u1eb7c 1, 5 ho\u1eb7c 2, 4 ho\u1eb7c 1, 2, 3 ; m\u1eabu x\u00f3a 18, 24 ho\u1eb7c 17, 25 ho\u1eb7c 19, 23 ho\u1eb7c 20, 22 ho\u1eb7c 11, 13, 18 ho\u1eb7c 12, 13, 17 ho\u1eb7c 11, 14, 17 ho\u1eb7c 11, 15, 16 ho\u1eb7c 12, 14, 16 ho\u1eb7c 13, 14, 15 th\u00ec t\u1eed x\u00f3a 7 ho\u1eb7c 1, 6 ho\u1eb7c 2, 5 ho\u1eb7c 3, 4 ho\u1eb7c 1, 2, 4 ; ... C\u00e1c b\u1ea1n h\u00e3y k\u1ec3 ti\u1ebfp th\u1eed xem \u0111\u01b0\u1ee3c bao nhi\u00eau c\u00e1ch n\u1eefa ? b) \u0110\u1ec3 gi\u00e1 tr\u1ecb ph\u00e2n s\u1ed1 kh\u00f4ng \u0111\u1ed5i, ta th\u00eam m\u1ed9t s\u1ed1 n\u00e0o \u0111\u00f3 v\u00e0o t\u1eed b\u1eb1ng 1\/6 s\u1ed1 th\u00eam v\u00e0o m\u1eabu. V\u1eady n\u1ebfu th\u00eam 2004 v\u00e0o m\u1eabu th\u00ec s\u1ed1 ph\u1ea3i th\u00eam v\u00e0o t\u1eed l\u00e0 : 2004 : 6 = 334. B\u00e0i 18 : Ng\u01b0\u1eddi ta l\u1ea5y t\u00edch c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 30 \u0111\u1ec3 chia cho 1000000. B\u1ea1n h\u00e3y cho bi\u1ebft : 1) Ph\u00e9p chia c\u00f3 d\u01b0 kh\u00f4ng ? 2) Th\u01b0\u01a1ng l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 bao nhi\u00eau ? B\u00e0i gi\u1ea3i : X\u00e9t t\u00edch A = 1 x 2 x 3 x ... x 29 x 30, trong \u0111\u00f3 c\u00e1c th\u1eeba s\u1ed1 chia h\u1ebft cho 5 l\u00e0 5, 10, 15, 20, 25, 30 ; m\u00e0 25 = 5 x 5 do \u0111\u00f3 c\u00f3 th\u1ec3 coi l\u00e0 c\u00f3 7 th\u1eeba s\u1ed1 chia h\u1ebft cho 5. M\u1ed7i th\u1eeba s\u1ed1 n\u00e0y nh\u00e2n v\u1edbi m\u1ed9t s\u1ed1 ch\u1eb5n cho ta m\u1ed9t s\u1ed1 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 s\u1ed1 0. Trong t\u00edch A c\u00f3 c\u00e1c th\u1eeba s\u1ed1 l\u00e0 s\u1ed1 ch\u1eb5n v\u00e0 kh\u00f4ng chia h\u1ebft cho 5 l\u00e0 : 2, 4, 6, 8, 12, . . . , 26, 28 (c\u00f3 12 s\u1ed1). Nh\u01b0 v\u1eadt trong t\u00edch A c\u00f3 \u00edt nh\u1ea5t 7 c\u1eb7p s\u1ed1 c\u00f3 t\u00edch t\u1eadn c\u00f9ng l\u00e0 0, do \u0111\u00f3 t\u00edch A c\u00f3 t\u1eadn c\u00f9ng l\u00e0 7 ch\u1eef s\u1ed1 0. S\u1ed1 1 000 000 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 6 ch\u1eef s\u1ed1 0 n\u00ean A chia h\u1ebft cho 1 000 000 v\u00e0 th\u01b0\u01a1ng l\u00e0 s\u1ed1 t\u1ef1 57","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 nhi\u00ean c\u00f3 t\u1eadn c\u00f9ng l\u00e0 ch\u1eef s\u1ed1 0. B\u00e0i 19 : Ba b\u1ea1n To\u00e1n, Tu\u1ed5i v\u00e0 Th\u01a1 c\u00f3 m\u1ed9t s\u1ed1 v\u1edf. N\u1ebfu l\u1ea5y 40% s\u1ed1 v\u1edf c\u1ee7a To\u00e1n chia \u0111\u1ec1u cho Tu\u1ed5i v\u00e0 Th\u01a1 th\u00ec s\u1ed1 v\u1edf c\u1ee7a ba b\u1ea1n b\u1eb1ng nhau. Nh\u01b0ng n\u1ebfu To\u00e1n b\u1edbt \u0111i 5 quy\u1ec3n th\u00ec s\u1ed1 v\u1edf c\u1ee7a To\u00e1n b\u1eb1ng t\u1ed5ng s\u1ed1 v\u1edf c\u1ee7a Tu\u1ed5i v\u00e0 Th\u01a1. H\u1ecfi m\u1ed7i b\u1ea1n c\u00f3 bao nhi\u00eau quy\u1ec3n v\u1edf ? B\u00e0i gi\u1ea3i : \u0110\u1ed5i 40% = 2\/5. N\u1ebfu l\u1ea5y 2\/5 s\u1ed1 v\u1edf c\u1ee7a To\u00e1n chia \u0111\u1ec1u cho Tu\u1ed5i v\u00e0 Th\u01a1 th\u00ec m\u1ed7i b\u1ea1n Tu\u1ed5i hay Th\u01a1 \u0111\u1ec1u \u0111\u01b0\u1ee3c th\u00eam 2\/5 : 2 = 1\/5 (s\u1ed1 v\u1edf c\u1ee7a To\u00e1n) S\u1ed1 v\u1edf c\u00f2n l\u1ea1i c\u1ee7a To\u00e1n sau khi cho l\u00e0 : 1 - 2\/5 = 3\/5 (s\u1ed1 v\u1edf c\u1ee7a To\u00e1n) Do \u0111\u00f3 l\u00fac \u0111\u1ea7u Tu\u1ed5i hay Th\u01a1 c\u00f3 s\u1ed1 v\u1edf l\u00e0 : 3\/5 - 1\/5 = 2\/5 (s\u1ed1 v\u1edf c\u1ee7a To\u00e1n) T\u1ed5ng s\u1ed1 v\u1edf c\u1ee7a Tu\u1ed5i v\u00e0 Th\u01a1 l\u00fac \u0111\u1ea7u l\u00e0 : 2\/5 x 2 = 4\/5 (s\u1ed1 v\u1edf c\u1ee7a To\u00e1n) M\u1eb7t kh\u00e1c theo \u0111\u1ec1 b\u00e0i n\u1ebfu To\u00e1n b\u1edbt \u0111i 5 quy\u1ec3n th\u00ec s\u1ed1 v\u1edf c\u1ee7a To\u00e1n b\u1eb1ng t\u1ed5ng s\u1ed1 v\u1edf c\u1ee7a Tu\u1ed5i v\u00e0 Th\u01a1, do \u0111\u00f3 5 quy\u1ec3n \u1ee9ng v\u1edbi : 1 - 4\/5 = 1\/5 (s\u1ed1 v\u1edf c\u1ee7a To\u00e1n) S\u1ed1 v\u1edf c\u1ee7a To\u00e1n l\u00e0 : 5 : 1\/5 = 25 (quy\u1ec3n) S\u1ed1 v\u1edf c\u1ee7a Tu\u1ed5i hay Th\u01a1 l\u00e0 : 25 x 2\/5 = 10 (quy\u1ec3n) B\u00e0i 20 : Hai s\u1ed1 t\u1ef1 nhi\u00ean A v\u00e0 B, bi\u1ebft A < B v\u00e0 hai s\u1ed1 c\u00f3 chung nh\u1eefng \u0111\u1eb7c \u0111i\u1ec3m sau : - L\u00e0 s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1. - Hai ch\u1eef s\u1ed1 trong m\u1ed7i s\u1ed1 gi\u1ed1ng nhau. - Kh\u00f4ng chia h\u1ebft cho 2 ; 3 v\u00e0 5. a) T\u00ecm 2 s\u1ed1 \u0111\u00f3. b) T\u1ed5ng c\u1ee7a 2 s\u1ed1 \u0111\u00f3 chia h\u1ebft cho s\u1ed1 t\u1ef1 nhi\u00ean n\u00e0o ? B\u00e0i gi\u1ea3i : V\u00ec A v\u00e0 B \u0111\u1ec1u kh\u00f4ng chia h\u1ebft cho 2 v\u00e0 5 n\u00ean A v\u00e0 B ch\u1ec9 c\u00f3 th\u1ec3 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 1 ; 58","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 3 ; 7 ; 9. V\u00ec 3 + 3 = 6 v\u00e0 9 + 9 = 18 l\u00e0 2 s\u1ed1 chia h\u1ebft cho 3 n\u00ean lo\u1ea1i tr\u1eeb s\u1ed1 33 v\u00e0 99. A < B n\u00ean A = 11 v\u00e0 B = 77. b) T\u1ed5ng c\u1ee7a hai s\u1ed1 \u0111\u00f3 l\u00e0 : 11 + 77 = 88. Ta c\u00f3 : 88 = 1 x 88 = 2 x 44 = 4 x 22 = 8 x 11. ABCD. H\u00e3y di\u1ec7n t\u00edch c\u00f2n V\u1eady t\u1ed5ng 2 s\u1ed1 chia h\u1ebft cho c\u00e1c s\u1ed1 : 1 ; 2 ; 4 ; 8 ; 11 ; 22 ; 44 ; 88. B\u00e0i 21 : Cho m\u1ea3nh b\u00eca h\u00ecnh vu\u00f4ng c\u1eaft t\u1eeb m\u1ea3nh b\u00eca \u0111\u00f3 m\u1ed9t h\u00ecnh vu\u00f4ng sao cho l\u1ea1i b\u1eb1ng di\u1ec7n t\u00edch c\u1ee7a m\u1ea3nh b\u00eca \u0111\u00e3 cho. B\u00e0i gi\u1ea3i : Theo \u0111\u1ea7u b\u00e0i th\u00ec h\u00ecnh vu\u00f4ng ABCD \u0111\u01b0\u1ee3c gh\u00e9p b\u1edfi 2 h\u00ecnh vu\u00f4ng nh\u1ecf v\u00e0 4 tam gi\u00e1c (trong \u0111\u00f3 c\u00f3 2 tam gi\u00e1c to, 2 tam gi\u00e1c con). Ta th\u1ea5y c\u00f3 th\u1ec3 gh\u00e9p 4 tam gi\u00e1c con \u0111\u1ec3 \u0111\u01b0\u1ee3c tam gi\u00e1c to \u0111\u1ed3ng th\u1eddi c\u0169ng gh\u00e9p 4 tam gi\u00e1c con \u0111\u1ec3 \u0111\u01b0\u1ee3c 1 h\u00ecnh vu\u00f4ng nh\u1ecf. V\u1eady di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh vu\u00f4ng ABCD ch\u00ednh l\u00e0 di\u1ec7n t\u00edch c\u1ee7a 2 + 2 x 4 + 2 x 4 = 18 (tam gi\u00e1c con). Do \u0111\u00f3 di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh vu\u00f4ng ABCD l\u00e0 : 18 x (10 x 10) \/ 2 = 900 (cm2) B\u00e0i 22 : Hai b\u1ea1n Xu\u00e2n v\u00e0 H\u1ea1 c\u00f9ng m\u1ed9t l\u00fac r\u1eddi nh\u00e0 c\u1ee7a m\u00ecnh \u0111i \u0111\u1ebfn nh\u00e0 b\u1ea1n. H\u1ecd g\u1eb7p nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m c\u00e1ch nh\u00e0 Xu\u00e2n 50 m. Bi\u1ebft r\u1eb1ng Xu\u00e2n \u0111i t\u1eeb nh\u00e0 m\u00ecnh \u0111\u1ebfn nh\u00e0 H\u1ea1 m\u1ea5t 12 ph\u00fat c\u00f2n H\u1ea1 \u0111i \u0111\u1ebfn nh\u00e0 Xu\u00e2n ch\u1ec9 m\u1ea5t 10 ph\u00fat. H\u00e3y t\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng gi\u1eefa nh\u00e0 hai b\u1ea1n. B\u00e0i gi\u1ea3i : Tr\u00ean c\u00f9ng m\u1ed9t qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec t\u1ec9 s\u1ed1 th\u1eddi gian \u0111i c\u1ee7a Xu\u00e2n v\u00e0 H\u1ea1 l\u00e0 : 12 : 10 = 6\/5. Th\u1eddi gian t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi v\u1eadn t\u1ed1c n\u00ean t\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c c\u1ee7a Xu\u00e2n v\u00e0 H\u1ea1 l\u00e0 5\/6. Nh\u01b0 v\u1eady Xu\u00e2n v\u00e0 H\u1ea1 c\u00f9ng xu\u1ea5t ph\u00e1t th\u00ec \u0111\u1ebfn khi g\u1eb7p nhau th\u00ec qu\u00e3ng \u0111\u01b0\u1eddng Xu\u00e2n \u0111i \u0111\u01b0\u1ee3c b\u1eb1ng 5\/6 qu\u00e3ng \u0111\u01b0\u1eddng H\u1ea1 \u0111i \u0111\u01b0\u1ee3c. Do \u0111\u00f3 qu\u00e3ng \u0111\u01b0\u1eddng H\u1ea1 \u0111i \u0111\u01b0\u1ee3c l\u00e0 : 50 : 5\/6 = 60 (m). 59","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 Qu\u00e3ng \u0111\u01b0\u1eddng gi\u1eefa nh\u00e0 Xu\u00e2n v\u00e0 H\u1ea1 l\u00e0 : 50 + 60 = 110 (m). B\u00e0i 23 : A l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 2004 ch\u1eef s\u1ed1. A l\u00e0 s\u1ed1 chia h\u1ebft cho 9 ; B l\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a A ; C l\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a B ; D l\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a C. T\u00ecm D. B\u00e0i gi\u1ea3i : V\u00ec A l\u00e0 s\u1ed1 chia h\u1ebft cho 9 m\u00e0 B l\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a A n\u00ean B chia h\u1ebft cho 9. T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 C, D c\u0169ng chia h\u1ebft cho 9 v\u00e0 \u0111\u01b0\u01a1ng nhi\u00ean kh\u00e1c 0. V\u00ec A g\u1ed3m 2004 ch\u1eef s\u1ed1 m\u00e0 m\u1ed7i ch\u1eef s\u1ed1 kh\u00f4ng v\u01b0\u1ee3t qu\u00e1 9 n\u00ean B kh\u00f4ng v\u01b0\u1ee3t qu\u00e1 9 x 2004 = 18036. Do \u0111\u00f3 B c\u00f3 kh\u00f4ng qu\u00e1 5 ch\u1eef s\u1ed1 v\u00e0 C < 9 x 5 = 45. Nh\u01b0ng C l\u00e0 s\u1ed1 chia h\u1ebft cho 9 v\u00e0 kh\u00e1c 0 n\u00ean C ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 9 ; 18 ; 27 ; 36. D\u00f9 tr\u01b0\u1eddng h\u1ee3p n\u00e0o x\u1ea3y ra th\u00ec ta c\u0169ng c\u00f3 D = 9. B\u00e0i 24 : M\u1ed9t khu v\u01b0\u1eddn h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 120 m. Ng\u01b0\u1eddi ta m\u1edf r\u1ed9ng khu v\u01b0\u1eddn nh\u01b0 h\u00ecnh v\u1ebd \u0111\u1ec3 \u0111\u01b0\u1ee3c m\u1ed9t v\u01b0\u1eddn h\u00ecnh ch\u1eef nh\u1eadt l\u1edbn h\u01a1n. T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n m\u1edbi m\u1edf th\u00eam. B\u00e0i gi\u1ea3i : N\u1ebfu ta \u201cd\u1ecbch chuy\u1ec3n\u201d khu v\u01b0\u1eddn c\u0169 ABCD v\u00e0o m\u1ed9t g\u00f3c c\u1ee7a khu v\u01b0\u1eddn m\u1edbi EFHD ta \u0111\u01b0\u1ee3c h\u00ecnh v\u1ebd b\u00ean. K\u00e9o d\u00e0i EF v\u1ec1 ph\u00eda F l\u1ea5y M sao cho FM = BC th\u00ec di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt BKHC \u0111\u00fang b\u1eb1ng di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt FMNK. Do \u0111\u00f3 ph\u1ea7n di\u1ec7n t\u00edch m\u1edbi m\u1edf th\u00eam ch\u00ednh l\u00e0 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt EMNA. Ta c\u00f3 AN = AB + KN + BK v\u00ec AB + KN = 120 : 2 = 60 (m) ; BK = 10 m n\u00ean AN = 70 m. V\u1eady di\u1ec7n t\u00edch ph\u1ea7n m\u1edbi m\u1edf th\u00eam l\u00e0 : 70 x 10 = 700 (m2) B\u00e0i 25 : Bao nhi\u00eau gi\u1edd ? Khi \u0111i g\u1eb7p n\u01b0\u1edbc ng\u01b0\u1edbc d\u00f2ng Kh\u00f3 kh\u0103n \u0111\u1ebfn b\u1ebfn m\u1ea5t tong t\u00e1m gi\u1edd Khi v\u1ec1 t\u1eeb l\u00fac xu\u1ed1ng \u0111\u00f2 \u0110\u1ebfn khi c\u1eadp b\u1ebfn b\u1ed1n gi\u1edd nh\u1eb9 veo H\u1ecfi r\u1eb1ng ri\u00eang m\u1ed9t kh\u00f3m b\u00e8o Bao nhi\u00eau gi\u1edd \u0111\u1ec3 tr\u00f4i theo ta v\u1ec1 ? B\u00e0i gi\u1ea3i : 60","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 C\u00e1ch 1 : V\u00ec \u0111\u00f2 \u0111i ng\u01b0\u1ee3c d\u00f2ng \u0111\u1ebfn b\u1ebfn m\u1ea5t 8 gi\u1edd n\u00ean trong 1 gi\u1edd \u0111\u00f2 \u0111i \u0111\u01b0\u1ee3c 1\/8 qu\u00e3ng s\u00f4ng \u0111\u00f3. \u0110\u00f2 \u0111i xu\u00f4i d\u00f2ng tr\u1edf v\u1ec1 m\u1ea5t 4 gi\u1edd n\u00ean trong 1 gi\u1edd \u0111\u00f2 \u0111i \u0111\u01b0\u1ee3c 1\/4 qu\u00e3ng s\u00f4ng \u0111\u00f3. V\u1eadn t\u1ed1c \u0111\u00f2 xu\u00f4i d\u00f2ng h\u01a1n v\u1eadn t\u1ed1c \u0111\u00f2 ng\u01b0\u1ee3c d\u00f2ng l\u00e0 : 1\/4 - 1\/8 = 1\/8 (qu\u00e3ng s\u00f4ng \u0111\u00f3). V\u00ec hi\u1ec7u v\u1eadn t\u1ed1c \u0111\u00f2 xu\u00f4i d\u00f2ng v\u00e0 v\u1eadn t\u1ed1c \u0111\u00f2 ng\u01b0\u1ee3c d\u00f2ng ch\u00ednh l\u00e0 2 l\u1ea7n v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc n\u00ean m\u1ed9t gi\u1edd kh\u00f3m b\u00e8o tr\u00f4i \u0111\u01b0\u1ee3c l\u00e0 : 1\/8 : 2 = 1\/16 (qu\u00e3ng s\u00f4ng \u0111\u00f3). Th\u1eddi gian \u0111\u1ec3 kh\u00f3m b\u00e8o tr\u00f4i theo \u0111\u00f2 v\u1ec1 l\u00e0 : 1 : 1\/16 = 16 (gi\u1edd). C\u00e1ch 2 : T\u1ec9 s\u1ed1 gi\u1eefa th\u1eddi gian \u0111\u00f2 xu\u00f4i d\u00f2ng v\u00e0 th\u1eddi gian \u0111\u00f2 ng\u01b0\u1ee3c d\u00f2ng l\u00e0 :4 : 8 = 1\/2 Tr\u00ean c\u00f9ng m\u1ed9t qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec v\u1eadn t\u1ed1c v\u00e0 th\u1eddi gian c\u1ee7a m\u1ed9t chuy\u1ec3n \u0111\u1ed9ng t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi nhau n\u00ean t\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c \u0111\u00f2 xu\u00f4i d\u00f2ng v\u00e0 v\u1eadn t\u1ed1c \u0111\u00f2 ng\u01b0\u1ee3c d\u00f2ng l\u00e0 2. V\u1eadn t\u1ed1c \u0111\u00f2 xu\u00f4i d\u00f2ng h\u01a1n v\u1eadn t\u1ed1c \u0111\u00f2 ng\u01b0\u1ee3c d\u00f2ng ch\u00ednh l\u00e0 2 l\u1ea7n v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc. Ta c\u00f3 s\u01a1 \u0111\u1ed3 : Theo s\u01a1 \u0111\u1ed3 ta c\u00f3 v\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng g\u1ea5p 2 l\u1ea7n v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc n\u00ean th\u1eddi gian \u0111\u1ec3 c\u1ee5m b\u00e8o tr\u00f4i theo \u0111\u00f2 v\u1ec1 g\u1ea5p 2 l\u1ea7n th\u1eddi gian ng\u01b0\u1ee3c d\u00f2ng. V\u1eady th\u1eddi gian c\u1ee5m b\u00e8o tr\u00f4i theo \u0111\u00f2 v\u1ec1 l\u00e0 : 8 x 2 = 16 (gi\u1edd). B\u00e0i 26 : M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p 4 l\u1ea7n chi\u1ec1u r\u1ed9ng. N\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 45 m th\u00ec \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi c\u00f3 chi\u1ec1u d\u00e0i v\u1eabn g\u1ea5p 4 l\u1ea7n chi\u1ec1u r\u1ed9ng. