460 bai toan vui luyen tri thong minh

["cho: 2 d\u01b0 1, 3 d\u01b0 2, 4 d\u01b0 3, 5 d\u01b0 4, 6 d\u01b0 5, 7 d\u01b0 6, 8 d\u01b0 7, 9 d\u01b0 8, 10 d\u01b0 9. V\u1eady gia \u0111\u00ecnh \u00f4ng Hai L\u00faa \u0111\u00e3 thu ho\u1ea1ch \u0111\u01b0\u1ee3c bao nhi\u00eau qu\u1ea3 d\u01b0a h\u1ea5u? a. 5.220 qu\u1ea3 b. 5.219 qu\u1ea3 c. 2.520 qu\u1ea3 d. 2.519 qu\u1ea3. C\u00e2u 8. N\u1ebfu ng\u00e0y h\u00f4m qua l\u00e0 ng\u00e0y mai c\u1ee7a th\u1ee9 Ba v\u00e0 ng\u00e0y mai l\u00e0 ng\u00e0y h\u00f4m qua c\u1ee7a th\u1ee9 B\u1ea3y th\u00ec ng\u00e0y h\u00f4m nay l\u00e0 th\u1ee9 m\u1ea5y? C\u00e2u 9. T\u00ecm l\u1ed7i sai c\u1ee7a b\u00e0i sau: C\u00f3: 1\u0111 = 1 (\u0111) x 1 (\u0111) = 10 (xu) x 10 (xu) = 100 xu = 10\u0111 V\u1eady: 1\u0111 = 10\u0111","B\u00e0i 36 C\u00e2u 1. \u0110i\u1ec1n s\u1ed1 v\u00e0o d\u1ea5u (?). 19 14 16 32 4824 3327 52? 4 C\u00e2u 2. C\u00f3 m\u1ed9t \u0111\u1ed1ng ti\u1ec1n xu, An l\u1ea5y \u0111i m\u1ed9t n\u1eafm, th\u00eam m\u1ed9t n\u1eafm n\u1eefa, n\u1eeda n\u1eafm n\u1eefa v\u00e0 1\/4 n\u1eefa, c\u1ed9ng 1 b\u1eb1ng 100. H\u1ecfi m\u1ed7i n\u1eafm l\u00e0 bao nhi\u00eau?. C\u00e2u 3. Cha m\u1eb9 c\u1ee7a H\u00f9ng c\u00f3 ba ng\u01b0\u1eddi con, m\u1ed9t ng\u01b0\u1eddi t\u00ean l\u00e0 T\u00e2m, m\u1ed9t ng\u01b0\u1eddi t\u00ean l\u00e0 C\u01b0\u1eddng. H\u1ecfi ng\u01b0\u1eddi th\u1ee9 ba t\u00ean l\u00e0 g\u00ec? C\u00e2u 4. M\u1ed9t ng\u01b0\u1eddi b\u00e1n ba lo\u1ea1i chanh g\u1ed3m: - 9kg chanh lo\u1ea1i I - 11kg chanh lo\u1ea1i II - 7kg chanh lo\u1ea1i III th\u00ec \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 69.200\u0111. Gi\u00e1 1kg chanh lo\u1ea1i I \u0111\u1eaft h\u01a1n lo\u1ea1i II l\u00e0 800\u0111, h\u01a1n lo\u1ea1i III l\u00e0 1.200\u0111. T\u00ednh gi\u00e1 ti\u1ec1n 1kg chanh m\u1ed7i lo\u1ea1i?","C\u00e2u 5. Ph\u00e2n t\u00edch m\u1ed7i s\u1ed1 sau \u0111\u00e2y Th\u00e0nh ra m\u1ed9t T\u00edch, \u0111i\u1ec1u n\u00e0y nh\u1edb ghi C\u00e1c th\u1eeba s\u1ed1 ph\u1ea3i gi\u1ed1ng y Gi\u1ecfi hay nh\u01b0 b\u1ea1n, kh\u00f3 chi \u0111\u00e1p li\u1ec1n Nhi \u0111\u1ed3ng thi v\u1edbi thi\u1ebfu ni\u00ean Ai \u0111\u00e1p nhanh \u0111\u00fang, \u01b0u ti\u00ean th\u01b0\u1edfng qu\u00e0 a\/ 8 b\/ 27 c\/ 16 d\/ 100 C\u00e2u 6. T\u00ecm s\u1ed1 \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng: 14 3 12 20 6 9 84 10 18 ? 6 C\u00e2u 7. Bi\u1ebft r\u1eb1ng c\u1ee9 ba th\u00f9ng m\u1eadt ong th\u00ec \u0111\u1ef1ng \u0111\u01b0\u1ee3c 27 l\u00edt. Trong kho c\u00f3 12 th\u00f9ng, ngo\u00e0i c\u1eeda h\u00e0ng c\u00f3 5 th\u00f9ng. H\u1ecfi t\u1ea5t c\u1ea3 c\u00f3 bao nhi\u00eau l\u00edt m\u1eadt ong? C\u00e2u 8. 