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u. B\u00e0i gi\u1ea3i : Khi t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 45 m th\u00ec khi \u0111\u00f3 chi\u1ec1u r\u1ed9ng s\u1ebd tr\u1edf th\u00e0nh chi\u1ec1u d\u00e0i c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi, c\u00f2n chi\u1ec1u d\u00e0i ban \u0111\u1ea7u s\u1ebd tr\u1edf th\u00e0nh chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi. Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 s\u01a1 \u0111\u1ed3 : Do \u0111\u00f3 45 m \u1ee9ng v\u1edbi s\u1ed1 ph\u1ea7n l\u00e0 : 16 - 1 = 15 (ph\u1ea7n) Chi\u1ec1u r\u1ed9ng ban \u0111\u1ea7u l\u00e0 : 61","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 45 : 15 = 3 (m) Chi\u1ec1u d\u00e0i ban \u0111\u1ea7u l\u00e0 : 3 x 4 = 12 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u00e0 : 3 x 12 = 36 (m2) B\u00e0i 27: B\u1ea1n An \u0111\u00e3 c\u00f3 m\u1ed9t s\u1ed1 b\u00e0i ki\u1ec3m tra, b\u1ea1n \u0111\u00f3 t\u00ednh r\u1eb1ng : N\u1ebfu \u0111\u01b0\u1ee3c th\u00eam ba \u0111i\u1ec3m 10 v\u00e0 ba \u0111i\u1ec3m 9 n\u1eefa th\u00ec \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c b\u00e0i s\u1ebd l\u00e0 8. N\u1ebfu \u0111\u01b0\u1ee3c th\u00eam m\u1ed9t \u0111i\u1ec3m 9 v\u00e0 hai \u0111i\u1ec3m 10 n\u1eefa th\u00ec \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c b\u00e0i l\u00e0 7,5. H\u1ecfi b\u1ea1n An \u0111\u00e3 c\u00f3 t\u1ea5t c\u1ea3 m\u1ea5y b\u00e0i ki\u1ec3m tra ? B\u00e0i gi\u1ea3i : N\u1ebfu \u0111\u01b0\u1ee3c th\u00eam ba \u0111i\u1ec3m 10 v\u00e0 ba \u0111i\u1ec3m 9 n\u1eefa th\u00ec s\u1ed1 \u0111i\u1ec3m \u0111\u01b0\u1ee3c th\u00eam l\u00e0 : 10 x 3 + 9 x 3 = 57 (\u0111i\u1ec3m) \u0110\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c b\u00e0i l\u00e0 8 th\u00ec s\u1ed1 \u0111i\u1ec3m ph\u1ea3i b\u00f9 th\u00eam v\u00e0o cho c\u00e1c b\u00e0i \u0111\u00e3 ki\u1ec3m tra l\u00e0 : 57 - 8 x (3 + 3) = 9 (\u0111i\u1ec3m) N\u1ebfu \u0111\u01b0\u1ee3c th\u00eam m\u1ed9t \u0111i\u1ec3m 9 v\u00e0 hai \u0111i\u1ec3m 10 n\u1eefa th\u00ec s\u1ed1 \u0111i\u1ec3m \u0111\u01b0\u1ee3c th\u00eam l\u00e0 : 9 x 1 + 10 x 2 = 28 (\u0111i\u1ec3m) \u0110\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c b\u00e0i l\u00e0 7,5 th\u00ec s\u1ed1 \u0111i\u1ec3m ph\u1ea3i b\u00f9 th\u00eam v\u00e0o cho c\u00e1c b\u00e0i \u0111\u00e3 ki\u1ec3m tra l\u00e0 : 29 - 7,5 x (1 + 2) = 6,5 (\u0111i\u1ec3m) Nh\u01b0 v\u1eady khi t\u0103ng \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c b\u00e0i t\u1eeb 7,5 l\u00ean 8 th\u00ec t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u00e1c b\u00e0i \u0111\u00e3 ki\u1ec3m tra s\u1ebd t\u0103ng l\u00ean l\u00e0 : 9 - 6,5 = 2,5 (\u0111i\u1ec3m) Hi\u1ec7u hai \u0111i\u1ec3m trung b\u00ecnh l\u00e0 : 8 - 7,5 = 0,5 (\u0111i\u1ec3m) 62","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 V\u1eady s\u1ed1 b\u00e0i \u0111\u00e3 ki\u1ec3m tra c\u1ee7a b\u1ea1n An l\u00e0 : 2,5 : 0,5 = 5 (b\u00e0i) b\u1eb1ng 5 \/ 8 di\u1ec7n B\u00e0i 28 : B\u1ea1n h\u00e3y c\u1eaft m\u1ed9t h\u00ecnh vu\u00f4ng c\u00f3 di\u1ec7n t\u00edch t\u00edch c\u1ee7a m\u1ed9t t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng cho tr\u01b0\u1edbc. B\u00e0i gi\u1ea3i : Chia c\u1ea1nh t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng cho tr\u01b0\u1edbc l\u00e0m 4 ph\u1ea7n b\u1eb1ng nhau (b\u1eb1ng c\u00e1ch g\u1ea5p \u0111\u00f4i li\u00ean ti\u1ebfp). Sau \u0111\u00f3 c\u1eaft theo c\u00e1c \u0111\u01b0\u1eddng AB, BC, CD, DA. C\u00e1c mi\u1ebfng b\u00eca AMB, BNC, CPD, DQA x\u1ebfp tr\u00f9ng kh\u00edt l\u00ean nhau n\u00ean AB = BC = CD = DA (c\u00f3 th\u1ec3 ki\u1ec3m tra b\u1eb1ng th\u01b0\u1edbc \u0111o). D\u00f9ng \u00eake ki\u1ec3m tra c\u00e1c g\u00f3c c\u1ee7a t\u1ea5m b\u00eca ABCD ta th\u1ea5y c\u00e1c g\u00f3c l\u00e0 vu\u00f4ng. N\u1ebfu k\u1ebb b\u1eb1ng b\u00fat ch\u00ec c\u00e1c \u0111\u01b0\u1eddng chia t\u1ea5m b\u00eca ban \u0111\u1ea7u th\u00e0nh nh\u1eefng \u00f4 vu\u00f4ng nh\u01b0 h\u00ecnh v\u1ebd th\u00ec ta c\u00f3 th\u1ec3 th\u1ea5y : + Di\u1ec7n t\u00edch t\u1ea5m b\u00eca MNPQ l\u00e0 16 \u00f4 vu\u00f4ng (gh\u00e9p 2 h\u00ecnh tam gi\u00e1c v\u1edbi nhau th\u00ec \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt g\u1ed3m 3 h\u00ecnh vu\u00f4ng). Do \u0111\u00f3 di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD l\u00e0 16 \u2013 6 = 10 (\u00f4 vu\u00f4ng) n\u00ean di\u1ec7n t\u00edch \u00f4 vu\u00f4ng ABCD b\u1eb1ng 10 \/ 16 = 5 \/ 8 di\u1ec7n t\u00edch t\u1ea5m b\u00eca ban \u0111\u1ea7u. B\u00e0i 29 : M\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c chia th\u00e0nh 4 h\u00ecnh ch\u1eef nh\u1eadt nh\u1ecf h\u01a1n c\u00f3 di\u1ec7n t\u00edch \u0111\u01b0\u1ee3c ghi nh\u01b0 h\u00ecnh v\u1ebd. B\u1ea1n c\u00f3 bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt c\u00f2n l\u1ea1i c\u00f3 di\u1ec7n t\u00edch l\u00e0 bao nhi\u00eau hay kh\u00f4ng ? B\u00e0i gi\u1ea3i : Hai h\u00ecnh ch\u1eef nh\u1eadt AMOP v\u00e0 MBQO c\u00f3 chi\u1ec1u r\u1ed9ng b\u1eb1ng nhau v\u00e0 c\u00f3 di\u1ec7n t\u00edch h\u00ecnh MBQO g\u1ea5p 3 l\u1ea7n di\u1ec7n t\u00edch h\u00ecnh AMOP (24 : 8 = 3 (l\u1ea7n)), do \u0111\u00f3 chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt MBQO g\u1ea5p 3 l\u1ea7n chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt AMOP (OQ = PO x 3). (1) Hai h\u00ecnh ch\u1eef nh\u1eadt POND v\u00e0 OQCN c\u00f3 chi\u1ec1u r\u1ed9ng b\u1eb1ng nhau v\u00e0 c\u00f3 chi\u1ec1u d\u00e0i h\u00ecnh OQCN g\u1ea5p 3 l\u1ea7n chi\u1ec1u d\u00e0i h\u00ecnh POND (1). Do \u0111\u00f3 di\u1ec7n t\u00edch h\u00ecnh OQCN g\u1ea5p 3 l\u1ea7n di\u1ec7n t\u00edch h\u00ecnh POND. V\u1eady di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt OQCD l\u00e0 : 16 x 3 = 48 (cm2). B\u00e0i 30 : Cho A = 2004 x 2004 x ... x 2004 (A g\u1ed3m 2003 th\u1eeba s\u1ed1) v\u00e0 B = 2003 x 2003 x ... 63","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 x 2003 (B g\u1ed3m 2004 th\u1eeba s\u1ed1). H\u00e3y cho bi\u1ebft A + B c\u00f3 chia h\u1ebft cho 5 hay kh\u00f4ng ? V\u00ec sao ? B\u00e0i gi\u1ea3i : A = (2004 x 2004 x ... x 2004) x 2004 = C x 2004 (C c\u00f3 2002 th\u1eeba s\u1ed1 2004). C c\u00f3 t\u1eadn c\u00f9ng l\u00e0 6 nh\u00e2n v\u1edbi 2004 n\u00ean A c\u00f3 t\u1eadn c\u00f9ng l\u00e0 4 (v\u00ec 6 x 4 = 24). B = 2003 x 2003 x ... x 2003 (g\u1ed3m 2004 th\u1eeba s\u1ed1) = (2003 x 2003 x 2003 x 2003) x ... x (2003 x 2003 x 2003 x 2003). V\u00ec 2004 : 4 = 501 (nh\u00f2m) n\u00ean B c\u00f3 501 nh\u00f3m, m\u1ed7i nh\u00f3m g\u1ed3m 4 th\u1eeba s\u1ed1 2003. T\u1eadn c\u00f9ng c\u1ee7a m\u1ed7i nh\u00f3m l\u00e0 1 (v\u00ec 3 x 3 = 9 ; 9 x 3 = 27 ; 27 x 3 = 81). V\u1eady t\u1eadn c\u00f9ng c\u1ee7a A + B l\u00e0 4 + 1 = 5. Do \u0111\u00f3 A + B chia h\u1ebft cho 5. B\u00e0i 31 : Bi\u1ebft r\u1eb1ng s\u1ed1 A ch\u1ec9 vi\u1ebft b\u1edfi c\u00e1c ch\u1eef s\u1ed1 9. H\u00e3y t\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf nh\u1ea5t m\u00e0 c\u1ed9ng s\u1ed1 n\u00e0y v\u1edbi A ta \u0111\u01b0\u1ee3c s\u1ed1 chia h\u1ebft cho 45. B\u00e0i gi\u1ea3i : C\u00e1ch 1 : A ch\u1ec9 vi\u1ebft b\u1edfi c\u00e1c ch\u1eef s\u1ed1 9 n\u00ean: V\u1eady A chia cho 45 d\u01b0 9. M\u1ed9t s\u1ed1 nh\u1ecf nh\u1ea5t m\u00e0 c\u1ed9ng v\u1edbi A \u0111\u1ec3 \u0111\u01b0\u1ee3c s\u1ed1 chia h\u1ebft cho 45 th\u00ec s\u1ed1 \u0111\u00f3 c\u1ed9ng v\u1edbi 9 ph\u1ea3i b\u1eb1ng 45. V\u1eady s\u1ed1 \u0111\u00f3 l\u00e0 : 45 - 9 = 36. C\u00e1ch 2 : G\u1ecdi s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf nh\u1ea5t c\u1ed9ng v\u00e0o A l\u00e0 m. Ta c\u00f3 A + m l\u00e0 s\u1ed1 chia h\u1ebft cho 45 hay chia h\u1ebft cho 5 v\u00e0 9 (v\u00ec 5 x 9 = 45 ; 5 v\u00e0 9 kh\u00f4ng c\u00f9ng chia h\u1ebft cho m\u1ed9t s\u1ed1 s\u1ed1 n\u00e0o \u0111\u00f3 kh\u00e1c 1). V\u00ec A vi\u1ebft b\u1edfi c\u00e1c ch\u1eef s\u1ed1 9 n\u00ean A chia h\u1ebft cho 9, do \u0111\u00f3 m chia h\u1ebft cho 9. A + m chia h\u1ebft cho 5 khi A + m c\u00f3 t\u1eadn c\u00f9ng l\u00e0 0 ho\u1eb7c 5 m\u00e0 A c\u00f3 t\u1eadn c\u00f9ng l\u00e0 9 n\u00ean m c\u00f3 t\u1eadn c\u00f9ng l\u00e0 1 ho\u1eb7c 6. S\u1ed1 nh\u1ecf nh\u1ea5t c\u00f3 t\u1eadn c\u00f9ng l\u00e0 1 ho\u1eb7c 6 m\u00e0 chia h\u1ebft cho 9 l\u00e0 36. V\u1eady m = 36. B\u00e0i 32 : Cho m\u1ed9t h\u00ecnh thang vu\u00f4ng c\u00f3 \u0111\u00e1y l\u1edbn b\u1eb1ng 3 m, \u0111\u00e1y nh\u1ecf v\u00e0 chi\u1ec1u cao b\u1eb1ng 2 m. H\u00e3y chia h\u00ecnh thang \u0111\u00f3 th\u00e0nh 5 h\u00ecnh tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. H\u00e3y t\u00ecm c\u00e1c ki\u1ec3u chia kh\u00e1c nhau sao cho s\u1ed1 \u0111o chi\u1ec1u cao c\u0169ng nh\u01b0 s\u1ed1 \u0111o \u0111\u00e1y c\u1ee7a tam gi\u00e1c \u0111\u1ec1u l\u00e0 nh\u1eefng s\u1ed1 t\u1ef1 nhi\u00ean. 64","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i gi\u1ea3i : Di\u1ec7n t\u00edch h\u00ecnh thang l\u00e0 : (3 + 2) x 2 : 2 = 5 (m2) Chia h\u00ecnh thang \u0111\u00f3 th\u00e0nh 5 tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau th\u00ec di\u1ec7n t\u00edch m\u1ed9t tam gi\u00e1c l\u00e0 : 5 : 5 = 1 (m2). C\u00e1c tam gi\u00e1c n\u00e0y c\u00f3 chi\u1ec1u cao v\u00e0 s\u1ed1 \u0111o \u0111\u00e1y l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean n\u00ean n\u1ebfu chi\u1ec1u cao l\u00e0 1m th\u00ec \u0111\u00e1y l\u00e0 2 m. N\u1ebfu chi\u1ec1u cao l\u00e0 2 m th\u00ec \u0111\u00e1y l\u00e0 1 m. C\u00f3 nhi\u1ec1u c\u00e1ch chia, TTT ch\u1ec9 n\u00eau m\u1ed9t s\u1ed1 c\u00e1ch chia sau : B\u00e0i 33 : B\u1ea1n h\u00e3y t\u00ednh chu vi c\u1ee7a h\u00ecnh c\u00f3 t\u1eeb m\u1ed9t h\u00ecnh vu\u00f4ng b\u1ecb c\u1eaft m\u1ea5t \u0111i m\u1ed9t ph\u1ea7n b\u1edfi m\u1ed9t \u0111\u01b0\u1eddng g\u1ea5p kh\u00fac g\u1ed3m c\u00e1c \u0111o\u1ea1n song song v\u1edbi c\u1ea1nh h\u00ecnh vu\u00f4ng. B\u00e0i gi\u1ea3i : Ta k\u00ed hi\u1ec7u c\u00e1c \u0111i\u1ec3m nh\u01b0 h\u00ecnh v\u1ebd sau : Nh\u00ecn h\u00ecnh v\u1ebd ta th\u1ea5y : CE + GH + KL + MD = CE + EI = CI. EG + HK + LM + DA = ID + DA = IA. T\u1eeb \u0111\u00f3 chu vi c\u1ee7a h\u00ecnh t\u00f4 m\u00e0u ch\u00ednh l\u00e0 : AB + BC + CE + EG + GH + HK + KL + LM + MD + DA = AB + BC + (CE + GH + KL + MD) + (EG + HK + LM + DA) = AB + BC + CI + IA = AB x 4. 65","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 V\u1eady chu vi c\u1ee7a h\u00ecnh t\u00f4 m\u00e0u l\u00e0 : 10 x 4 = 40 (cm). B\u00e0i 34 : Cho b\u0103ng gi\u1ea5y g\u1ed3m 13 \u00f4 v\u1edbi s\u1ed1 \u1edf \u00f4 th\u1ee9 hai l\u00e0 112 v\u00e0 s\u1ed1 \u1edf \u00f4 th\u1ee9 b\u1ea3y l\u00e0 215. Bi\u1ebft r\u1eb1ng t\u1ed5ng c\u1ee7a ba s\u1ed1 \u1edf ba \u00f4 li\u00ean ti\u1ebfp lu\u00f4n b\u1eb1ng 428. T\u00ednh t\u1ed5ng c\u1ee7a c\u00e1c ch\u1eef s\u1ed1 tr\u00ean b\u0103ng gi\u1ea5y \u0111\u00f3. B\u00e0i gi\u1ea3i : Ta chia c\u00e1c \u00f4 th\u00e0nh c\u00e1c nh\u00f3m 3 \u00f4, m\u1ed7i nh\u00f3m \u0111\u00e1nh s\u1ed1 th\u1ee9 t\u1ef1 nh\u01b0 sau : T\u1ed5ng c\u00e1c s\u1ed1 c\u1ee7a m\u1ed7i nh\u00f3m 3 \u00f4 li\u00ean ti\u1ebfp l\u00e0 428. Nh\u01b0 v\u1eady ta th\u1ea5y c\u00e1c s\u1ed1 vi\u1ebft \u1edf \u00f4 s\u1ed1 1 l\u00e0 215, \u1edf \u00f4 s\u1ed1 2 l\u00e0 112, \u1edf \u00f4 s\u1ed1 3 l\u00e0 : 428 - (215 + 112) = 101. Ta c\u00f3 b\u0103ng gi\u1ea5y ghi s\u1ed1 nh\u01b0 sau : T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a m\u1ed7i nh\u00f3m 3 \u00f4 l\u00e0 : 2 + 1 + 5 + 1 + 1 + 2 + 1 + 0 + 1 = 14. C\u00f3 t\u1ea5t c\u1ea3 4 nh\u00f3m 3 \u00f4 v\u00e0 m\u1ed9t s\u1ed1 \u1edf \u00f4 s\u1ed1 1 n\u00ean t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 tr\u00ean b\u0103ng gi\u1ea5y l\u00e0 : 14 x 4 + 2 + 1 + 5 = 64. B\u00e0i 35 : Tu\u1ed5i c\u1ee7a em t\u00f4i hi\u1ec7n nay b\u1eb1ng 4 l\u1ea7n tu\u1ed5i c\u1ee7a n\u00f3 khi tu\u1ed5i c\u1ee7a anh t\u00f4i b\u1eb1ng tu\u1ed5i c\u1ee7a em t\u00f4i hi\u1ec7n nay. \u0110\u1ebfn khi tu\u1ed5i c\u1ee7a em t\u00f4i b\u1eb1ng tu\u1ed5i c\u1ee7a anh t\u00f4i hi\u1ec7n nay th\u00ec t\u1ed5ng s\u1ed1 tu\u1ed5i c\u1ee7a hai anh em l\u00e0 51. H\u1ecfi hi\u1ec7n nay anh t\u00f4i, em t\u00f4i bao nhi\u00eau tu\u1ed5i ? B\u00e0i gi\u1ea3i : Hi\u1ec7u s\u1ed1 tu\u1ed5i c\u1ee7a hai anh em l\u00e0 m\u1ed9t s\u1ed1 kh\u00f4ng \u0111\u1ed5i. di\u1ec5n s\u1ed1 tu\u1ed5i c\u1ee7a hai anh em \u1edf Ta c\u00f3 s\u01a1 \u0111\u1ed3 bi\u1ec3u (T\u0110), hi\u1ec7n nay (HN), sau n\u00e0y c\u00e1c th\u1eddi \u0111i\u1ec3m : Tr\u01b0\u1edbc \u0111\u00e2y 66","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 (SN) : Gi\u00e1 tr\u1ecb m\u1ed9t ph\u1ea7n l\u00e0 : 51 : (7 + 10) = 3 (tu\u1ed5i) Tu\u1ed5i em hi\u1ec7n nay l\u00e0 : 3 x 4 = 12 (tu\u1ed5i) Tu\u1ed5i anh hi\u1ec7n nay l\u00e0 : 3 x 7 = 21 (tu\u1ed5i) B\u00e0i 36 : Tham gia SEA Games 22 m\u00f4n b\u00f3ng \u0111\u00e1 nam v\u00f2ng lo\u1ea1i \u1edf b\u1ea3ng B c\u00f3 b\u1ed1n \u0111\u1ed9i thi \u0111\u1ea5u theo th\u1ec3 th\u1ee9c \u0111\u1ea5u v\u00f2ng tr\u00f2n m\u1ed9t l\u01b0\u1ee3t v\u00e0 t\u00ednh \u0111i\u1ec3m theo quy \u0111\u1ecbnh hi\u1ec7n h\u00e0nh. K\u1ebft th\u00fac v\u00f2ng lo\u1ea1i, t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u00e1c \u0111\u1ed9i \u1edf b\u1ea3ng B l\u00e0 17 \u0111i\u1ec3m. H\u1ecfi \u1edf b\u1ea3ng B m\u00f4n b\u00f3ng \u0111\u00e1 nam c\u00f3 m\u1ea5y tr\u1eadn h\u00f2a ? B\u00e0i gi\u1ea3i : B\u1ea3ng B c\u00f3 4 \u0111\u1ed9i thi \u0111\u1ea5u v\u00f2ng tr\u00f2n n\u00ean s\u1ed1 tr\u1eadn \u0111\u1ea5u l\u00e0 : 4 x 3 : 2 = 6 (tr\u1eadn) M\u1ed7i tr\u1eadn th\u1eafng th\u00ec \u0111\u1ed9i th\u1eafng \u0111\u01b0\u1ee3c 3 \u0111i\u1ec3m \u0111\u1ed9i thua th\u00ec \u0111\u01b0\u1ee3c 0 \u0111i\u1ec3m n\u00ean t\u1ed5ng s\u1ed1 \u0111i\u1ec3m l\u00e0 : 3 + 0 = 3 (\u0111i\u1ec3m). M\u1ed7i tr\u1eadn h\u00f2a th\u00ec m\u1ed7i \u0111\u1ed9i \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m n\u00ean t\u1ed5ng s\u1ed1 \u0111i\u1ec3m l\u00e0 : 1 + 1 = 2 (\u0111i\u1ec3m). C\u00e1ch 1 : Gi\u1ea3 s\u1eed 6 tr\u1eadn \u0111\u1ec1u th\u1eafng th\u00ec t\u1ed5ng s\u1ed1 \u0111i\u1ec3m l\u00e0 : 6 x 3 = 18 (\u0111i\u1ec3m). S\u1ed1 \u0111i\u1ec3m d\u00f4i ra l\u00e0 : 18 - 17 = 1 (\u0111i\u1ec3m). S\u1edf d\u0129 d\u00f4i ra 1 \u0111i\u1ec3m l\u00e0 v\u00ec m\u1ed9t tr\u1eadn th\u1eafng h\u01a1n m\u1ed9t tr\u1eadn h\u00f2a l\u00e0 : 3 - 2 = 1 (\u0111i\u1ec3m). V\u1eady s\u1ed1 tr\u1eadn h\u00f2a l\u00e0 : 1 : 1 = 1 (tr\u1eadn) C\u00e1ch 2 : Gi\u1ea3 s\u1eed 6 tr\u1eadn \u0111\u1ec1u h\u00f2a th\u00ec s\u1ed1 \u0111i\u1ec3m \u1edf b\u1ea3ng B l\u00e0 : 6 x 2 = 12 (\u0111i\u1ec3m). S\u1ed1 \u0111i\u1ec3m \u1edf b\u1ea3ng B b\u1ecb h\u1ee5t \u0111i : 17 - 12 = 5 (\u0111i\u1ec3m). S\u1edf d\u0129 b\u1ecb h\u1ee5t \u0111i 5 \u0111i\u1ec3m l\u00e0 v\u00ec m\u1ed7i tr\u1eadn h\u00f2a k\u00e9m m\u1ed7i tr\u1eadn th\u1eafng l\u00e0 : 3 - 2 = 1 (\u0111i\u1ec3m). V\u1eady s\u1ed1 tr\u1eadn th\u1eafng l\u00e0 : 5 : 1 = 5 (tr\u1eadn). S\u1ed1 tr\u1eadn h\u00f2a l\u00e0 : 6 - 5 = 1 (tr\u1eadn). B\u00e0i 37 : M\u1ed9t c\u1eeda h\u00e0ng c\u00f3 ba th\u00f9ng A, B, C \u0111\u1ec3 \u0111\u1ef1ng d\u1ea7u. Trong \u0111\u00f3 th\u00f9ng A \u0111\u1ef1ng \u0111\u1ea7y 67","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 d\u1ea7u c\u00f2n th\u00f9ng B v\u00e0 C th\u00ec \u0111ang \u0111\u1ec3 kh\u00f4ng. N\u1ebfu \u0111\u1ed5 d\u1ea7u \u1edf th\u00f9ng A v\u00e0o \u0111\u1ea7y th\u00f9ng B th\u00ec th\u00f9ng A c\u00f2n 2\/5 th\u00f9ng. N\u1ebfu \u0111\u1ed5 d\u1ea7u \u1edf th\u00f9ng A v\u00e0o \u0111\u1ea7y th\u00f9ng C th\u00ec th\u00f9ng A c\u00f2n 5\/9 th\u00f9ng. Mu\u1ed1n \u0111\u1ed5 d\u1ea7u \u1edf th\u00f9ng A v\u00e0o \u0111\u1ea7y c\u1ea3 th\u00f9ng B v\u00e0 th\u00f9ng C th\u00ec ph\u1ea3i th\u00eam 4 l\u00edt n\u1eefa. H\u1ecfi m\u1ed7i th\u00f9ng ch\u1ee9a bao nhi\u00eau l\u00edt d\u1ea7u ? B\u00e0i gi\u1ea3i : So v\u1edbi th\u00f9ng A th\u00ec th\u00f9ng B c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c s\u1ed1 d\u1ea7u l\u00e0 : 1 - 2\/5 = 3\/5 (th\u00f9ng A). Th\u00f9ng C c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c s\u1ed1 d\u1ea7u l\u00e0 : 1 - 5\/9 = 4\/9 (th\u00f9ng A). C\u1ea3 2 th\u00f9ng c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c s\u1ed1 d\u1ea7u nhi\u1ec1u h\u01a1n th\u00f9ng A l\u00e0 : (3\/5 + 4\/9) - 1 = 2\/45 (th\u00f9ng A). 2\/45 s\u1ed1 d\u1ea7u th\u00f9ng A ch\u00ednh l\u00e0 4 l\u00edt d\u1ea7u. Do \u0111\u00f3 s\u1ed1 d\u1ea7u \u1edf th\u00f9ng A l\u00e0 : 4 : 2\/45 = 90 (l\u00edt). Th\u00f9ng B c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c l\u00e0 : 90 x 3\/5 = 54 (l\u00edt). Th\u00f9ng C c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c l\u00e0 : 90 x 4\/9 = 40 (l\u00edt). B\u00e0i 38 : H\u1ea3i h\u1ecfi D\u01b0\u01a1ng : \u201cAnh ph\u1ea3i h\u01a1n 30 tu\u1ed5i ph\u1ea3i kh\u00f4ng ?\u201d. Anh D\u01b0\u01a1ng n\u00f3i : \u201cSao gi\u00e0 th\u1ebf ! N\u1ebfu tu\u1ed5i c\u1ee7a anh nh\u00e2n v\u1edbi 6 th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1, hai ch\u1eef s\u1ed1 cu\u1ed1i ch\u00ednh l\u00e0 tu\u1ed5i anh\u201d. C\u00e1c b\u1ea1n c\u00f9ng H\u1ea3i t\u00ednh tu\u1ed5i c\u1ee7a anh D\u01b0\u01a1ng nh\u00e9. B\u00e0i gi\u1ea3i : C\u00e1ch 1 : Tu\u1ed5i c\u1ee7a anh D\u01b0\u01a1ng kh\u00f4ng qu\u00e1 30, khi nh\u00e2n v\u1edbi 6 s\u1ebd l\u00e0 s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1. V\u1eady ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m c\u1ee7a t\u00edch l\u00e0 1. Hai ch\u1eef s\u1ed1 cu\u1ed1i c\u1ee7a s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 ch\u00ednh l\u00e0 tu\u1ed5i anh. V\u1eady tu\u1ed5i anh D\u01b0\u01a1ng khi nh\u00e2n v\u1edbi 6 h\u01a1n tu\u1ed5i anh D\u01b0\u01a1ng l\u00e0 100 tu\u1ed5i. 68","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 Ta c\u00f3 s\u01a1 \u0111\u1ed3 : Tu\u1ed5i c\u1ee7a anh D\u01b0\u01a1ng l\u00e0 : 100 : (6 - 1) = 20 (tu\u1ed5i) C\u00e1ch 2 : G\u1ecdi tu\u1ed5i c\u1ee7a anh D\u01b0\u01a1ng l\u00e0 (a > 0, a, b l\u00e0 ch\u1eef s\u1ed1) V\u00ec kh\u00f4ng qu\u00e1 30 n\u00ean khi nh\u00e2n v\u1edbi 6 s\u1ebd \u0111\u01b0\u1ee3c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 m\u00e0 ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m l\u00e0 1. Ta c\u00f3 ph\u00e9p t\u00ednh : V\u1eady tu\u1ed5i c\u1ee7a anh D\u01b0\u01a1ng l\u00e0 20. B\u00e0i 39 : \u1edf SEA Games 22 v\u1eeba qua, ch\u1ecb Nguy\u1ec5n Th\u1ecb T\u0129nh gi\u00e0nh Huy ch\u01b0\u01a1ng v\u00e0ng \u1edf c\u1ef1 li 200 m. Bi\u1ebft r\u1eb1ng ch\u1ecb ch\u1ea1y 200 m ch\u1ec9 m\u1ea5t gi\u00e2y. B\u1ea1n h\u00e3y cho bi\u1ebft ch\u1ecb ch\u1ea1y 400 m h\u1ebft bao nhi\u00eau gi\u00e2y ? B\u00e0i gi\u1ea3i : K\u1ebft qu\u1ea3 thi \u0111\u1ea5u \u1edf SEA Games 22 \u0111\u00e3 cho bi\u1ebft : Ch\u1ecb Nguy\u1ec5n Th\u1ecb T\u0129nh ch\u1ea1y c\u1ef1 li 400 m v\u1edbi th\u1eddi gian l\u00e0 51 gi\u00e2y 82. Nh\u1eadn x\u00e9t : D\u1ee5ng \u00fd c\u1ee7a ng\u01b0\u1eddi ra \u0111\u1ec1 l\u00e0 mu\u1ed1n c\u00e1c b\u1ea1n gi\u1ea3i to\u00e1n l\u01b0u \u00fd \u0111\u1ebfn t\u00ednh th\u1ef1c t\u1ebf c\u1ee7a \u0111\u1ec1 to\u00e1n. \u0110\u1ec1 to\u00e1n \u0111\u1ecdc l\u00ean c\u1ee9 nh\u01b0 l\u00e0 lo\u1ea1i to\u00e1n v\u1ec1 t\u01b0\u01a1ng quan t\u1ec9 l\u1ec7 thu\u1eadn. \u0110a s\u1ed1 c\u00e1c b\u1ea1n \u0111\u1ec1u t\u01b0\u1edfng nh\u01b0 v\u1eady n\u00ean \u0111\u00e3 gi\u1ea3i sai, ra \u0111\u00e1p s\u1ed1 l\u00e0 gi\u00e2y (!). B\u00e0i 40 : H\u00e3y kh\u00e1m ph\u00e1 \u201cb\u00ed m\u1eadt\u201d c\u1ee7a h\u00ecnh vu\u00f4ng r\u1ed3i \u0111i\u1ec1n n\u1ed1t b\u1ed1n s\u1ed1 t\u1ef1 nhi\u00ean c\u00f2n thi\u1ebfu v\u00e0o \u00f4 tr\u1ed1ng. B\u00e0i gi\u1ea3i : \u201cB\u00ed m\u1eadt\u201d c\u1ee7a h\u00ecnh vu\u00f4ng l\u00e0 t\u1ed5ng c\u00e1c s\u1ed1 h\u00e0ng ngang, h\u00e0ng d\u1ecdc v\u00e0 \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a h\u00ecnh vu\u00f4ng \u0111\u1ec1u b\u1eb1ng 34 (c\u00e1c b\u1ea1n t\u1ef1 ki\u1ec3m tra l\u1ea1i). G\u1ecdi c\u00e1c s\u1ed1 c\u1ea7n t\u00ecm \u1edf 4 g\u00f3c c\u1ee7a h\u00ecnh vu\u00f4ng l\u00e0 a, b, c, d. \u1edf h\u00e0ng ngang \u0111\u1ea7u ti\u00ean, ta c\u00f3 : a + 3 69","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 + 2 + b = 34, t\u1eeb \u0111\u00f3 a + b = 34 - 5 = 29 (1). \u1edf c\u1ed9t d\u1ecdc \u0111\u1ea7u ti\u00ean ta c\u00f3 : a + 5 + 9 + d = 34, t\u1eeb \u0111\u00f3 a + d = 34 - 14 = 20 (2). T\u1eeb (1) v\u00e0 (2) ta c\u00f3 : a + b - (a + d) = 29 - 20 = 9 hay b - d = 9 (3). \u1edf m\u1ed9t \u0111\u01b0\u1eddng ch\u00e9o, ta l\u1ea1i c\u00f3 : b + 6 + 11 + d = 34, t\u1eeb \u0111\u00f3 b + d = 34 - 17 = 17 (4). T\u1eeb (3) v\u00e0 (4) ta c\u00f3 : (b - d) + (b + d) = 9 + 17 hay b + b = 26 ; b = 13. V\u00ec b + d = 17 n\u00ean d = 17 - 13 = 4. V\u00ec a + b = 29 n\u00ean a = 29 - 13 = 16. \u1edf \u0111\u01b0\u1eddng ch\u00e9o th\u1ee9 hai, ta c\u00f3 a + 10 + 7 + c = 34 hay a + c = 34 - 17 = 17. T\u1eeb \u0111\u00f3 c = 17 - 16 = 1. Thay a, b, c, d b\u1eb1ng c\u00e1c s\u1ed1 v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c ta c\u00f3 h\u00ecnh vu\u00f4ng sau : Nh\u1eadn x\u00e9t : H\u00ecnh vu\u00f4ng tr\u00ean g\u1ecdi l\u00e0 h\u00ecnh vu\u00f4ng k\u00ec \u1ea3o (ho\u1eb7c ma ph\u01b0\u01a1ng) c\u1ea5p 4. Ng\u01b0\u1eddi ta \u0111\u00e3 nh\u00ecn th\u1ea5y n\u00f3 l\u1ea7n \u0111\u1ea7u ti\u00ean trong b\u1ea3n kh\u1eafc c\u1ee7a h\u1ecda s\u0129 \u0110uy-r\u01a1 n\u0103m 1514. C\u00e1c b\u1ea1n c\u00f3 th\u1ec3 th\u1ea5y : T\u1ed5ng b\u1ed1n s\u1ed1 trong b\u1ed1n \u00f4 \u1edf b\u1ed1n g\u00f3c c\u0169ng b\u1eb1ng 34. B\u00e0i 41 : B\u1ea1n c\u00f3 th\u1ec3 c\u1eaft h\u00ecnh n\u00e0y : th\u00e0nh 16 h\u00ecnh: B\u1ea1n h\u00e3y n\u00f3i r\u00f5 c\u00e1ch c\u1eaft nh\u00e9 ! B\u00e0i gi\u1ea3i : T\u1ed5ng s\u1ed1 \u00f4 vu\u00f4ng l\u00e0 : 8 x 8 = 64 (\u00f4) Khi ta c\u1eaft h\u00ecnh vu\u00f4ng ban \u0111\u1ea7u th\u00e0nh c\u00e1c ph\u1ea7n nh\u1ecf (h\u00ecnh ch\u1eef T), m\u1ed7i ph\u1ea7n g\u1ed3m 4 \u00f4 vu\u00f4ng th\u00ec s\u1ebd \u0111\u01b0\u1ee3c s\u1ed1 h\u00ecnh l\u00e0 : 64 : 4 = 16 (h\u00ecnh) 70","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 m\u1ed9t c\u00e1ch c\u1eaft nh\u01b0 Ta c\u00f3 th\u1ec3 c\u1eaft theo nhi\u1ec1u c\u00e1ch kh\u00e1c nhau. Xin n\u00eau sau : B\u00e0i 42 : Cho h\u00ecnh vu\u00f4ng nh\u01b0 h\u00ecnh v\u1ebd. Em h\u00e3y thay c\u00e1c ch\u1eef b\u1edfi c\u00e1c s\u1ed1 th\u00edch h\u1ee3p sao cho t\u1ed5ng c\u00e1c s\u1ed1 \u1edf c\u00e1c \u00f4 thu\u1ed9c h\u00e0ng ngang, c\u1ed9t d\u1ecdc, \u0111\u01b0\u1eddng ch\u00e9o \u0111\u1ec1u b\u1eb1ng nhau. B\u00e0i gi\u1ea3i : V\u00ec t\u1ed5ng c\u00e1c s\u1ed1 \u1edf h\u00e0ng ngang, c\u1ed9t d\u1ecdc, \u0111\u01b0\u1eddng ch\u00e9o \u0111\u1ec1u b\u1eb1ng nhau n\u00ean ta c\u00f3 : a + 35 + b = a + 9 + d hay 26 + b = d (c\u00f9ng tr\u1eeb 2 v\u1ebf \u0111i a v\u00e0 9). Do \u0111\u00f3 d - b = 26. b + g + d = 35 + g + 13 hay b + d = 48. V\u1eady b = (48 - 26 ) : 2 = 11, d = 48 - 11 = 37. d + 13 + c = d + 9 + a hay 4 + c = a (c\u00f9ng tr\u1eeb 2 v\u1ebf \u0111i d v\u00e0 9). Do \u0111\u00f3 a - c = 4, a + g + c = 9 + g +39 hay a + c = 9 + 39 (c\u00f9ng tr\u1eeb 2 v\u1ebf \u0111i g), do \u0111\u00f3 a + c = 48. V\u1eady c = (48 - 4) : 2 = 22, a = 22 + 4 = 26. 