5 quy\u1ec3n v\u1edf v\u00e0 3 quy\u1ec3n s\u00e1ch gi\u00e1 43.500\u0111","1 quy\u1ec3n s\u00e1ch \u0111\u1eaft h\u01a1n 1 quy\u1ec3n v\u1edf 5.500\u0111. T\u00ednh gi\u00e1 ti\u1ec1n m\u1ed7i quy\u1ec3n?","C\u00e2u 1. S\u1ed1 hai ch\u1eef s\u1ed1 vi\u1ebft ra Ch\u1ecdn sao \u01a1i b\u1ea1n nh\u1edb l\u00e0 nh\u01b0 sau: H\u00e0ng ch\u1ee5c, \u0111\u01a1n v\u1ecb b\u1eb1ng nhau \u0110\u1ed1 ai, ai bi\u1ebft, vi\u1ebft mau kh\u00f3 g\u00ec Kh\u00f3 g\u00ec, nhanh k\u1ec3 ra \u0111i M\u1eddi em, m\u1eddi b\u1ea1n thi\u1ebfu nhi thi t\u00e0i. C\u00e2u 2. S\u1eed d\u1ee5ng ch\u00eca kh\u00f3a (key) d\u01b0\u1edbi \u0111\u00e2y l\u00e0m theo ph\u00e9p t\u00ednh J, K, L v\u1edbi m\u1ed7i gi\u00e1 tr\u1ecb bao quanh c\u00e1c ch\u1eef c\u00e1i Key: ! = 4, @ = 6, { = 8 L @@ @ {! ! { ! {! @{ @{@ J=?K=?L=? C\u00e2u 3. H\u00ecnh b\u00ean d\u01b0\u1edbi c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c chia ra l\u00e0 hai b\u1edfi m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng m\u00e0 khi h\u1ee3p hai ph\u1ea7n l\u1ea1i th\u00e0nh m\u1ed9t h\u00ecnh vu\u00f4ng ho\u00e0n ch\u1ec9nh. H\u1ecfi \u0111\u01b0\u1eddng \u0111\u00f3","k\u1ebft n\u1ed1i hai \u0111i\u1ec3m n\u00e0o? a. 3 v\u00e0 12 b. 4 v\u00e0 11 c. 5 v\u00e0 16 d. 6 v\u00e0 18 e. 7 v\u00e0 14 f. 8 v\u00e0 17 g. 9 v\u00e0 3 h. 10 v\u00e0 4 i. 11 v\u00e0 5 j. 12 v\u00e0 2 C\u00e2u 4. M\u1ed9t \u0111\u1ed9i c\u00f4ng nh\u00e2n \u0111\u1eafp \u0111\u00ea ph\u00f2ng l\u1ee5t, 4 ng\u00e0y \u0111\u1ea7u m\u1ed7i ng\u00e0y \u0111\u1eafp \u0111\u01b0\u1ee3c 115 m3, 6 ng\u00e0y sau m\u1ed7i ng\u00e0y \u0111\u1eafp \u0111\u01b0\u1ee3c 140 m3. H\u1ecfi trung b\u00ecnh m\u1ed7i ng\u00e0y \u0111\u1ed9i d\u00e2n c\u00f4ng \u0111\u00f3 \u0111\u1eafp \u0111\u01b0\u1ee3c bao nhi\u00eau m\u00e9t kh\u1ed1i \u0111\u1ea5t? C\u00e2u 5. B\u1ed9 \u0111\u1ed9i s\u1ed1 ch\u00fa 900 Chia th\u00e0nh ti\u1ec3u \u0111\u1ed9i 15 ng\u01b0\u1eddi n\u00e0y a. \u0110\u01b0\u1ee3c m\u1ea5y ti\u1ec3u \u0111\u1ed9i \u0111\u00e2y gh\u00e9p v\u00e0o B\u1ea1n \u01a1i! c\u00f3 bi\u1ebft ch\u0103ng n\u00e0o. \u0110\u1ea1i \u0111\u1ed9i: 3 trung \u0111\u1ed9i gh\u00e9p v\u00e0o th\u00e0nh ngay","\u0110\u1ed1 ai, ai bi\u1ebft, ai hay b. \u0110\u01b0\u1ee3c m\u1ea5y \u0111\u1ea1i \u0111\u1ed9i, \u0111\u00e1p ngay, kh\u00f3 g\u00ec! C\u00e2u 6. C\u00f3 17 xe v\u1eeba xe taxi v\u1eeba xe lam. Xe taxi c\u00f3 4 b\u00e1nh, xe lam c\u00f3 3 b\u00e1nh. H\u1ecfi c\u00f3 m\u1ea5y xe taxi, m\u1ea5y xe lam, bi\u1ebft r\u1eb1ng c\u00f3 t\u1ea5t c\u1ea3 62 b\u00e1nh? C\u00e2u 7. M\u1ed9t anh ch\u00e0ng \u0111i c\u00e2u c\u00e1. Khi tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi: \\\"Anh c\u00e2u \u0111\u01b0\u1ee3c bao nhi\u00eau c\u00e1?