35 + g + 13 = a + 35 + b = 26 + 35 + 11 = 72. Do \u0111\u00f3 48 + g = 72 ; g = 72 - 48 = 24. Thay a = 26, b = 11, c = 22, d =37 , g = 24 v\u00e0o h\u00ecnh v\u1ebd ta c\u00f3 : B\u00e0i 43 : S\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u1ee7a m\u1ed9t quy\u1ec3n s\u00e1ch b\u1eb1ng \u0111\u00fang 2 l\u1ea7n s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3. H\u1ecfi cu\u1ed1n s\u00e1ch \u0111\u00f3 c\u00f3 bao nhi\u00eau trang ? B\u00e0i gi\u1ea3i : \u0110\u1ec3 s\u1ed1 ch\u1eef s\u1ed1 b\u1eb1ng \u0111\u00fang 2 l\u1ea7n s\u1ed1 trang quy\u1ec3n s\u00e1ch th\u00ec trung b\u00ecnh m\u1ed7i trang ph\u1ea3i d\u00f9ng hai ch\u1eef s\u1ed1. T\u1eeb trang 1 \u0111\u1ebfn trang 9 c\u00f3 9 trang g\u1ed3m m\u1ed9t ch\u1eef s\u1ed1, n\u00ean c\u00f2n thi\u1ebfu 9 ch\u1eef s\u1ed1. T\u1eeb trang 10 \u0111\u1ebfn trang 99 c\u00f3 90 trang, m\u1ed7i trang \u0111\u1ee7 hai ch\u1eef s\u1ed1. T\u1eeb trang 100 tr\u1edf \u0111i m\u1ed7i trang c\u00f3 3 ch\u1eef s\u1ed1, m\u1ed7i trang th\u1eeba m\u1ed9t ch\u1eef s\u1ed1, n\u00ean ph\u1ea3i c\u00f3 9 trang \u0111\u1ec3 \u201cb\u00f9\u201d \u0111\u1ee7 cho 9 trang g\u1ed3m m\u1ed9t ch\u1eef s\u1ed1. 71","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 V\u1eady quy\u1ec3n s\u00e1ch c\u00f3 s\u1ed1 trang l\u00e0 : 9 + 90 + 9 = 108 (trang). B\u00e0i 44 : Ng\u01b0\u1eddi ta ng\u0103n th\u1eeda \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh 2 m\u1ea3nh, m\u1ed9t m\u1ea3nh h\u00ecnh vu\u00f4ng, m\u1ed9t m\u1ea3nh h\u00ecnh ch\u1eef nh\u1eadt. Bi\u1ebft chu vi ban \u0111\u1ea7u h\u01a1n chu vi m\u1ea3nh \u0111\u1ea5t h\u00ecnh vu\u00f4ng l\u00e0 28 m. Di\u1ec7n t\u00edch c\u1ee7a th\u1eeda \u0111\u1ea5t ban \u0111\u1ea7u h\u01a1n di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u00e0 224 m2. T\u00ednh di\u1ec7n t\u00edch th\u1eeda \u0111\u1ea5t ban \u0111\u1ea7u. B\u00e0i gi\u1ea3i : N\u1eeda chu vi h\u00ecnh ABCD h\u01a1n n\u1eeda chu vi h\u00ecnh AMND l\u00e0 : 28 : 2 = 14 (m). N\u1eeda chu vi h\u00ecnh ABCD l\u00e0 AD + AB. N\u1eeda chu vi h\u00ecnh AMND l\u00e0 AD + AM. Do \u0111\u00f3 : MB = AB - AM = 14 (m). Chi\u1ec1u r\u1ed9ng BC c\u1ee7a h\u00ecnh ABCD l\u00e0 : 224 : 14 = 16 (m) Chi\u1ec1u d\u00e0i AB c\u1ee7a h\u00ecnh ABCD l\u00e0 : 16 + 14 = 30 (m) Di\u1ec7n t\u00edch h\u00ecnh ABCD l\u00e0 : 30 x 16 = 480 (m2). B\u00e0i 45 : Trong m\u1ed9t h\u1ed9i ngh\u1ecb c\u00f3 100 ng\u01b0\u1eddi tham d\u1ef1, trong \u0111\u00f3 c\u00f3 10 ng\u01b0\u1eddi kh\u00f4ng bi\u1ebft ti\u1ebfng Nga v\u00e0 ti\u1ebfng Anh, c\u00f3 75 ng\u01b0\u1eddi bi\u1ebft ti\u1ebfng Nga v\u00e0 83 ng\u01b0\u1eddi bi\u1ebft Ti\u1ebfng Anh. H\u1ecfi trong h\u1ed9i ngh\u1ecb c\u00f3 bao nhi\u00eau ng\u01b0\u1eddi bi\u1ebft c\u1ea3 2 th\u1ee9 ti\u1ebfng Nga v\u00e0 Anh ? B\u00e0i gi\u1ea3i : C\u00e1ch 1 : S\u1ed1 ng\u01b0\u1eddi bi\u1ebft \u00edt nh\u1ea5t 1 trong 2 th\u1ee9 ti\u1ebfng Nga v\u00e0 Anh l\u00e0 : 100 - 10 = 90 (ng\u01b0\u1eddi). S\u1ed1 ng\u01b0\u1eddi ch\u1ec9 bi\u1ebft ti\u1ebfng Anh l\u00e0 : 90 - 75 = 15 (ng\u01b0\u1eddi) S\u1ed1 ng\u01b0\u1eddi bi\u1ebft c\u1ea3 ti\u1ebfng Nga v\u00e0 ti\u1ebfng Anh l\u00e0 : 83 - 15 = 68 (ng\u01b0\u1eddi) C\u00e1ch 2 : S\u1ed1 ng\u01b0\u1eddi bi\u1ebft \u00edt nh\u1ea5t m\u1ed9t trong 2 th\u1ee9 ti\u1ebfng l\u00e0 : 100 - 10 = 90 (ng\u01b0\u1eddi). S\u1ed1 ng\u01b0\u1eddi ch\u1ec9 bi\u1ebft ti\u1ebfng Nga l\u00e0 : 90 - 83 = 7 (ng\u01b0\u1eddi). S\u1ed1 ng\u01b0\u1eddi ch\u1ec9 bi\u1ebft ti\u1ebfng Anh l\u00e0 : 90 - 75 = 15 (ng\u01b0\u1eddi). S\u1ed1 ng\u01b0\u1eddi bi\u1ebft c\u1ea3 2 th\u1ee9 ti\u1ebfng Nga v\u00e0 Anh l\u00e0 : 90 - (7 + 15) = 68 (ng\u01b0\u1eddi) B\u00e0i 46 : M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00e3 b\u1ecb c\u1eaft \u0111i m\u1ed9t h\u00ecnh vu\u00f4ng \u1edf m\u1ed9t g\u00f3c. Ch\u1ec9 c\u1ea7n m\u1ed9t nh\u00e1t c\u1eaft 72","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 th\u1eb3ng, b\u1ea1n h\u00e3y chia ph\u1ea7n c\u00f2n l\u1ea1i th\u00e0nh 2 ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Gi\u1ea3i : Ch\u1ec9 c\u1ea7n c\u00e1c b\u1ea1n bi\u1ebft \u0111\u01b0\u1ee3c t\u00ednh ch\u1ea5t: M\u1ecdi \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua t\u00e2m c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt \u0111\u1ec3 chia h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh hai h\u00ecnh c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. C\u00f3 th\u1ec3 chia \u0111\u01b0\u1ee3c b\u1eb1ng nhi\u1ec1u c\u00e1ch: B\u00e0i 47 : Cho bi\u1ebft : 4 x 396 x 0,25 : (x + 0,75) = 1,32. H\u00e3y t\u00ecm c\u00e1ch \u0111\u1eb7t th\u00eam m\u1ed9t d\u1ea5u ph\u1ea9y v\u00e0o ch\u1ed7 n\u00e0o \u0111\u00f3 trong \u0111\u1eb3ng th\u1ee9c tr\u00ean \u0111\u1ec3 gi\u00e1 tr\u1ecb c\u1ee7a x gi\u1ea3m 297 \u0111\u01a1n v\u1ecb. B\u00e0i gi\u1ea3i : Theo \u0111\u1ec1 b\u00e0i : 4 x 396 x 0,25 : (x + 0,75) = 1,32 ; v\u00ec 4 x 0,25 = 1 n\u00ean ta c\u00f3 : 396 : (x + 0,75) = 1,32 hay x + 0,75 = 396 : 1,32 = 300. Khi x gi\u1ea3m \u0111i 297 \u0111\u01a1n v\u1ecb th\u00ec t\u1ed5ng x + 0,75 c\u0169ng gi\u1ea3m \u0111i 297 \u0111\u01a1n v\u1ecb, t\u1ee9c l\u00e0 x + 0,75 = 300 - 297 = 3 hay x = 3 - 0,75 = 2,25. Trong \u0111\u1eb3ng th\u1ee9c x + 0,75 = 396 : 1,32 ; \u0111\u1ec3 x = 2,25 th\u00ec ph\u1ea3i th\u00eam d\u1ea5u ph\u1ea9y v\u00e0o s\u1ed1 396 \u0111\u1ec3 c\u00f3 s\u1ed1 3,96. Nh\u01b0 v\u1eady c\u1ea7n \u0111\u1eb7t th\u00eam d\u1ea5u ph\u1ea9y v\u00e0o gi\u1eefa ch\u1eef s\u1ed1 3 v\u00e0 9 c\u1ee7a s\u1ed1 396 \u0111\u1ec3 x gi\u1ea3m \u0111i 297 \u0111\u01a1n v\u1ecb. C\u00e1c b\u1ea1n c\u00f3 th\u1ec3 th\u1eed l\u1ea1i. B\u00e0i 48 : \u0110i\u1ec1n \u0111\u1ee7 9 ch\u1eef s\u1ed1 : 1, 2, 3, 4, 5, 6, 7, 8, 9 v\u00e0o 9 \u00f4 tr\u1ed1ng sau \u0111\u1ec3 \u0111\u01b0\u1ee3c ph\u00e9p t\u00ednh \u0111\u00fang : B\u00e0i gi\u1ea3i : B\u00e0i to\u00e1n ch\u1ec9 c\u00f3 b\u1ed1n c\u00e1ch \u0111i\u1ec1n nh\u01b0 sau : 2 x 78 = 156 = 39 x 4 4 x 39 = 156 = 78 x 2 3 x 58 = 174 = 29 x 6 6 x 29 = 174 = 58 x 3 73","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i 49 : T\u00ednh tu\u1ed5i c\u1ee7a \u00f4ng bi\u1ebft: Th\u1eddi ni\u00ean thi\u1ebfu chi\u1ebfm 1\/5 qu\u00e3ng \u0111\u1eddi c\u1ee7a \u00f4ng, 1\/8 qu\u00e3ng \u0111\u1eddi c\u00f2n l\u1ea1i l\u00e0 tu\u1ed5i sinh vi\u00ean, 1\/7 s\u1ed1 tu\u1ed5i c\u00f2n l\u1ea1i \u00f4ng \u0111\u01b0\u1ee3c h\u1ecdc \u1edf tr\u01b0\u1eddng qu\u00e2n \u0111\u1ed9i. Ti\u1ebfp theo \u00f4ng \u0111\u01b0\u1ee3c r\u00e8n luy\u1ec7n 7 n\u0103m li\u1ec1n v\u00e0 sau \u0111\u00f3 \u0111\u01b0\u1ee3c vinh d\u1ef1 tr\u1ef1c ti\u1ebfp \u0111\u00e1nh M\u0129. Nh\u01b0 v\u1eady th\u1eddi gian \u0111\u00e1nh M\u0129 v\u1eeba tr\u00f2n 1\/2 qu\u00e3ng \u0111\u1eddi c\u1ee7a \u00f4ng. B\u00e0i gi\u1ea3i : Ph\u00e2n s\u1ed1 ch\u1ec9 s\u1ed1 tu\u1ed5i c\u00f2n l\u1ea1i sau th\u1eddi ni\u00ean thi\u1ebfu c\u1ee7a \u00f4ng l\u00e0 : 1- 1\/5 = 1\/4 (s\u1ed1 tu\u1ed5i \u00f4ng) Th\u1eddi sinh vi\u00ean c\u1ee7a \u00f4ng c\u00f3 s\u1ed1 n\u0103m l\u00e0 : 4\/5 x 1\/8 = 1\/10 (s\u1ed1 tu\u1ed5i \u00f4ng) S\u1ed1 n\u0103m c\u00f2n l\u1ea1i sau th\u1eddi sinh vi\u00ean c\u1ee7a \u00f4ng l\u00e0 : 4\/5 - 1\/10 = 7\/10 (s\u1ed1 tu\u1ed5i \u00f4ng) S\u1ed1 n\u0103m h\u1ecdc \u1edf tr\u01b0\u1eddng qu\u00e2n \u0111\u1ed9i c\u1ee7a \u00f4ng l\u00e0 : 7\/10 x 1\/7 = 1\/10 (s\u1ed1 tu\u1ed5i \u00f4ng) Do \u0111\u00f3: 7 n\u0103m r\u00e8n luy\u1ec7n c\u1ee7a \u00f4ng l\u00e0 : 1 - (1\/5 + 1\/10 + 1\/10 + 1\/2) = 1\/10 (s\u1ed1 tu\u1ed5i \u00f4ng) Suy ra s\u1ed1 tu\u1ed5i c\u1ee7a \u00f4ng l\u00e0 : 7: 1\/10 = 70 (tu\u1ed5i). B\u00e0i 50 : M\u1ed9t mi\u1ebfng b\u00eca h\u00ecnh ch\u1eef nh\u1eadt, c\u00f3 chi\u1ec1u r\u1ed9ng 30 cm, chi\u1ec1u d\u00e0i 40 cm. Ng\u01b0\u1eddi ta mu\u1ed1n c\u1eaft \u0111i m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt n\u1eb1m ch\u00ednh gi\u1eefa mi\u1ebfng b\u00eca tr\u00ean sao cho c\u1ea1nh c\u1ee7a hai h\u00ecnh ch\u1eef nh\u1eadt song song v\u00e0 c\u00e1ch \u0111\u1ec1u nhau, \u0111\u1ed3ng th\u1eddi di\u1ec7n t\u00edch c\u1eaft \u0111i b\u1eb1ng 1\/2 di\u1ec7n t\u00edch mi\u1ebfng b\u00eca ban \u0111\u1ea7u. H\u1ecfi hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng c\u1ee7a hai h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u v\u00e0 c\u1eaft \u0111i c\u00e1ch nhau bao nhi\u00eau ? B\u00e0i gi\u1ea3i : Chia mi\u1ebfng b\u00eca ABCD th\u00e0nh c\u00e1c \u00f4 vu\u00f4ng, m\u1ed7i \u00f4 vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 5 cm. S\u1ed1 \u00f4 vu\u00f4ng c\u1ee7a mi\u1ebfng b\u00eca \u0111\u00f3 l\u00e0 : 8 x 6 = 48 (\u00f4 vu\u00f4ng). S\u1ed1 \u00f4 vu\u00f4ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt MNPQ l\u00e0 : 6 x 4 = 24 (\u00f4 vu\u00f4ng) V\u00ec 48 : 24 = 2 (l\u1ea7n) n\u00ean h\u00ecnh ch\u1eef nh\u1eadt MNPQ c\u00f3 di\u1ec7n t\u00edch \u0111\u00fang b\u1eb1ng di\u1ec7n t\u00edch h\u00ecnh c\u1eaft \u0111i. M\u1eb7t kh\u00e1c c\u00e1c c\u1ea1nh c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt MNPQ song song v\u00e0 c\u00e1ch \u0111\u1ec1u c\u00e1c c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng c\u1ee7a mi\u1ebfng b\u00eca ABCD. V\u00ec v\u1eady h\u00ecnh MNPQ \u0111\u00fang l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt b\u1ecb c\u1eaft \u0111i. M\u1ed7i c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng c\u1ee7a h\u00ecnh ABCD v\u00e0 MNPQ c\u00e1ch nhau 5 cm. B\u00e0i 51 : T\u00ecm 4 s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 t\u1ed5ng b\u1eb1ng 2003. Bi\u1ebft r\u1eb1ng n\u1ebfu x\u00f3a b\u1ecf ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a s\u1ed1 th\u1ee9 nh\u1ea5t ta \u0111\u01b0\u1ee3c s\u1ed1 th\u1ee9 hai. N\u1ebfu x\u00f3a b\u1ecf ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a s\u1ed1 th\u1ee9 hai ta \u0111\u01b0\u1ee3c s\u1ed1 th\u1ee9 ba. N\u1ebfu x\u00f3a b\u1ecf ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a s\u1ed1 th\u1ee9 ba ta \u0111\u01b0\u1ee3c s\u1ed1 th\u1ee9 t\u01b0. B\u00e0i gi\u1ea3i : S\u1ed1 th\u1ee9 nh\u1ea5t kh\u00f4ng th\u1ec3 nhi\u1ec1u h\u01a1n 4 ch\u1eef s\u1ed1 v\u00ec t\u1ed5ng 4 s\u1ed1 b\u1eb1ng 2003. N\u1ebfu s\u1ed1 th\u1ee9 nh\u1ea5t c\u00f3 \u00edt h\u01a1n 4 ch\u1eef s\u1ed1 th\u00ec s\u1ebd kh\u00f4ng t\u1ed3n t\u1ea1i s\u1ed1 th\u1ee9 t\u01b0. V\u1eady s\u1ed1 th\u1ee9 nh\u1ea5t ph\u1ea3i c\u00f3 4 ch\u1eef s\u1ed1. G\u1ecdi s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 abcd (a > 0, a, b, c, d < 10). S\u1ed1 th\u1ee9 hai, s\u1ed1 th\u1ee9 ba, s\u1ed1 th\u1ee9 t\u01b0 l\u1ea7n l\u01b0\u1ee3t s\u1ebd l\u00e0 : abc ; ab ; a. Theo b\u00e0i ra ta c\u00f3 ph\u00e9p t\u00ednh : abcd + abc + ab + a = 2003. Theo ph\u00e2n t\u00edch c\u1ea5u t\u1ea1o s\u1ed1 ta c\u00f3 : aaaa + bbb + cc + d = 2003 (*) T\u1eeb ph\u00e9p t\u00ednh (*) ta c\u00f3 a < 2, n\u00ean a = 1. Thay a = 1 v\u00e0o (*) ta \u0111\u01b0\u1ee3c : 1111 + bbb + cc + d = 2003. bbb + cc + d = 2003 - 1111 bbb + cc + d = 892 (**) 74","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 b > 7 v\u00ec n\u1ebfu b nh\u1ecf h\u01a1n ho\u1eb7c b\u1eb1ng 7 th\u00ec bbb + cc + d nh\u1ecf h\u01a1n 892 ; b < 9 v\u00ec n\u1ebfu b = 9 th\u00ec bbb = 999 > 892. Suy ra b ch\u1ec9 c\u00f3 th\u1ec3 b\u1eb1ng 8. Thay b = 8 v\u00e0o (**) ta \u0111\u01b0\u1ee3c : 888 + cc + d = 892 cc + d = 892 - 888 cc + d = 4 T\u1eeb \u0111\u00e2y suy ra c ch\u1ec9 c\u00f3 th\u1ec3 b\u1eb1ng 0 v\u00e0 d = 4. V\u1eady s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 1804, s\u1ed1 th\u1ee9 hai l\u00e0 180, s\u1ed1 th\u1ee9 ba l\u00e0 18 v\u00e0 s\u1ed1 th\u1ee9 t\u01b0 l\u00e0 1. Th\u1eed l\u1ea1i : 1804 + 180 + 18 + 1 = 2003 (\u0111\u00fang) B\u00e0i 52 : M\u1ed9t ng\u01b0\u1eddi mang ra ch\u1ee3 5 gi\u1ecf t\u00e1o g\u1ed3m hai lo\u1ea1i. S\u1ed1 t\u00e1o trong m\u1ed7i gi\u1ecf l\u1ea7n l\u01b0\u1ee3t l\u00e0 : 20 ; 25 ; 30 ; 35 v\u00e0 40. M\u1ed7i gi\u1ecf ch\u1ec9 \u0111\u1ef1ng m\u1ed9t lo\u1ea1i t\u00e1o. Sau khi b\u00e1n h\u1ebft m\u1ed9t gi\u1ecf t\u00e1o n\u00e0o \u0111\u00f3, ng\u01b0\u1eddi \u1ea5y th\u1ea5y r\u1eb1ng : S\u1ed1 t\u00e1o lo\u1ea1i 2 c\u00f2n l\u1ea1i \u0111\u00fang b\u1eb1ng n\u1eeda s\u1ed1 t\u00e1o lo\u1ea1i 1. H\u1ecfi s\u1ed1 t\u00e1o lo\u1ea1i 2 c\u00f2n l\u1ea1i l\u00e0 bao nhi\u00eau ? B\u00e0i gi\u1ea3i : S\u1ed1 t\u00e1o ng\u01b0\u1eddi \u0111\u00f3 mang ra ch\u1ee3 l\u00e0 : 20 + 25 + 30 + 35 + 40 = 150 (qu\u1ea3) V\u00ec s\u1ed1 t\u00e1o lo\u1ea1i 2 c\u00f2n l\u1ea1i \u0111\u00fang b\u1eb1ng n\u1eeda s\u1ed1 t\u00e1o lo\u1ea1i 1 n\u00ean sau khi b\u00e1n, s\u1ed1 t\u00e1o c\u00f2n l\u1ea1i ph\u1ea3i chia h\u1ebft cho 3. V\u00ec t\u1ed5ng s\u1ed1 t\u00e1o mang ra ch\u1ee3 l\u00e0 150 qu\u1ea3 chia h\u1ebft cho 3 n\u00ean s\u1ed1 t\u00e1o \u0111\u00e3 b\u00e1n ph\u1ea3i chia h\u1ebft cho 3. Trong c\u00e1c s\u1ed1 20, 25, 30, 35, 40 ch\u1ec9 c\u00f3 30 chia h\u1ebft cho 3. Do v\u1eady ng\u01b0\u1eddi \u1ea5y \u0111\u00e3 b\u00e1n gi\u1ecf t\u00e1o \u0111\u1ef1ng 30 qu\u1ea3. T\u1ed5ng s\u1ed1 t\u00e1o c\u00f2n l\u1ea1i l\u00e0 : 150 - 30 = 120 (qu\u1ea3) Ta c\u00f3 s\u01a1 \u0111\u1ed3 bi\u1ec3u di\u1ec5n s\u1ed1 t\u00e1o c\u1ee7a lo\u1ea1i 1 v\u00e0 lo\u1ea1i 2 c\u00f2n l\u1ea1i : S\u1ed1 t\u00e1o lo\u1ea1i 2 c\u00f2n l\u1ea1i l\u00e0 : 120 : (2 + 1) = 40 (qu\u1ea3) V\u1eady ng\u01b0\u1eddi \u1ea5y c\u00f2n l\u1ea1i gi\u1ecf \u0111\u1ef1ng 40 qu\u1ea3 ch\u00ednh l\u00e0 s\u1ed1 t\u00e1o lo\u1ea1i 2 c\u00f2n l\u1ea1i. \u0110\u00e1p s\u1ed1 : 40 qu\u1ea3 B\u00e0i 53 : Kh\u00f4ng \u0111\u01b0\u1ee3c thay \u0111\u1ed5i v\u1ecb tr\u00ed c\u1ee7a c\u00e1c ch\u1eef s\u1ed1 \u0111\u00e3 vi\u1ebft tr\u00ean b\u1ea3ng : 8 7 6 5 4 3 2 1 m\u00e0 ch\u1ec9 \u0111\u01b0\u1ee3c vi\u1ebft th\u00eam c\u00e1c d\u1ea5u c\u1ed9ng (+), b\u1ea1n c\u00f3 th\u1ec3 cho \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 c\u1ee7a d\u00e3y ph\u00e9p t\u00ednh l\u00e0 90 \u0111\u01b0\u1ee3c kh\u00f4ng ? B\u00e0i gi\u1ea3i : C\u00f3 hai c\u00e1ch \u0111i\u1ec1n : 8 + 7 + 65 + 4 + 3 + 2 + 1 = 90 75","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 8 + 7 + 6 + 5 + 43 + 21 = 90 \u0110\u1ec3 t\u00ecm \u0111\u01b0\u1ee3c hai c\u00e1ch \u0111i\u1ec1n n\u00e0y ta c\u00f3 th\u1ec3 c\u00f3 nh\u1eadn x\u00e9t sau : T\u1ed5ng 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 ; 90 - 36 = 54. Nh\u01b0 v\u1eady mu\u1ed1n c\u00f3 t\u1ed5ng 90 th\u00ec trong c\u00e1c s\u1ed1 h\u1ea1ng ph\u1ea3i c\u00f3 m\u1ed9t ho\u1eb7c hai s\u1ed1 l\u00e0 s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1. N\u1ebfu s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 \u0111\u00f3 l\u00e0 87 ho\u1eb7c 76 m\u00e0 87 > 54, 76 > 54 n\u00ean kh\u00f4ng th\u1ec3 \u0111\u01b0\u1ee3c. N\u1ebfu s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 l\u00e0 65 ; 65 + 36 - 6 - 5 = 90, ta c\u00f3 th\u1ec3 \u0111i\u1ec1n : 8 + 7 + 65 + 4 + 3 + 2 + 1 - 90. N\u1ebfu s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 l\u00e0 54 th\u00ec c\u0169ng kh\u00f4ng th\u1ec3 c\u00f3 t\u1ed5ng l\u00e0 90 \u0111\u01b0\u1ee3c v\u00ec 54 + 36 - 5 - 4 < 90. N\u1ebfu s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 l\u00e0 43 ; 43 < 54 n\u00ean c\u0169ng kh\u00f4ng th\u1ec3 \u0111\u01b0\u1ee3c. N\u1ebfu trong t\u1ed5ng c\u00f3 2 s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 l\u00e0 43 v\u00e0 21 th\u00ec ta c\u00f3 43 + 21 - (4 + 3 + 2 + 1) = 54. Nh\u01b0 v\u1eady ta c\u00f3 th\u1ec3 \u0111i\u1ec1n : 8 + 7 + 6 + 5 + 43 + 21 = 90. B\u00e0i 54 : Cho ph\u00e2n s\u1ed1 M = (1 + 2 +... + 9)\/(11 + 12 +... +19). H\u00e3y b\u1edbt m\u1ed9t s\u1ed1 h\u1ea1ng \u1edf t\u1eed s\u1ed1 v\u00e0 m\u1ed9t s\u1ed1 h\u1ea1ng \u1edf m\u1eabu s\u1ed1 sao cho gi\u00e1 tr\u1ecb ph\u00e2n s\u1ed1 kh\u00f4ng thay \u0111\u1ed5i. T\u00f3m t\u1eaft b\u00e0i gi\u1ea3i : M = (1 + 2 +... + 9)\/(11 + 12 +... +19) = 45\/135 = 1\/3. Theo t\u00ednh ch\u1ea5t c\u1ee7a hai t\u1ec9 s\u1ed1 b\u1eb1ng nhau th\u00ec 45\/135 = (45 - k)\/(135 - kx3)(k l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf h\u01a1n 45). Do \u0111\u00f3 \u1edf t\u1eed s\u1ed1 c\u1ee7a M b\u1edbt \u0111i 4 ; 5 ; 6 th\u00ec t\u01b0\u01a1ng \u1ee9ng \u1edf m\u1eabu s\u1ed1 ph\u1ea3i b\u1edbt \u0111i 12 ; 15 ; 18. suy ra B x B = 289. V\u1eady B = 17 (v\u00ec 17 x 17 = 289). B\u00e0i 57 : C\u1ea3 l\u1edbp 4A ph\u1ea3i l\u00e0m m\u1ed9t b\u00e0i ki\u1ec3m tra to\u00e1n g\u1ed3m c\u00f3 3 b\u00e0i to\u00e1n. Gi\u00e1o vi\u00ean ch\u1ee7 nhi\u1ec7m l\u1edbp b\u00e1o c\u00e1o v\u1edbi nh\u00e0 tr\u01b0\u1eddng r\u1eb1ng : c\u1ea3 l\u1edbp m\u1ed7i em \u0111\u1ec1u l\u00e0m \u0111\u01b0\u1ee3c \u00edt nh\u1ea5t m\u1ed9t b\u00e0i, trong l\u1edbp c\u00f3 20 em gi\u1ea3i \u0111\u01b0\u1ee3c b\u00e0i to\u00e1n th\u1ee9 nh\u1ea5t, 14 em gi\u1ea3i \u0111\u01b0\u1ee3c b\u00e0i to\u00e1n th\u1ee9 hai, 10 em gi\u1ea3i \u0111\u01b0\u1ee3c b\u00e0i to\u00e1n th\u1ee9 ba, 5 em gi\u1ea3i \u0111\u01b0\u1ee3c b\u00e0i to\u00e1n th\u1ee9 hai v\u00e0 th\u1ee9 ba, 2 em gi\u1ea3i \u0111\u01b0\u1ee3c b\u00e0i to\u00e1n th\u1ee9 nh\u1ea5t v\u00e0 th\u1ee9 hai, c\u00f3 m\u1ed7i m\u1ed9t em \u0111\u01b0\u1ee3c 10 \u0111i\u1ec3m v\u00ec \u0111\u00e3 gi\u1ea3i \u0111\u01b0\u1ee3c c\u1ea3 ba b\u00e0i. H\u1ecfi r\u1eb1ng l\u1edbp h\u1ecdc \u0111\u00f3 c\u00f3 bao nhi\u00eau em t\u1ea5t c\u1ea3 ? B\u00e0i gi\u1ea3i : M\u1ed7i h\u00ecnh tr\u00f2n \u0111\u1ec3 ghi s\u1ed1 b\u1ea1n gi\u1ea3i \u0111\u00fang m\u1ed9t b\u00e0i n\u00e0o \u0111\u00f3. V\u00ec ch\u1ec9 c\u00f3 m\u1ed9t b\u1ea1n gi\u1ea3i \u0111\u00fang 3 b\u00e0i n\u00ean \u0111i\u1ec1n s\u1ed1 1 v\u00e0o ph\u1ea7n chung c\u1ee7a 3 h\u00ecnh tr\u00f2n. S\u1ed1 b\u1ea1n gi\u1ea3i \u0111\u00fang b\u00e0i I v\u00e0 b\u00e0i II l\u00e0 2 n\u00ean ph\u1ea7n chung c\u1ee7a hai h\u00ecnh tr\u00f2n n\u00e0y m\u00e0 kh\u00f4ng chung v\u1edbi h\u00ecnh tr\u00f2n c\u00f2n l\u1ea1i s\u1ebd \u0111\u01b0\u1ee3c ghi 76","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 s\u1ed1 1 (v\u00ec 2 - 1 = 1). T\u01b0\u01a1ng t\u1ef1, ta ghi \u0111\u01b0\u1ee3c c\u00e1c s\u1ed1 v\u00e0o c\u00e1c ph\u1ea7n c\u00f2n l\u1ea1i. S\u1ed1 h\u1ecdc sinh l\u1edbp 4A ch\u00ednh l\u00e0 t\u1ed5ng c\u00e1c s\u1ed1 \u0111\u00e3 \u0111i\u1ec1n v\u00e0o c\u00e1c ph\u1ea7n : 13 + 5 + 1 + 1 + 4 + 8 + 0 = 32 (HS) B\u00e0i 59 : S = 1\/2 + 1\/3 + 1\/4 + 1\/5 + 1\/6 + 1\/7 + 1\/8 c\u00f3 ph\u1ea3i l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean kh\u00f4ng ? V\u00ec sao ? B\u00e0i gi\u1ea3i : C\u00e1c b\u1ea1n \u0111\u00e3 gi\u1ea3i theo 3 h\u01b0\u1edbng sau \u0111\u00e2y : H\u01b0\u1edbng 1 : T\u00ednh S = 1 201\/280 H\u01b0\u1edbng 2 : Khi qui \u0111\u1ed3ng m\u1eabu s\u1ed1 \u0111\u1ec3 t\u00ednh S th\u00ec m\u1eabu s\u1ed1 chung l\u00e0 s\u1ed1 ch\u1eb5n. V\u1edbi m\u1eabu s\u1ed1 chung n\u00e0y th\u00ec 1\/2 ; 1\/3 ; 1\/4 ; 1\/5 ; 1\/6 ; 1\/7 s\u1ebd tr\u1edf th\u00e0nh c\u00e1c ph\u00e2n s\u1ed1 m\u00e0 t\u1eed s\u1ed1 l\u00e0 s\u1ed1 ch\u1eb5n, ch\u1ec9 c\u00f3 1\/8 l\u00e0 tr\u1edf th\u00e0nh ph\u00e2n s\u1ed1 m\u00e0 t\u1eed s\u1ed1 l\u00e0 s\u1ed1 l\u1ebb. V\u1eady S l\u00e0 m\u1ed9t ph\u00e2n s\u1ed1 c\u00f3 t\u1eed s\u1ed1 l\u00e0 s\u1ed1 l\u1ebb v\u00e0 m\u1eabu s\u1ed1 l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean S kh\u00f4ng ph\u1ea3i l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean. H\u01b0\u1edbng 3 : Ch\u1ee9ng minh 5\/4 < S < 2 Th\u1eadt v\u1eady 1\/3 + 1\/4 + 1\/5 + 1\/6 + 1\/7 + 1\/8 > 6 x 1\/8 = 3\/4 n\u00ean S > 3\/4 + 1\/2 = 5\/4 M\u1eb7t kh\u00e1c : 1\/4 + 1\/5 + 1\/6 + 1\/7 < 4 x 1\/4 = 1 n\u00ean S < 1 + 1\/2 + 1\/3 + 1\/8 = 1 + 1\/2 + 11\/24 <2 V\u00ec 5\/4 < S < 2 n\u00ean S kh\u00f4ng ph\u1ea3i l\u00e0 s\u1ed1 t\u1ef1 nhi\u00ean. B\u00e0i 60 : Cho hai h\u00ecnh vu\u00f4ng ABCD v\u00e0 MNPQ nh\u01b0 trong h\u00ecnh v\u1ebd. Bi\u1ebft BD = 12 cm. H\u00e3y t\u00ednh di\u1ec7n t\u00edch ph\u1ea7n g\u1ea1ch ch\u00e9o. B\u00e0i gi\u1ea3i : Di\u1ec7n t\u00edch tam gi\u00e1c ABD l\u00e0 : Di\u1ec7n t\u00edch h\u00ecnh (12 x (12 : 2))\/2 = 36 (cm2) vu\u00f4ng ABCD l\u00e0 : 36 x 2 = 72 (cm2) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng AEOK l\u00e0 : 72 : 4 = 18 (cm2) Do \u0111\u00f3 : OE x OK = 18 (cm2) r x r = 18 (cm2) Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n t\u00e2m O l\u00e0 : 18 x 3,14 = 56,92 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c MON = r x r : 2 = 18 : 2 = 9 (cm2) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng MNPQ l\u00e0 : 9 x 4 = 36 (cm2) V\u1eady di\u1ec7n t\u00edch ph\u1ea7n g\u1ea1ch ch\u00e9o l\u00e0 : 77","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 56,52 - 36 = 20,52 (cm2) B\u00e0i 61 : B\u1ea1n To\u00e0n nh\u00e2n m\u1ed9t s\u1ed1 v\u1edbi 2002 nh\u01b0ng \u201c\u0111\u00e3ng tr\u00ed\u201d qu\u00ean vi\u1ebft 2 ch\u1eef s\u1ed1 0 c\u1ee7a s\u1ed1 2002 n\u00ean k\u1ebft qu\u1ea3 \u201cb\u1ecb\u201d gi\u1ea3m \u0111i 3965940 \u0111\u01a1n v\u1ecb. To\u00e0n \u0111\u00e3 \u0111\u1ecbnh nh\u00e2n s\u1ed1 n\u00e0o v\u1edbi 2002 ? B\u00e0i gi\u1ea3i : V\u00ec \\\"\u0111\u00e3ng tr\u00ed\\\" n\u00ean b\u1ea1n To\u00e0n \u0111\u00e3 nh\u00e2n nh\u1ea7m s\u1ed1 \u0111\u00f3 v\u1edbi 22. Th\u1eeba s\u1ed1 th\u1ee9 hai b\u1ecb gi\u1ea3m \u0111i s\u1ed1 \u0111\u01a1n v\u1ecb l\u00e0 : 2002 - 22 = 1980 (\u0111\u01a1n v\u1ecb). Do \u0111\u00f3 k\u1ebft qu\u1ea3 b\u1ecb gi\u1ea3m \u0111i 1980 l\u1ea7n th\u1eeba s\u1ed1 th\u1ee9 nh\u1ea5t, v\u00e0 b\u1eb1ng 3965940 \u0111\u01a1n v\u1ecb. V\u1eady th\u1eeba s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 : 3965940 : 1980 = 2003. B\u00e0i 62 : Ng\u01b0\u1eddi ta c\u1ed9ng 5 s\u1ed1 v\u00e0 chia cho 5 th\u00ec \u0111\u01b0\u1ee3c 138. N\u1ebfu x\u1ebfp c\u00e1c s\u1ed1 theo th\u1ee9 t\u1ef1 l\u1edbn d\u1ea7n th\u00ec c\u1ed9ng 3 s\u1ed1 \u0111\u1ea7u ti\u00ean v\u00e0 chia cho 3 s\u1ebd \u0111\u01b0\u1ee3c 127, c\u1ed9ng 3 s\u1ed1 cu\u1ed1i v\u00e0 chia cho 3 s\u1ebd \u0111\u01b0\u1ee3c 148. B\u1ea1n c\u00f3 bi\u1ebft s\u1ed1 \u0111\u1ee9ng gi\u1eefa theo th\u1ee9 t\u1ef1 tr\u00ean l\u00e0 s\u1ed1 n\u00e0o kh\u00f4ng ? B\u00e0i gi\u1ea3i : 138 l\u00e0 trung b\u00ecnh c\u1ed9ng c\u1ee7a 5 s\u1ed1, n\u00ean t\u1ed5ng 5 s\u1ed1 l\u00e0 : 138 x 5 = 690. T\u1ed5ng c\u1ee7a ba s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 : 127 x 3 = 381. T\u1ed5ng c\u1ee7a ba s\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 : 148 x 3 = 444. T\u1ed5ng c\u1ee7a hai s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 : 690 - 444 = 246. S\u1ed1 \u1edf gi\u1eefa l\u00e0 s\u1ed1 \u0111\u1ee9ng th\u1ee9 ba, n\u00ean s\u1ed1 \u1edf gi\u1eefa l\u00e0 : 381 - 246 = 135. B\u00e0i 70: T\u00e2m gi\u00fap b\u00e1n cam trong ba ng\u00e0y, Ng\u00e0y th\u1ee9 hai: s\u1ed1 cam b\u00e1n \u0111\u01b0\u1ee3c t\u0103ng 10% so v\u1edbi ng\u00e0y th\u1ee9 nh\u1ea5t. Ng\u00e0y th\u1ee9 ba: s\u1ed1 cam b\u00e1n \u0111\u01b0\u1ee3c gi\u1ea3m 10% so v\u1edbi ng\u00e0y th\u1ee9 hai. B\u1ea1n c\u00f3 bi\u1ebft trong ng\u00e0y th\u1ee9 nh\u1ea5t v\u00e0 ng\u00e0y th\u1ee9 ba th\u00ec ng\u00e0y n\u00e0o T\u00e2m b\u00e1n \u0111\u01b0\u1ee3c nhi\u1ec1u cam h\u01a1n kh\u00f4ng ? B\u00e0i gi\u1ea3i: Bi\u1ec3u th\u1ecb s\u1ed1 cam b\u00e1n ng\u00e0y th\u1ee9 nh\u1ea5t l\u00e0 100% th\u00ec s\u1ed1 b\u00e1n ng\u00e0y th\u1ee9 hai l\u00e0: 100% + 10% = 110% (s\u1ed1 cam ng\u00e0y th\u1ee9 nh\u1ea5t) Bi\u1ec3u th\u1ecb s\u1ed1 cam b\u00e1n ng\u00e0y th\u1ee9 hai l\u00e0 100% th\u00ec s\u1ed1 b\u00e1n ng\u00e0y th\u1ee9 hai l\u00e0: 100% - 10% = 90% (s\u1ed1 cam ng\u00e0y th\u1ee9 hai) So v\u1edbi ng\u00e0y th\u1ee9 nh\u1ea5t th\u00ec s\u1ed1 cam ng\u00e0y th\u1ee9 ba b\u00e1n l\u00e0: 110% x 90% = 99% (s\u1ed1 cam ng\u00e0y th\u1ee9 nh\u1ea5t) V\u00ec 100% > 99% n\u00ean ng\u00e0y th\u1ee9 nh\u1ea5t b\u00e1n \u0111\u01b0\u1ee3c nhi\u1ec1u cam h\u01a1n ng\u00e0y th\u1ee9 ba. B\u00e0i 71: Cu T\u00ed ch\u1ecdn 4 ch\u1eef s\u1ed1 li\u00ean ti\u1ebfp nhau v\u00e0 d\u00f9ng 4 ch\u1eef s\u1ed1 n\u00e0y \u0111\u1ec3 vi\u1ebft ra 3 s\u1ed1 g\u1ed3m 4 78","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 ch\u1eef s\u1ed1 kh\u00e1c nhau. Bi\u1ebft r\u1eb1ng s\u1ed1 th\u1ee9 nh\u1ea5t vi\u1ebft c\u00e1c ch\u1eef s\u1ed1 theo th\u1ee9 t\u1ef1 t\u0103ng d\u1ea7n, s\u1ed1 th\u1ee9 hai vi\u1ebft c\u00e1c ch\u1eef s\u1ed1 theo th\u1ee9 t\u1ef1 gi\u1ea3m d\u1ea7n v\u00e0 s\u1ed1 th\u1ee9 ba vi\u1ebft c\u00e1c ch\u1eef s\u1ed1 theo th\u1ee9 t\u1ef1 n\u00e0o \u0111\u00f3. Khi c\u1ed9ng ba s\u1ed1 v\u1eeba vi\u1ebft th\u00ec \u0111\u01b0\u1ee3c t\u1ed5ng l\u00e0 12300. B\u1ea1n h\u00e3y cho bi\u1ebft c\u00e1c s\u1ed1 m\u00e0 cu T\u00ed \u0111\u00e3 vi\u1ebft. B\u00e0i gi\u1ea3i : G\u1ecdi 4 s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb nh\u1ecf \u0111\u1ebfn l\u1edbn l\u00e0 a, b, c, d. S\u1ed1 th\u1ee9 nh\u1ea5t cu T\u00ed vi\u1ebft l\u00e0 abcd, s\u1ed1 th\u1ee9 hai cu T\u00ed vi\u1ebft l\u00e0 dcba. Ta x\u00e9t c\u00e1c ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn c\u1ee7a ba s\u1ed1 c\u00f3 t\u1ed5ng l\u00e0 12300: a l\u00e0 s\u1ed1 l\u1edbn h\u01a1n 1 v\u00ec n\u1ebfu a = 1 th\u00ec d = 4, khi \u0111\u00f3 s\u1ed1 th\u1ee9 ba c\u00f3 ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn l\u1edbn nh\u1ea5t l\u00e0 4 v\u00e0 t\u1ed5ng c\u1ee7a ba ch\u1eef s\u1ed1 n\u00e0y l\u1edbn nh\u1ea5t l\u00e0: 1 + 4 + 4 = 9 < 12; nh\u01b0 v\u1eady t\u1ed5ng c\u1ee7a ba s\u1ed1 nh\u1ecf h\u01a1n 12300. a l\u00e0 s\u1ed1 nh\u1ecf h\u01a1n 5 v\u00ec n\u1ebfu a = 5 th\u00ec d = 8 v\u00e0 a + d = 13 > 12; nh\u01b0 v\u1eady t\u1ed5ng c\u1ee7a ba s\u1ed1 l\u1edbn h\u01a1n 12300. a ch\u1ec9 c\u00f3 th\u1ec3 nh\u1eadn 3 gi\u00e1 tr\u1ecb l\u00e0 2, 3, 4. - N\u1ebfu a = 2 th\u00ec s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 2345, s\u1ed1 th\u1ee9 hai l\u00e0 5432. S\u1ed1 th\u1ee9 ba l\u00e0: 12300 - (2345 + 5432) = 4523 (\u0111\u00fang, v\u00ec s\u1ed1 n\u00e0y c\u00f3 c\u00e1c ch\u1eef s\u1ed1 l\u00e0 2, 3, 4, 5). - N\u1ebfu a = 3 th\u00ec s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 3456, s\u1ed1 th\u1ee9 hai l\u00e0 6543. S\u1ed1 th\u1ee9 ba l\u00e0 : 12300 - (3456 + 6543) = 2301 (lo\u1ea1i, v\u00ec s\u1ed1 n\u00e0y c\u00f3 c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c v\u1edbi 3, 4, 5, 6). - N\u1ebfu a = 4 th\u00ec s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 4567, s\u1ed1 th\u1ee9 hai l\u00e0 7654. S\u1ed1 th\u1ee9 ba l\u00e0: 12300 - (4567 + 7654) = 79 (lo\u1ea1i). V\u1eady c\u00e1c s\u1ed1 m\u00e0 cu T\u00ed \u0111\u00e3 vi\u1ebft l\u00e0 : 2345, 5432, 4523. B\u00e0i 55 : Ch\u1ec9 c\u00f3 m\u1ed9t chi\u1ebfc ca \u0110\u1ef1ng \u0111\u1ea7y v\u1eeba m\u1ed9t l\u00edt B\u1ea1n h\u00e3y mau cho bi\u1ebft \u0110ong n\u1eeda l\u00edt th\u1ebf n\u00e0o ? B\u00e0i gi\u1ea3i : 79","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 Ai kh\u00e9o tay tinh m\u1eaft Nghi\u00eang ca nh\u01b0 h\u00ecnh tr\u00ean S\u1ebd \u0111\u1ea1t y\u00eau c\u1ea7u li\u1ec1n Trong ca : \u0111\u00fang n\u1eeda l\u00edt ! B\u00e0i 56 : \u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p theo m\u1eabu : B\u00e0i gi\u1ea3i : B\u00e0i n\u00e0y c\u00f3 hai c\u00e1ch \u0111i\u1ec1n : C\u00e1ch 1 : Theo h\u00ecnh 1, ta c\u00f3 4 l\u00e0 trung b\u00ecnh c\u1ed9ng c\u1ee7a 3 v\u00e0 5 (v\u00ec (3 + 5) : 2 = 4). Khi \u0111\u00f3 \u1edf h\u00ecnh 2, g\u1ecdi A l\u00e0 s\u1ed1 c\u1ea7n \u0111i\u1ec1n, ta c\u00f3 A l\u00e0 trung b\u00ecnh c\u1ed9ng c\u1ee7a 5 v\u00e0 13. Do \u0111\u00f3 A = (5 + 13) : 2 = 9. \u1edf h\u00ecnh 3, g\u1ecdi B l\u00e0 s\u1ed1 c\u1ea7n \u0111i\u1ec1n, ta c\u00f3 15 l\u00e0 trung b\u00ecnh c\u1ed9ng c\u1ee7a 8 v\u00e0 B. Do \u0111\u00f3 8 + B = 15 x 2. T\u1eeb \u0111\u00f3 t\u00ecm \u0111\u01b0\u1ee3c B = 22. C\u00e1ch 2 : Theo h\u00ecnh 1, ta c\u00f3 3 x 3 + 4 x 4 = 5 x 5. Khi \u0111\u00f3 \u1edf h\u00ecnh 2 ta c\u00f3 : 5 x 5 + A x A = 13 x 13. suy ra A x A = 144. V\u1eady A = 12 (v\u00ec 12 x 12 = 144). \u1edf h\u00ecnh 3 ta c\u00f3 : 8 x 8 + 15 x 15 = B x B. B\u00e0i 58 : B\u1ea1n h\u00e3y \u0111i\u1ec1n c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 9 v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng \u0111\u1ec3 c\u00e1c ph\u00e9p t\u00ednh \u0111\u1ec1u th\u1ef1c hi\u1ec7n \u0111\u00fang (c\u1ea3 h\u00e0ng d\u1ecdc v\u00e0 h\u00e0ng ngang). B\u00e0i gi\u1ea3i : Ta \u0111\u1eb7t t\u00ean cho c\u00e1c s\u1ed1 ph\u1ea3i t\u00ecm nh\u01b0 trong b\u1ea3ng. C\u00e1c s\u1ed1 \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 c\u00e1c s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1 n\u00ean t\u1ed5ng c\u00e1c s\u1ed1 l\u1edbn nh\u1ea5t ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 17. 80","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 \u1edf c\u1ed9t 1, c\u00f3 A + D : H = 6, n\u00ean H ch\u1ec9 c\u00f3 th\u1ec3 l\u1edbn nh\u1ea5t l\u00e0 2. C\u1ed9t 5 c\u00f3 C + G : M = 5 n\u00ean M ch\u1ec9 c\u00f3 th\u1ec3 l\u1edbn nh\u1ea5t l\u00e0 3. * N\u1ebfu H = 1 th\u00ec A + D = 6 = 2 + 4, do \u0111\u00f3 M = 3 v\u00e0 H + K = 2 x 3 = 6 = 1 + 5. K = 5 th\u00ec B x E = 4 + 5 = 9, nh\u01b0 th\u1ebf ch\u1ec9 c\u00f3 th\u1ec3 B ho\u1eb7c E b\u1eb1ng 1, \u0111i\u1ec1u \u0111\u00f3 ch\u1ee9ng t\u1ecf H kh\u00f4ng th\u1ec3 b\u1eb1ng 1. * N\u1ebfu H = 2 th\u00ec M ph\u1ea3i b\u1eb1ng 1 ho\u1eb7c 3; n\u1ebfu M = 1 th\u00ec H + K = 2, nh\u01b0 v\u1eady K = 0, \u0111i\u1ec1u n\u00e0y c\u0169ng kh\u00f4ng th\u1ec3 \u0111\u01b0\u1ee3c. V\u1eady M = 3 ; H + K = 6 th\u00ec K = 4. H = 2 th\u00ec A + D = 12 = 5 + 7 ; nh\u01b0 v\u1eady A = 5, D = 7 ho\u1eb7c D = 5, A = 7. K = 4 th\u00ec B x E = 4 + 4 = 8 = 1 x 8 ; nh\u01b0 v\u1eady B = 1, E = 8 ho\u1eb7c E = 1, B = 8. M = 3 th\u00ec C + G = 15 = 6 + 9 ; nh\u01b0 v\u1eady C = 6, G = 9 ho\u1eb7c G = 6, C = 9 ; G ch\u1ec9 c\u00f3 th\u1ec3 b\u1eb1ng 9 v\u00ec n\u1ebfu G = 6 th\u00ec D + E = 10, m\u00e0 trong c\u00e1c s\u1ed1 1, 5, 7, 8 kh\u00f4ng c\u00f3 hai s\u1ed1 n\u00e0o c\u00f3 t\u1ed5ng b\u1eb1ng 10. V\u1eady C = 6 v\u00e0 A + B = 8, nh\u01b0 v\u1eady B ch\u1ec9 c\u00f3 th\u1ec3 b\u1eb1ng 1, A = 7 th\u00ec D = 5 v\u00e0 E = 8. C\u00e1c s\u1ed1 \u0111i\u1ec1n v\u00e0o b\u1ea3ng nh\u01b0 h\u00ecnh sau. B\u00e0i 63 : Cho b\u1ea3ng \u00f4 vu\u00f4ng g\u1ed3m 10 d\u00f2ng v\u00e0 10 c\u1ed9t. Hai b\u1ea1n T\u00edn v\u00e0 Nhi t\u00f4 m\u00e0u c\u00e1c \u00f4, m\u1ed7i \u00f4 m\u1ed9t m\u00e0u trong 3 m\u00e0u : xanh, \u0111\u1ecf, t\u00edm. B\u1ea1n T\u00edn b\u1ea3o : \\\"L\u1ea7n n\u00e0o t\u00f4 xong h\u1ebft c\u00e1c \u00f4 c\u0169ng c\u00f3 2 d\u00f2ng m\u00e0 tr\u00ean 2 d\u00f2ng \u0111\u00f3 c\u00f3 m\u1ed9t m\u00e0u t\u00f4 s\u1ed1 \u00f4 d\u00f2ng n\u00e0y b\u1eb1ng t\u00f4 s\u1ed1 \u00f4 d\u00f2ng kia\\\". B\u1ea1n Nhi b\u1ea3o : \\\"T\u1edb ph\u00e1t hi\u1ec7n ra bao gi\u1edd c\u0169ng c\u00f3 2 c\u1ed9t \u0111\u01b0\u1ee3c t\u00f4 nh\u01b0 th\u1ebf\\\". N\u00e0o, b\u1ea1n h\u00e3y cho bi\u1ebft ai \u0111\u00fang, ai sai ? B\u00e0i gi\u1ea3i : Gi\u1ea3 s\u1eed s\u1ed1 \u00f4 t\u00f4 m\u00e0u \u0111\u1ecf \u1edf t\u1ea5t c\u1ea3 c\u00e1c d\u00f2ng \u0111\u1ec1u kh\u00e1c nhau m\u00e0 m\u1ed7i d\u00f2ng c\u00f3 10 \u00f4 n\u00ean s\u1ed1 \u00f4 \u0111\u01b0\u1ee3c t\u00f4 m\u00e0u \u0111\u1ecf \u00edt nh\u1ea5t l\u00e0 : 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (\u00f4). L\u00ed lu\u1eadn t\u01b0\u01a1ng t\u1ef1 v\u1edbi m\u00e0u xanh, m\u00e0u t\u00edm ta c\u0169ng c\u00f3 k\u1ebft qu\u1ea3 nh\u01b0 v\u1eady. Do \u0111\u00f3 b\u1ea3ng s\u1ebd c\u00f3 \u00edt nh\u1ea5t 45 + 45 + 45 = 135 (\u00f4). \u0110i\u1ec1u n\u00e0y m\u00e2u thu\u1eabn v\u1edbi b\u1ea3ng ch\u1ec9 c\u00f3 100 \u00f4. Ch\u1ee9ng t\u1ecf \u00edt nh\u1ea5t ph\u1ea3i c\u00f3 2 d\u00f2ng m\u00e0 s\u1ed1 \u00f4 t\u00f4 b\u1edfi c\u00f9ng m\u1ed9t m\u00e0u l\u00e0 nh\u01b0 nhau. \u0110\u1ed1i v\u1edbi c\u00e1c c\u1ed9t, ta c\u0169ng l\u1eadp lu\u1eadn t\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean. Do \u0111\u00f3 c\u1ea3 hai b\u1ea1n \u0111\u1ec1u n\u00f3i \u0111\u00fang. B\u00e0i 64 : B\u1ea1n h\u00e3y \u0111i\u1ec1n \u0111\u1ee7 c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 14 v\u00e0o c\u00e1c \u00f4 vu\u00f4ng sao cho t\u1ed5ng 4 s\u1ed1 \u1edf m\u1ed7i h\u00e0ng ngang hay t\u1ed5ng 5 s\u1ed1 \u1edf m\u1ed7i c\u1ed9t d\u1ecdc \u0111\u1ec1u l\u00e0 30. B\u00e0i gi\u1ea3i : T\u1ed5ng c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 14 l\u00e0 : (14 + 1) x 14 : 2 = 105. T\u1ed5ng c\u00e1c s\u1ed1 c\u1ee7a 4 h\u00e0ng l\u00e0 : 30 x 4 = 120. T\u1ed5ng b\u1ed1n s\u1ed1 \u1edf b\u1ed1n \u00f4 c\u00f3 d\u1ea5u * l\u00e0 : 120 - 105 = 15. C\u1eb7p b\u1ed1n s\u1ed1 \u1edf 81","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 b\u1ed1n \u00f4 c\u00f3 d\u1ea5u * l\u00e0 m\u1ed9t trong c\u00e1c tr\u01b0\u1eddng h\u1ee3p sau : 15 = 1 + 2 + 3 + 9 (1) =2+3+4 = 1 + 2 + 4 + 8 (2) T\u1eeb m\u1ed7i tr\u01b0\u1eddng = 1 + 2 + 5 + 7 (3) = 1 + 3 + 4 + 7 (4) = 1 + 3 + 5 + 7 (5) + 6 (6) h\u1ee3p n\u00e0y c\u00f3 th\u1ec3 t\u1ea1o n\u00ean nhi\u1ec1u c\u00e1ch s\u1eafp x\u1ebfp c\u00e1c s\u1ed1 kh\u00e1c nhau. B\u00e0i 65: C\u0103n ph\u00f2ng c\u00f3 4 b\u1ee9c t\u01b0\u1eddng, tr\u00ean m\u1ed7i b\u1ee9c t\u01b0\u1eddng treo 3 l\u00e1 c\u1edd m\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa 3 l\u00e1 c\u1edd tr\u00ean m\u1ed9t b\u1ee9c t\u01b0\u1eddng l\u00e0 nh\u01b0 nhau. B\u1ea1n c\u00f3 bi\u1ebft c\u0103n ph\u00f2ng treo m\u1ea5y l\u00e1 c\u1edd kh\u00f4ng ? B\u00e0i gi\u1ea3i: \u0110\u1ec3 \u0111\u01a1n gi\u1ea3n, ta s\u1ebd treo t\u1ea5t c\u1ea3 c\u00e1c l\u00e1 c\u1edd \u1edf \u0111\u1ed9 cao ngang nhau tr\u00ean c\u1ea3 4 b\u1ee9c t\u01b0\u1eddng. Khi \u0111\u00f3 c\u00e1ch treo c\u1edd s\u1ebd gi\u1ed1ng nh\u01b0 b\u00e0i to\u00e1n tr\u1ed3ng c\u00e2y. Ta c\u00f3 5 c\u00e1ch tr\u1ed3ng \u1ee9ng v\u1edbi s\u1ed1 l\u00e1 c\u1edd l\u00e0 8, 9, 10, 11, 12 l\u00e1 c\u1edd nh\u01b0 sau (coi m\u1ed7i l\u00e1 c\u1edd l\u00e0 m\u1ed9t \u0111i\u1ec3m ch\u1ea5m tr\u00f2n): N\u1ebfu c\u00e1c l\u00e1 c\u1edd \u0111\u01b0\u1ee3c treo \u1edf \u0111\u1ed9 cao kh\u00e1c nhau tr\u00ean m\u1ed7i b\u1ee9c t\u01b0\u1eddng th\u00ec v\u1ecb tr\u00ed 3 l\u00e1 c\u1edd tr\u00ean m\u1ed9t b\u1ee9c t\u01b0\u1eddng s\u1ebd t\u1ea1o th\u00e0nh 3 \u0111\u1ec9nh c\u1ee7a m\u1ed9t h\u00ecnh tam gi\u00e1c \u0111\u1ec1u. Khi \u0111\u00f3 ta s\u1ebd c\u00f3 c\u00e1c c\u00e1ch treo kh\u00e1c \u1ee9ng v\u1edbi s\u1ed1 l\u00e1 c\u1edd l\u00e0 6,] 7, 8, 9, 10, 11, 12 l\u00e1 c\u1edd. Xin n\u00eau ra 2 c\u00e1ch treo \u1ee9ng v\u1edbi s\u1ed1 l\u00e1 c\u1edd l\u00e0 6 l\u00e1 v\u00e0 7 l\u00e1 nh\u01b0 sau: V\u1eady s\u1ed1 l\u00e1 c\u1edd trong c\u0103n ph\u00f2ng c\u00f3 th\u1ec3 t\u1eeb 6 \u0111\u1ebfn 12 l\u00e1 c\u1edd. B\u00e0i 66: L\u1ecd Lem chia m\u1ed9t qu\u1ea3 d\u01b0a (d\u01b0a \u0111\u1ecf) th\u00e0nh 9 ph\u1ea7n cho 9 c\u1ee5 gi\u00e0. Nh\u01b0ng khi c\u00e1c c\u1ee5 \u0103n xong, L\u1ecd Lem th\u1ea5y c\u00f3 10 mi\u1ebfng v\u1ecf d\u01b0a. L\u1ecd Lem chia d\u01b0a ki\u1ec3u g\u00ec \u1ea5y nh\u1ec9 ? B\u00e0i gi\u1ea3i: C\u00f3 nhi\u1ec1u c\u00e1ch b\u1ed5 d\u01b0a, Lo Lem \u0111\u00e3 b\u1ed5 d\u01b0a nh\u01b0 sau: 82","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 C\u1eaft ngang qu\u1ea3 d\u01b0a l\u00e0m 3 ph\u1ea7n, sau \u0111\u00f3 l\u1ea1i b\u1ed5 d\u1ecdc qu\u1ea3 d\u01b0a l\u00e0m 3 ph\u1ea7n s\u1ebd \u0111\u01b0\u1ee3c 9 mi\u1ebfng d\u01b0a (nh\u01b0 h\u00ecnh v\u1ebd) chia cho 9 c\u1ee5, sau khi \u0103n xong s\u1ebd c\u00f3 10 mi\u1ebfng v\u1ecf d\u01b0a. V\u00ec ri\u00eang mi\u1ebfng s\u1ed1 5 c\u00f3 v\u1ecf \u1edf 2 \u0111\u1ea7u, n\u00ean khi \u0103n xong s\u1ebd c\u00f3 2 mi\u1ebfng v\u1ecf. B\u00e0i 67: B\u1ea1n h\u00e3y \u0111i\u1ec1n \u0111\u1ee7 c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 10 v\u00e0o c\u00e1c \u00f4 vu\u00f4ng sao cho t\u1ed5ng c\u00e1c s\u1ed1 \u1edf n\u00e9t d\u1ecdc (1 n\u00e9t) c\u0169ng nh\u01b0 \u1edf n\u00e9t ngang (3 n\u00e9t) \u0111\u1ec1u l\u00e0 16. B\u00e0i gi\u1ea3i: T\u1ea5t c\u1ea3 c\u00e1c b\u1ea1n \u0111\u1ec1u nh\u1eadn ra m\u1ed9t ph\u01b0\u01a1ng \u00e1n \u0111i\u1ec1n s\u1ed1: a = 1; b = 9; c = 5; d = 4; e = i = 1; k = 8; l = 7. T\u1eeb \u0111\u00f3 s\u1ebd c\u00f3 c\u00e1c 6; g = 10; h = 3; b\u1eb1ng c\u00e1ch: ph\u01b0\u01a1ng \u00e1n kh\u00e1c 1) \u0110\u1ed5i c\u00e1c \u00f4 b v\u00e0 c. 2) \u0110\u1ed5i c\u00e1c \u00f4 k v\u00e0 l. 3) \u0110\u1ed5i c\u00e1c \u00f4 d v\u00e0 h. 4) \u0110\u1ed5i \u0111\u1ed3ng th\u1eddi c\u1ea3 3 \u00f4 a, b, c cho 3 \u00f4 i, k, l. Nh\u01b0 v\u1eady c\u00e1c b\u1ea1n s\u1ebd c\u00f3 16 c\u00e1ch \u0111i\u1ec1n s\u1ed1 kh\u00e1c nhau. B\u00e0i 68: Trong m\u1ed9t cu\u1ed9c thi t\u00e0i To\u00e1n Tu\u1ed5i th\u01a1 c\u00f3 51 b\u1ea1n tham d\u1ef1. Lu\u1eadt cho \u0111i\u1ec3m nh\u01b0 sau: + M\u1ed7i b\u00e0i l\u00e0m \u0111\u00fang \u0111\u01b0\u1ee3c 4 \u0111i\u1ec3m. + M\u1ed7i b\u00e0i l\u00e0m sai ho\u1eb7c kh\u00f4ng l\u00e0m s\u1ebd b\u1ecb tr\u1eeb 1 \u0111i\u1ec3m. B\u1ea1n ch\u1ee9ng t\u1ecf r\u1eb1ng t\u00ecm \u0111\u01b0\u1ee3c 11 b\u1ea1n c\u00f3 s\u1ed1 \u0111i\u1ec3m b\u1eb1ng nhau. B\u00e0i gi\u1ea3i: Thi t\u00e0i gi\u1ea3i To\u00e1n Tu\u1ed5i th\u01a1 c\u00f3 5 b\u00e0i. S\u1ed1 \u0111i\u1ec3m c\u1ee7a 51 b\u1ea1n thi c\u00f3 th\u1ec3 x\u1ebfp theo 5 lo\u1ea1i \u0111i\u1ec3m sau \u0111\u00e2y: 83","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 + L\u00e0m \u0111\u00fang 5 b\u00e0i \u0111\u01b0\u1ee3c: 4 x 5 = 20 (\u0111i\u1ec3m). + L\u00e0m \u0111\u00fang 4 b\u00e0i \u0111\u01b0\u1ee3c: 4 x 4 - 1 x 1 = 15 (\u0111i\u1ec3m). + L\u00e0m \u0111\u00fang 3 b\u00e0i \u0111\u01b0\u1ee3c: 4 x 3 - 1 x 2 = 10 (\u0111i\u1ec3m). + L\u00e0m \u0111\u00fang 2 b\u00e0i \u0111\u01b0\u1ee3c: 4 x 2 - 1 x 3 = 5 (\u0111i\u1ec3m). + L\u00e0m \u0111\u00fang 1 b\u00e0i \u0111\u01b0\u1ee3c: 4 x 1 - 1 x 4 = 0 (\u0111i\u1ec3m). V\u00ec 51 : 5 = 10 (d\u01b0 1) n\u00ean ph\u1ea3i c\u00f3 \u00edt nh\u1ea5t 11 b\u1ea1n c\u00f3 s\u1ed1 \u0111i\u1ec3m b\u1eb1ng nhau. B\u00e0i 69: V\u0169 H\u1eefu c\u00f9ng v\u1edbi L\u01b0\u01a1ng Th\u1ebf Vinh Hai nh\u00e0 to\u00e1n h\u1ecdc, m\u1ed9t n\u0103m sinh Th\u1ef1c h\u00e0nh, t\u00ednh to\u00e1n \u0111\u1ec1u th\u00f4ng th\u1ea1o V\u1ebb vang d\u00e2n t\u1ed9c n\u01b0\u1edbc non m\u00ecnh N\u0103m sinh c\u1ee7a hai \u00f4ng l\u00e0 m\u1ed9t s\u1ed1 c\u00f3 b\u1ed1n ch\u1eef s\u1ed1, t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 b\u1eb1ng 10. N\u1ebfu vi\u1ebft n\u0103m sinh theo th\u1ee9 t\u1ef1 ng\u01b0\u1ee3c l\u1ea1i th\u00ec n\u0103m sinh kh\u00f4ng \u0111\u1ed5i. B\u1ea1n \u0111\u00e3 bi\u1ebft n\u0103m sinh c\u1ee7a hai \u00f4ng ch\u01b0a? B\u00e0i gi\u1ea3i: G\u1ecdi n\u0103m sinh c\u1ee7a hai \u00f4ng l\u00e0 abba (a \u2260 0, a < 3, b <10). Ta c\u00f3: a + b + b + a = 10 hay (a + b) x 2 = 10. Do \u0111\u00f3 a + b = 5. V\u00ec a \u2260 0 v\u00e0 a < 3 n\u00ean a = 1 ho\u1eb7c 2. * N\u1ebfu a = 1 th\u00ec b = 5 - 1 = 4. Khi \u0111\u00f3 n\u0103m sinh c\u1ee7a hai \u00f4ng l\u00e0 1441 (\u0111\u00fang). * N\u1ebfu a = 2 th\u00ec b = 5 - 2 = 3. Khi \u0111\u00f3 n\u0103m sinh c\u1ee7a hai \u00f4ng l\u00e0 2332 (lo\u1ea1i). 84","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 V\u1eady hai \u00f4ng V\u0169 H\u1eefu v\u00e0 L\u01b0\u01a1ng Th\u1ebf Vinh sinh n\u0103m 1441. B\u00e0i 72: V\u1edbi 4 ch\u1eef s\u1ed1 2 v\u00e0 c\u00e1c d\u1ea5u ph\u00e9p t\u00ednh b\u1ea1n c\u00f3 th\u1ec3 vi\u1ebft \u0111\u01b0\u1ee3c m\u1ed9t bi\u1ec3u th\u1ee9c \u0111\u1ec3 c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 9 \u0111\u01b0\u1ee3c kh\u00f4ng? T\u00f4i \u0111\u00e3 c\u1ed1 g\u1eafng vi\u1ebft m\u1ed9t bi\u1ec3u th\u1ee9c \u0111\u1ec3 c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 7 nh\u01b0ng ch\u01b0a \u0111\u01b0\u1ee3c. C\u00f2n b\u1ea1n? B\u1ea1n th\u1eed s\u1ee9c xem n\u00e0o! B\u00e0i gi\u1ea3i: V\u1edbi b\u1ed1n ch\u1eef s\u1ed1 2 ta vi\u1ebft \u0111\u01b0\u1ee3c bi\u1ec3u th\u1ee9c c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng 9 l\u00e0: 22 : 2 - 2 = 9. nh\u01b0 h\u00ecnh d\u01b0\u1edbi. Kh\u00f4ng th\u1ec3 d\u00f9ng b\u1ed1n ch\u1eef s\u1ed1 2 \u0111\u1ec3 vi\u1ebft \u0111\u01b0\u1ee3c bi\u1ec3u th\u1ee9c c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 7. B\u00e0i 73: V\u1edbi 36 que di\u00eam \u0111\u00e3 \u0111\u01b0\u1ee3c x\u1ebfp 1) B\u1ea1n \u0111\u1ebfm \u0111\u01b0\u1ee3c bao nhi\u00eau h\u00ecnh vu\u00f4ng? 2) B\u1ea1n h\u00e3y nh\u1ea5c ra 4 que di\u00eam \u0111\u1ec3 ch\u1ec9 c\u00f2n 4 h\u00ecnh vu\u00f4ng \u0111\u01b0\u1ee3c kh\u00f4ng? B\u00e0i gi\u1ea3i : 1) Nh\u00ecn v\u00e0o h\u00ecnh v\u1ebd, ta th\u1ea5y c\u00f3 2 lo\u1ea1i h\u00ecnh vu\u00f4ng, h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 1 que di\u00eam v\u00e0 h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 2 que di\u00eam. H\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 1 que di\u00eam g\u1ed3m c\u00f3 13 h\u00ecnh, h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 2 que di\u00eam g\u1ed3m c\u00f3 4 h\u00ecnh. V\u1eady c\u00f3 t\u1ea5t c\u1ea3 l\u00e0 17 h\u00ecnh vu\u00f4ng. 2) M\u1ed7i que di\u00eam c\u00f3 th\u1ec3 n\u1eb1m tr\u00ean c\u1ea1nh c\u1ee7a nhi\u1ec1u nh\u1ea5t l\u00e0 3 h\u00ecnh vu\u00f4ng, n\u1ebfu nh\u1eb7t ra 4 que di\u00eam th\u00ec ta b\u1edbt \u0111i nhi\u1ec1u nh\u1ea5t l\u00e0 : 4 x 3 = 12 (h\u00ecnh vu\u00f4ng), c\u00f2n l\u1ea1i 17 - 12 = 5 (h\u00ecnh vu\u00f4ng). Nh\u01b0 v\u1eady kh\u00f4ng th\u1ec3 nh\u1eb7t ra 4 que di\u00eam \u0111\u1ec3 c\u00f2n l\u1ea1i 4 h\u00ecnh vu\u00f4ng \u0111\u01b0\u1ee3c. B\u00e0i 74: C\u00f3 7 th\u00f9ng \u0111\u1ef1ng \u0111\u1ea7y d\u1ea7u, 7 th\u00f9ng ch\u1ec9 c\u00f2n n\u1eeda th\u00f9ng d\u1ea7u v\u00e0 7 v\u1ecf th\u00f9ng. L\u00e0m sao c\u00f3 th\u1ec3 chia cho 3 ng\u01b0\u1eddi \u0111\u1ec3 m\u1ecdi ng\u01b0\u1eddi \u0111\u1ec1u c\u00f3 l\u01b0\u1ee3ng d\u1ea7u nh\u01b0 nhau v\u00e0 s\u1ed1 th\u00f9ng nh\u01b0 nhau ? B\u00e0i gi\u1ea3i: G\u1ecdi th\u00f9ng \u0111\u1ea7y d\u1ea7u l\u00e0 A, th\u00f9ng c\u00f3 n\u1eeda th\u00f9ng d\u1ea7u l\u00e0 B, th\u00f9ng kh\u00f4ng c\u00f3 d\u1ea7u l\u00e0 C. 85","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 C\u00e1ch 1: Kh\u00f4ng ph\u1ea3i \u0111\u1ed5 d\u1ea7u t\u1eeb th\u00f9ng n\u00e0y sang th\u00f9ng kia. Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t nh\u1eadn: 3A, 1B, 3C. Ng\u01b0\u1eddi th\u1ee9 hai nh\u1eadn: 2A, 3B, 2C. Ng\u01b0\u1eddi th\u1ee9 ba nh\u1eadn: 2A, 3B, 2C. C\u00e1ch 2: Kh\u00f4ng ph\u1ea3i \u0111\u1ed5 d\u1ea7u t\u1eeb th\u00f9ng n\u00e0y sang th\u00f9ng kia. Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t nh\u1eadn: 3A, 1B, 3C. Ng\u01b0\u1eddi th\u1ee9 hai nh\u1eadn: 3A, 1B, 3C. Ng\u01b0\u1eddi th\u1ee9 ba nh\u1eadn: 1A, 5B, 1C. C\u00e1ch 3: \u0110\u1ed5 d\u1ea7u t\u1eeb th\u00f9ng n\u00e0y sang th\u00f9ng kia. L\u1ea5y 4 th\u00f9ng ch\u1ee9a n\u1eeda th\u00f9ng d\u1ea7u (4B) \u0111\u1ed5 \u0111\u1ea7y sang 2 th\u00f9ng kh\u00f4ng (2C) \u0111\u1ec3 \u0111\u01b0\u1ee3c 2 th\u00f9ng \u0111\u1ea7y d\u1ea7u (2A). Khi \u0111\u00f3 c\u00f3 9A, 3B, 9C v\u00e0 m\u1ed7i ng\u01b0\u1eddi s\u1ebd nh\u1eadn \u0111\u01b0\u1ee3c nh\u01b0 nhau l\u00e0 3A, 1B, 3C. B\u00e0i 75: H\u00e3y v\u1ebd 4 \u0111o\u1ea1n th\u1eb3ng \u0111i qua 9 \u0111i\u1ec3m \u1edf h\u00ecnh b\u00ean m\u00e0 kh\u00f4ng \u0111\u01b0\u1ee3c nh\u1ea5c b\u00fat hay t\u00f4 l\u1ea1i. B\u00e0i gi\u1ea3i: C\u00e1i kh\u00f3 \u1edf b\u00e0i to\u00e1n n\u00e0y l\u00e0 ch\u1ec9 \u0111\u01b0\u1ee3c v\u1ebd 4 \u0111o\u1ea1n th\u1eb3ng v\u00e0 ch\u1ec9 \u0111\u01b0\u1ee3c v\u1ebd b\u1eb1ng m\u1ed9t n\u00e9t n\u00ean c\u1ea7n ph\u1ea3i \u201ct\u1ea1o th\u00eam\u201d hai \u0111i\u1ec3m \u1edf b\u00ean ngo\u00e0i 9 \u0111i\u1ec3m th\u00ec m\u1edbi th\u1ef1c hi\u1ec7n \u0111\u01b0\u1ee3c y\u00eau c\u1ea7u c\u1ee7a \u0111\u1ec1 b\u00e0i. Xin n\u00eau ra m\u1ed9t c\u00e1ch v\u1ebd v\u1edbi hai \u201c\u0111\u01b0\u1eddng \u0111i\u201d kh\u00e1c nhau (b\u1eaft \u0111\u1ea7u t\u1eeb \u0111i\u1ec3m 1 v\u00e0 k\u1ebft th\u00fac \u1edf \u0111i\u1ec3m 2 v\u1edbi \u0111\u01b0\u1eddng \u0111i theo chi\u1ec1u m\u0169i t\u00ean) nh\u01b0 sau: 86","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 Khi xoay ho\u1eb7c l\u1eadt hai h\u00ecnh tr\u00ean ta s\u1ebd c\u00f3 c\u00e1c c\u00e1ch v\u1ebd kh\u00e1c. B\u00e0i 76: Chi\u1ebfc b\u00e1nh trung thu Nh\u00e2n tr\u00f2n \u1edf gi\u1eefa H\u00e3y c\u1eaft 4 l\u1ea7n Th\u00e0nh 12 mi\u1ebfng Nh\u01b0ng nh\u1edb \u0111i\u1ec1u ki\u1ec7n C\u00e1c mi\u1ebfng b\u1eb1ng nhau V\u00e0 l\u1ea7n c\u1eaft n\u00e0o C\u0169ng qua gi\u1eefa b\u00e1nh B\u00e0i gi\u1ea3i: C\u00f3 nhi\u1ec1u c\u00e1ch c\u1eaft \u0111\u01b0\u1ee3c c\u00e1c b\u1ea1n \u0111\u1ec1 xu\u1ea5t. Xin gi\u1edbi thi\u1ec7u 3 c\u00e1ch. C\u00e1ch 1: Nh\u00e1t th\u1ee9 nh\u1ea5t chia \u0111\u00f4i theo b\u1ec1 d\u1ea7y c\u1ee7a chi\u1ebfc b\u00e1nh v\u00e0 \u0111\u1ec3 nguy\u00ean v\u1ecb tr\u00ed n\u00e0y c\u1eaft th\u00eam 3 nh\u00e1t (nh\u01b0 h\u00ecnh v\u1ebd). L\u01b0u \u00fd l\u00e0 AM = BN = DQ = CP = 1\/6 AB v\u00e0 IA = ID = KB = KC = 1\/2 AB. C\u00e1c b\u1ea1n c\u00f3 th\u1ec3 d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c 12 mi\u1ebfng b\u00e1nh l\u00e0 b\u1eb1ng nhau v\u00e0 c\u1ea3 3 nh\u00e1t c\u1eaft \u0111\u1ec1u \u0111i qua \u0111\u00fang ... t\u00e2m b\u00e1nh. C\u00e1ch 2: C\u1eaft 2 nh\u00e1t theo 2 \u0111\u01b0\u1eddng ch\u00e9o \u0111\u1ec3 \u0111\u01b0\u1ee3c 4 mi\u1ebfng r\u1ed3i ch\u1ed3ng 4 mi\u1ebfng n\u00e0y l\u00ean nhau c\u1eaft 2 nh\u00e1t \u0111\u1ec3 chia m\u1ed7i mi\u1ebfng th\u00e0nh 3 ph\u1ea7n b\u1eb1ng nhau (l\u01b0u \u00fd: BM = MN = NC). 87","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 C\u00e1ch 3: Nh\u00e1t th\u1ee9 nh\u1ea5t c\u1eaft nh\u01b0 c\u00e1ch 1 v\u00e0 \u0111\u1ec3 nguy\u00ean v\u1ecb tr\u00ed n\u00e0y \u0111\u1ec3 c\u1eaft th\u00eam 3 nh\u00e1t nh\u01b0 h\u00ecnh v\u1ebd. L\u01b0u \u00fd: AN = AM = CQ = CP = 1\/2 AB. B\u00e0i 77: M\u1ed7i \u0111\u1ec9nh c\u1ee7a m\u1ed9t t\u1ea5m b\u00eca h\u00ecnh tam gi\u00e1c \u0111\u01b0\u1ee3c \u0111\u00e1nh s\u1ed1 l\u1ea7n l\u01b0\u1ee3t l\u00e0 1; 2; 3. Ng\u01b0\u1eddi ta ch\u1ed3ng c\u00e1c tam gi\u00e1c n\u00e0y l\u00ean nhau sao cho kh\u00f4ng c\u00f3 ch\u1eef s\u1ed1 n\u00e0o b\u1ecb che l\u1ea5p. M\u1ed9t b\u1ea1n c\u1ed9ng t\u1ea5t c\u1ea3 c\u00e1c ch\u1eef s\u1ed1 nh\u00ecn th\u1ea5y th\u00ec \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 2002. Li\u1ec7u b\u1ea1n \u0111\u00f3 c\u00f3 t\u00ednh nh\u1ea7m kh\u00f4ng? B\u00e0i gi\u1ea3i: T\u1ed5ng c\u00e1c s\u1ed1 tr\u00ean ba \u0111\u1ec9nh c\u1ee7a m\u1ed7i h\u00ecnh tam gi\u00e1c l\u00e0 1 + 2 + 3 = 6. T\u1ed5ng n\u00e0y l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho 6. Khi ch\u1ed3ng c\u00e1c h\u00ecnh tam gi\u00e1c n\u00e0y l\u00ean nhau sao cho kh\u00f4ng c\u00f3 ch\u1eef s\u1ed1 n\u00e0o b\u1ecb che l\u1ea5p, r\u1ed3i t\u00ednh t\u1ed5ng t\u1ea5t c\u1ea3 c\u00e1c ch\u1eef s\u1ed1 nh\u00ecn th\u1ea5y \u0111\u01b0\u1ee3c ph\u1ea3i c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 s\u1ed1 chia h\u1ebft cho 6. V\u00ec s\u1ed1 2002 kh\u00f4ng chia h\u1ebft cho 6 n\u00ean b\u1ea1n \u0111\u00f3 \u0111\u00e3 t\u00ednh sai. B\u00e0i 78: B\u1ea1n h\u00e3y \u0111i\u1ec1n \u0111\u1ee7 12 s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 12, m\u1ed7i s\u1ed1 v\u00e0o m\u1ed9t \u00f4 vu\u00f4ng sao cho t\u1ed5ng 4 s\u1ed1 c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t c\u1ed9t hay m\u1ed9t h\u00e0ng \u0111\u1ec1u nh\u01b0 nhau. B\u00e0i gi\u1ea3i: T\u1ed5ng c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 12 l\u00e0: (12+1) x 12 : 2 = 78 V\u00ec t\u1ed5ng 4 s\u1ed1 c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t c\u1ed9t hay m\u1ed9t h\u00e0ng \u0111\u1ec1u nh\u01b0 nhau n\u00ean t\u1ed5ng s\u1ed1 c\u1ee7a 4 h\u00e0ng v\u00e0 c\u1ed9t ph\u1ea3i l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho 4. \u0110\u1eb7t c\u00e1c ch\u1eef c\u00e1i A, B, C, D v\u00e0o c\u00e1c \u00f4 vu\u00f4ng \u1edf gi\u1eefa (h\u00ecnh v\u1ebd). Khi t\u00ednh t\u1ed5ng s\u1ed1 c\u1ee7a 4 h\u00e0ng v\u00e0 c\u1ed9t th\u00ec c\u00e1c s\u1ed1 \u1edf c\u00e1c \u00f4 A, B, C, D \u0111\u01b0\u1ee3c t\u00ednh hai l\u1ea7n. Do \u0111\u00f3 \u0111\u1ec3 t\u1ed5ng 4 h\u00e0ng, c\u1ed9t chia h\u1ebft cho 4 th\u00ec t\u1ed5ng 4 s\u1ed1 c\u1ee7a 4 \u00f4 A, B, C, D ph\u1ea3i chia cho 4 d\u01b0 2 (v\u00ec 78 chia cho 4 d\u01b0 2). Ta th\u1ea5y t\u1ed5ng c\u1ee7a 4 s\u1ed1 c\u00f3 th\u1ec3 l\u00e0: 10, 14, 18, 22, 26, 30, 34, 38, 42. Ta x\u00e9t m\u1ed9t v\u00e0i tr\u01b0\u1eddng h\u1ee3p: 1) T\u1ed5ng c\u1ee7a 4 s\u1ed1 b\u00e9 nh\u1ea5t l\u00e0 10. Khi \u0111\u00f3 4 s\u1ed1 s\u1ebd l\u00e0 1, 2, 3, 4. Do \u0111\u00f3 t\u1ed5ng c\u1ee7a m\u1ed7i h\u00e0ng (hay m\u1ed7i c\u1ed9t) l\u00e0: (78 + 10) : 4 = 22. Xin n\u00eau ra m\u1ed9t c\u00e1ch \u0111i\u1ec1n nh\u01b0 h\u00ecnh d\u01b0\u1edbi: 88","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 2) T\u1ed5ng c\u1ee7a 4 s\u1ed1 l\u00e0 14. Ta c\u00f3: 14 = 1 + 2 + 3 + 8 = 1 + 2 + 4 + 7 = 1 + 3 + 4 + 6 = 2 + 3 + 4 + 5. Do \u0111\u00f3 t\u1ed5ng c\u1ee7a m\u1ed7i h\u00e0ng (hay m\u1ed7i c\u1ed9t) l\u00e0: (78 + 14) : 4 = 23. Xin n\u00eau ra m\u1ed9t c\u00e1ch \u0111i\u1ec1n nh\u01b0 h\u00ecnh sau: C\u00e1c tr\u01b0\u1eddng h\u1ee3p c\u00f2n l\u1ea1i s\u1ebd cho ta k\u1ebft qu\u1ea3 \u1edf m\u1ed7i h\u00e0ng (hay m\u1ed7i c\u1ed9t) l\u1ea7n l\u01b0\u1ee3t l\u00e0 24, 25, 26, 27, 28, 29, 30. C\u00f3 r\u1ea5t nhi\u1ec1u c\u00e1ch \u0111i\u1ec1n \u0111\u1ea5y! C\u00e1c b\u1ea1n th\u1eed t\u00ecm ti\u1ebfp xem sao? B\u00e0i 79: M\u1ed9t \u0111\u1ed9i tuy\u1ec3n tham d\u1ef1 k\u1ef3 thi h\u1ecdc sinh gi\u1ecfi 3 m\u00f4n V\u0103n, To\u00e1n, Ngo\u1ea1i ng\u1eef do th\u00e0nh ph\u1ed1 t\u1ed5 ch\u1ee9c \u0111\u1ea1t \u0111\u01b0\u1ee3c 15 gi\u1ea3i. H\u1ecfi \u0111\u1ed9i tuy\u1ec3n h\u1ecdc sinh gi\u1ecfi \u0111\u00f3 c\u00f3 bao nhi\u00eau h\u1ecdc sinh? Bi\u1ebft r\u1eb1ng: H\u1ecdc sinh n\u00e0o c\u0169ng c\u00f3 gi\u1ea3i. B\u1ea5t k\u1ef3 m\u00f4n n\u00e0o c\u0169ng c\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh ch\u1ec9 \u0111\u1ea1t 1 gi\u1ea3i. B\u1ea5t k\u1ef3 hai m\u00f4n n\u00e0o c\u0169ng c\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 hai m\u00f4n. C\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 3 m\u00f4n. T\u1ed5ng s\u1ed1 h\u1ecdc sinh \u0111\u1ea1t 3 gi\u1ea3i, 2 gi\u1ea3i, 1 gi\u1ea3i t\u0103ng d\u1ea7n. B\u00e0i gi\u1ea3i: G\u1ecdi s\u1ed1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 3 m\u00f4n l\u00e0 a (h\u1ecdc sinh) G\u1ecdi s\u1ed1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 2 m\u00f4n l\u00e0 b (h\u1ecdc sinh) G\u1ecdi s\u1ed1 h\u1ecdc sinh ch\u1ec9 \u0111\u1ea1t gi\u1ea3i 1 m\u00f4n l\u00e0 c (h\u1ecdc sinh) T\u1ed5ng s\u1ed1 gi\u1ea3i \u0111\u1ea1t \u0111\u01b0\u1ee3c l\u00e0: 89","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 3 x a + 2 x b + c = 15 (gi\u1ea3i). V\u00ec t\u1ed5ng s\u1ed1 h\u1ecdc sinh \u0111\u1ea1t 3 gi\u1ea3i, 2 gi\u1ea3i, 1 gi\u1ea3i t\u0103ng d\u1ea7n n\u00ean a < b < c. V\u00ec b\u1ea5t k\u1ef3 2 m\u00f4n n\u00e0o c\u0169ng c\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 2 m\u00f4n n\u00ean: - C\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 2 m\u00f4n V\u0103n v\u00e0 To\u00e1n. - C\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 2 m\u00f4n To\u00e1n v\u00e0 Ngo\u1ea1i Ng\u1eef. - C\u00f3 \u00edt nh\u1ea5t 1 h\u1ecdc sinh \u0111\u1ea1t gi\u1ea3i c\u1ea3 2 m\u00f4n V\u0103n v\u00e0 Ngo\u1ea1i Ng\u1eef. Do v\u1eady b= 3. Gi\u1ea3 s\u1eed a = 2 th\u00ec b b\u00e9 nh\u1ea5t l\u00e0 3, c b\u00e9 nh\u1ea5t l\u00e0 4; do \u0111\u00f3 t\u1ed5ng s\u1ed1 gi\u1ea3i b\u00e9 nh\u1ea5t l\u00e0: 3 x 2 + 2 x 3 + 4 = 16 > 15 (lo\u1ea1i). Do \u0111\u00f3 a < 2, n\u00ean a = 1. Ta c\u00f3: 3 x 1 + 2 x b + c = 15 suy ra: 2 x b + c = 12. N\u1ebfu b = 3 th\u00ec c = 12 - 2 x 3 = 6 (\u0111\u00fang). N\u1ebfu b = 4 th\u00ec c = 12 - 2 x 4 = 4 (lo\u1ea1i v\u00ec tr\u00e1i v\u1edbi \u0111i\u1ec1u ki\u1ec7n b < c) V\u1eady c\u00f3 1 b\u1ea1n \u0111\u1ea1t 3 gi\u1ea3i, 3 b\u1ea1n \u0111\u1ea1t 2 gi\u1ea3i, 6 b\u1ea1n \u0111\u1ea1t 1 gi\u1ea3i. \u0110\u1ed9i tuy\u1ec3n \u0111\u00f3 c\u00f3 s\u1ed1 h\u1ecdc sinh l\u00e0: 1 + 3 + 6 = 10 (b\u1ea1n). B\u00e0i 80: \u0110i\u1ec1n s\u1ed1 S\u1eed d\u1ee5ng c\u00e1c s\u1ed1 3, 5, 8, 10 v\u00e0 c\u00e1c d\u1ea5u +, - , x \u0111\u1ec3 \u0111i\u1ec1n v\u00e0o m\u1ed7i \u00f4 c\u00f2n tr\u1ed1ng \u1edf b\u1ea3ng sau: ( Ch\u1ec9 \u0111\u01b0\u1ee3c \u0111i\u1ec1n m\u1ed9t d\u1ea5u ho\u1eb7c m\u1ed9t s\u1ed1 v\u00e0o m\u1ed7i h\u00e0ng ho\u1eb7c m\u1ed7i c\u1ed9t. \u0110i\u1ec1n t\u1eeb tr\u00e1i sang ph\u1ea3i, t\u1eeb tr\u00ean xu\u1ed1ng d\u01b0\u1edbi) 90","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i gi\u1ea3i: B\u1ea1n \u0111\u1ecdc c\u00f3 th\u1ec3 x\u00e9t c\u00e1c t\u1ed5ng theo t\u1eebng h\u00e0ng, t\u1eebng c\u1ed9t v\u00e0 kh\u00f4ng kh\u00f3 kh\u0103n l\u1eafm s\u1ebd c\u00f3 k\u1ebft qu\u1ea3 sau: B\u00e0i 81: 20 Gi\u1ecf d\u01b0a h\u1ea5u Tr\u00ed v\u00e0 D\u0169ng gi\u00fap b\u1ed1 m\u1eb9 x\u1ebfp 65 qu\u1ea3 d\u01b0a h\u1ea5u m\u1ed7i qu\u1ea3 n\u1eb7ng 1kg, 35 qu\u1ea3 d\u01b0a h\u1ea5u m\u1ed7i qu\u1ea3 n\u1eb7ng 2kg v\u00e0 15 qu\u1ea3 d\u01b0a h\u1ea5u m\u1ed7i qu\u1ea3 n\u1eb7ng 3kg v\u00e0o trong 20 gi\u1ecf. M\u1ecdi ng\u01b0\u1eddi c\u00f9ng \u0111ang l\u00e0m vi\u1ec7c, Tr\u00ed ch\u1ea1y \u0111\u1ebfn b\u00e0n h\u1ecdc l\u1ea5y gi\u1ea5y b\u00fat ra ghi... ghi v\u00e0 Tr\u00ed la l\u00ean: \u201cC\u00f3 x\u1ebfp th\u1ebf n\u00e0o \u0111i ch\u0103ng n\u1eefa, ch\u00fang ta lu\u00f4n t\u00ecm \u0111\u01b0\u1ee3c 2 gi\u1ecf trong 20 gi\u1ecf n\u00e0y c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng b\u1eb1ng nhau\u201d. C\u00e1c b\u1ea1n h\u00e3y ch\u1ee9ng t\u1ecf l\u00e0 Tr\u00ed \u0111\u00e3 n\u00f3i \u0111\u00fang. B\u00e0i gi\u1ea3i: T\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng d\u01b0a l\u00e0: 1 x 65 + 2 x 35 + 3 x 15 = 180 (kg). Gi\u1ea3 s\u1eed kh\u1ed1i l\u01b0\u1ee3ng d\u01b0a \u1edf m\u1ed7i gi\u1ecf kh\u00e1c nhau th\u00ec t\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng d\u01b0a \u1edf 20 gi\u1ecf b\u00e9 nh\u1ea5t l\u00e0: 1 + 2 + 3 + ... + 19 + 20 = 210 (kg). V\u00ec 210 kg > 180 kg n\u00ean ch\u1eafc ch\u1eafn ph\u1ea3i c\u00f3 \u00edt nh\u1ea5t 2 gi\u1ecf trong 20 gi\u1ecf c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng b\u1eb1ng nhau. V\u1eady Tr\u00ed \u0111\u00e3 n\u00f3i \u0111\u00fang. B\u00e0i 82: Ho\u00e0ng mua 6 quy\u1ec3n v\u1edf, H\u00f9ng mua 3 quy\u1ec3n v\u1edf. Hai b\u1ea1n g\u00f3p s\u1ed1 v\u1edf c\u1ee7a m\u00ecnh v\u1edbi s\u1ed1 v\u1edf c\u1ee7a b\u1ea1n S\u01a1n, r\u1ed3i chia \u0111\u1ec1u cho nhau. S\u01a1n t\u00ednh r\u1eb1ng m\u00ecnh ph\u1ea3i tr\u1ea3 c\u00e1c b\u1ea1n \u0111\u00fang 800 \u0111\u1ed3ng. T\u00ednh gi\u00e1 ti\u1ec1n 1 quy\u1ec3n v\u1edf, bi\u1ebft r\u1eb1ng c\u1ea3 ba b\u1ea1n \u0111\u1ec1u mua c\u00f9ng m\u1ed9t lo\u1ea1i v\u1edf. B\u00e0i gi\u1ea3i: V\u00ec Ho\u00e0ng v\u00e0 H\u00f9ng g\u00f3p s\u1ed1 v\u1edf c\u1ee7a m\u00ecnh v\u1edbi s\u1ed1 v\u1edf c\u1ee7a S\u01a1n, r\u1ed3i chia \u0111\u1ec1u cho nhau, n\u00ean t\u1ed5ng s\u1ed1 v\u1edf c\u1ee7a ba b\u1ea1n l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho 3. S\u1ed1 v\u1edf c\u1ee7a Ho\u00e0ng v\u00e0 H\u00f9ng \u0111\u1ec1u chia h\u1ebft cho 3 n\u00ean s\u1ed1 v\u1edf c\u1ee7a S\u01a1n c\u0169ng l\u00e0 s\u1ed1 chia h\u1ebft cho 3. S\u1ed1 v\u1edf c\u1ee7a S\u01a1n ph\u1ea3i \u00edt h\u01a1n 6 v\u00ec n\u1ebfu s\u1ed1 v\u1edf c\u1ee7a S\u01a1n b\u1eb1ng ho\u1eb7c nhi\u1ec1u h\u01a1n s\u1ed1 v\u1edf c\u1ee7a Ho\u00e0ng (6 91","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 quy\u1ec3n) th\u00ec sau khi g\u00f3p v\u1edf l\u1ea1i chia \u0111\u1ec1u S\u01a1n s\u1ebd kh\u00f4ng ph\u1ea3i tr\u1ea3 th\u00eam 800 \u0111\u1ed3ng. S\u1ed1 v\u1edf c\u1ee7a S\u01a1n kh\u00e1c 0 (S\u01a1n ph\u1ea3i c\u00f3 v\u1edf c\u1ee7a m\u00ecnh th\u00ec m\u1edbi g\u00f3p chung v\u1edbi c\u00e1c b\u1ea1n \u0111\u01b0\u1ee3c ch\u1ee9!), nh\u1ecf h\u01a1n 6 v\u00e0 chia h\u1ebft cho 3 n\u00ean S\u01a1n c\u00f3 3 quy\u1ec3n v\u1edf. S\u1ed1 v\u1edf c\u1ee7a m\u1ed7i b\u1ea1n sau khi chia \u0111\u1ec1u l\u00e0: (6 + 3 + 3) : 3 = 4 (quy\u1ec3n) Nh\u01b0 v\u1eady S\u01a1n \u0111\u01b0\u1ee3c c\u00e1c b\u1ea1n \u0111\u01b0a th\u00eam: 4 - 3 = 1 (quy\u1ec3n) Gi\u00e1 ti\u1ec1n m\u1ed9t quy\u1ec3n v\u1edf l\u00e0 800 \u0111\u1ed3ng. B\u00e0i 83: H\u00e3y \u0111i\u1ec1n c\u00e1c s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 9 v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00e1c ph\u00e9p t\u00ednh \u0111\u00fang B\u00e0i gi\u1ea3i: \u0110\u1eb7t c\u00e1c ch\u1eef c\u00e1i v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng: Theo \u0111\u1ea7u b\u00e0i ta c\u00f3 c\u00e1c ch\u1eef c\u00e1i kh\u00e1c nhau bi\u1ec3u th\u1ecb c\u00e1c s\u1ed1 kh\u00e1c nhau. Do \u0111\u00f3: a \u2260 1; c \u2260 1; d \u2260 1; 9 = 1 x 9 = 3 x 3 n\u00ean b \u2260 9 v\u00e0 e \u2260 9; v\u00e0 7 = 1 b > 1; e > 1. V\u00ec e \u2260 7. x 7 n\u00ean b \u2260 7 v\u00e0 Do \u0111\u00f3: b = 6 v\u00e0 e = 8 ho\u1eb7c b = 8 v\u00e0 e = 6. V\u00ec 6 = 2 x 3 v\u00e0 8 = 2 x 4 n\u00ean a = b : c = e : d = 2. Trong c\u00e1c \u00f4 tr\u1ed1ng a, b, c, d, e \u0111\u00e3 c\u00f3 c\u00e1c s\u1ed1 2, 3, 4, 6, 8; do \u0111\u00f3 ch\u1ec9 c\u00f2n c\u00e1c s\u1ed1 1, 5, 7, 9 \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng g, h, i, k. * N\u1ebfu e = 6 th\u00ec g = 7 v\u00e0 h = 1. Do \u0111\u00f3 a = i - k = 9 - 5 = 42 (lo\u1ea1i). * N\u1ebfu e = 8 th\u00ec g = 9 v\u00e0 h = 1. Do \u0111\u00f3 a = i - k = 7 - 5 = 2 (\u0111\u00fang). Khi \u0111\u00f3: b = 6 v\u00e0 c = 3. K\u1ebft qu\u1ea3: 92","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 B\u00e0i 84: C\u00f3 13 t\u1ea5m b\u00eca, m\u1ed7i t\u1ea5m b\u00eca \u0111\u01b0\u1ee3c ghi m\u1ed9t ch\u1eef s\u1ed1 v\u00e0 x\u1ebfp theo th\u1ee9 t\u1ef1 sau: Kh\u00f4ng thay \u0111\u1ed5i th\u1ee9 t\u1ef1 c\u00e1c t\u1ea5m b\u00eca, h\u00e3y \u0111\u1eb7t gi\u1eefa ch\u00fang d\u1ea5u c\u00e1c ph\u00e9p t\u00ednh + , - , x v\u00e0 d\u1ea5u ngo\u1eb7c n\u1ebfu c\u1ea7n, sao cho k\u1ebft qu\u1ea3 l\u00e0 2002. B\u00e0i gi\u1ea3i: B\u00e0i to\u00e1n c\u00f3 r\u1ea5t nhi\u1ec1u c\u00e1ch \u0111\u1eb7t d\u1ea5u ph\u00e9p t\u00ednh v\u00e0 d\u1ea5u ngo\u1eb7c. Xin n\u00eau m\u1ed9t s\u1ed1 c\u00e1ch: C\u00e1ch 1: (123 + 4 x 5) x (6 + 7 - 8 + 9 + 1 - 2 - 3 + 4) = 2002 C\u00e1ch 2: (1 x 2 + 3 x 4) x (5 + 6) x [(7 + 8 + 9) - (1 + 2 x 3 + 4)] = 2002 C\u00e1ch 3: (1 + 2 + 3 + 4 x 5) x (6 x 7 + 8 + 9 - 1 + 23 - 4) = 2002 B\u00e0i 85: Hai b\u1ea1n Huy v\u00e0 Nam \u0111i mua 18 g\u00f3i b\u00e1nh v\u00e0 12 g\u00f3i k\u1eb9o \u0111\u1ec3 \u0111\u1ebfn l\u1edbp li\u00ean hoan. Huy \u0111\u01b0a cho c\u00f4 b\u00e1n h\u00e0ng 2 t\u1edd 100000 \u0111\u1ed3ng v\u00e0 \u0111\u01b0\u1ee3c tr\u1ea3 l\u1ea1i 72000 \u0111\u1ed3ng. Nam n\u00f3i: \u201cC\u00f4 t\u00ednh sai r\u1ed3i\u201d. B\u1ea1n h\u00e3y cho bi\u1ebft Nam n\u00f3i \u0111\u00fang hay sai? Gi\u1ea3i th\u00edch t\u1ea1i sao? B\u00e0i gi\u1ea3i: V\u00ec s\u1ed1 18 v\u00e0 s\u1ed1 12 \u0111\u1ec1u chia h\u1ebft cho 3, n\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n mua 18 g\u00f3i b\u00e1nh v\u00e0 12 g\u00f3i k\u1eb9o ph\u1ea3i l\u00e0 s\u1ed1 chia h\u1ebft cho 3. V\u00ec Huy \u0111\u01b0a cho c\u00f4 b\u00e1n h\u00e0ng 2 t\u1edd 100000 \u0111\u1ed3ng v\u00e0 \u0111\u01b0\u1ee3c tr\u1ea3 l\u1ea1i 72000 \u0111\u1ed3ng, n\u00ean s\u1ed1 ti\u1ec1n mua 18 g\u00f3i b\u00e1nh v\u00e0 12 g\u00f3i k\u1eb9o l\u00e0: 100000 x 2 - 72000 = 128000 (\u0111\u1ed3ng). V\u00ec s\u1ed1 128000 kh\u00f4ng chia h\u1ebft cho 3, n\u00ean b\u1ea1n Nam n\u00f3i \u201cC\u00f4 t\u00ednh sai r\u1ed3i\u201d l\u00e0 \u0111\u00fang. 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 93","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 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) 94","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 v\u1ecb c\u00f3 nh\u1edb 1 sang * Tr\u01b0\u1eddng h\u1ee3p 2: Ph\u00e9p c\u1ed9ng \u1edf h\u00e0ng \u0111\u01a1n b\u1ea3ng: h\u00e0ng ch\u1ee5c. Khi \u0111\u00f3: M + O = 12 v\u00e0 A + G = 9. Ta c\u00f3 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: 95","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 * Bi\u1ec3n s\u1ed1 39A 0452. Xin n\u00eau ra m\u1ed9t s\u1ed1 c\u00e1ch: (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 0 : (38 + 88) = 0 M\u1ed9t v\u00e0i c\u00e1ch kh\u00e1c: (9 - 8) + 0 - 8 : 8 = 0 8:8+8+0+0=9.... * Bi\u1ec3n s\u1ed1 52N 8233. Xin n\u00eau ra m\u1ed9t s\u1ed1 c\u00e1ch: 96","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 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.... 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. 97","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 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. 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 98","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 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: 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: 99","C\u00c1C \u0110\u1ec0 THI H\u1eccC SINH GI\u1eceI L\u1edaP 5 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: 100"]