\\\", anh ta n\u00f3i: \\\"M\u1ed9t n\u1eeda c\u1ee7a 8, s\u1ed1 6 kh\u00f4ng c\u00f3 \u0111\u1ea7u, s\u1ed1 9 kh\u00f4ng c\u00f3 \u0111u\u00f4i\\\". H\u1ecfi anh ch\u00e0ng \u0111\u00e3 c\u00e2u \u0111\u01b0\u1ee3c bao nhi\u00eau c\u00e1? C\u00e2u 8. B\u1ed1n c\u00e1i ch\u00e9n v\u00e0 1 c\u00e1i \u1ea5m n\u1eb7ng b\u1eb1ng 17 th\u1ecfi ch\u00ec. Ri\u00eang c\u00e1i \u1ea5m n\u1eb7ng b\u1eb1ng 1 c\u00e1i ch\u00e9n v\u00e0 7 th\u1ecfi ch\u00ec. H\u1ecfi c\u00e1i \u1ea5m c\u00e2n n\u1eb7ng b\u1eb1ng m\u1ea5y th\u1ecfi ch\u00ec? C\u00e2u 9. Hai s\u1ed1 n\u00e0o ti\u1ebfp theo d\u00e3y s\u1ed1 sau? 40 | 33 | 37 | 30 | ? | ?","B\u00e0i 38 C\u00e2u 1. Khi ng\u01b0\u1eddi ta h\u1ecfi con c\u00e1 b\u1eaft \u0111\u01b0\u1ee3c n\u1eb7ng bao nhi\u00eau, ng\u01b0\u1eddi \u0111\u00e1nh c\u00e1 tr\u1ea3 l\u1eddi: \\\"\u0110u\u00f4i n\u00f3 n\u1eb7ng 150g, \u0111\u1ea7u n\u00f3 n\u1eb7ng b\u1eb1ng \u0111u\u00f4i v\u00e0 1\/2 th\u00e2n, c\u00f2n th\u00e2n n\u1eb7ng b\u1eb1ng \u0111\u1ea7u v\u00e0 \u0111u\u00f4i\\\". Nh\u01b0 th\u1ebf con c\u00e1 c\u1ee7a anh ta n\u1eb7ng bao nhi\u00eau? C\u00e2u 2. T\u00ecm s\u1ed1 \u0111i\u1ec1n v\u00e0o d\u1ea5u (?). 2 | 4 | 3 | 12 | 9 | ? | ? C\u00e2u 3. S\u1ed1 xe taxi nhi\u1ec1u h\u01a1n s\u1ed1 xe lam l\u00e0 9 chi\u1ebfc. S\u1ed1 b\u00e1nh xe taxi nhi\u1ec1u h\u01a1n b\u00e1nh xe lam l\u00e0 42 c\u00e1i. T\u00ednh s\u1ed1 xe taxi v\u00e0 xe lam? C\u00e2u 4. S\u1ed1 n\u00e0o ti\u1ebfp theo d\u00e3y s\u1ed1 sau? 5, 10, 15, 20, 25, 30, 35 C\u00e2u 5. M\u1ed9t n\u00f4ng d\u00e2n mang 2 gi\u1ecf tr\u1ee9ng ra ch\u1ee3 b\u00e1n, m\u1ed7i gi\u1ecf c\u00f3 30 qu\u1ea3 tr\u1ee9ng. Trong gi\u1ecf tr\u1ee9ng b\u00e9, b\u00e0 d\u1ef1 \u0111\u1ecbnh s\u1ebd b\u00e1n v\u1edbi gi\u00e1 1 \u0111\u1ed3ng \u0111\u01b0\u1ee3c 3 qu\u1ea3. Gi\u1ecf tr\u1ee9ng to b\u00e0 s\u1ebd b\u00e1n 1 \u0111\u1ed3ng 2 qu\u1ea3. Tuy nhi\u00ean khi \u1edf ch\u1ee3 b\u00e0 thay \u0111\u1ed5i \u00fd \u0111\u1ecbnh, b\u00e0 \u0111\u1ec3 tr\u1ee9ng l\u1eabn l\u1ed9n v\u00e0 b\u00e1n v\u1edbi gi\u00e1 2 \u0111\u1ed3ng \u0111\u01b0\u1ee3c 5 qu\u1ea3. Nh\u01b0 th\u1ebf c\u00f3 l\u1ee3i cho","b\u00e0 so v\u1edbi \u00fd \u0111\u1ecbnh ban \u0111\u1ea7u kh\u00f4ng? C\u00e2u 6. C\u00f3 bao nhi\u00eau h\u00ecnh vu\u00f4ng trong h\u00ecnh d\u01b0\u1edbi \u0111\u00e2y? C\u00e2u 7. M\u1ed9t tr\u1ea1i ch\u0103n nu\u00f4i c\u00f3 408 con v\u1eeba tr\u00e2u v\u1eeba ng\u1ef1a v\u1eeba b\u00f2. Bi\u1ebft r\u1eb1ng: s\u1ed1 tr\u00e2u \u00edt h\u01a1n s\u1ed1 ng\u1ef1a 12 con, s\u1ed1 b\u00f2 g\u1ea5p \u0111\u00f4i s\u1ed1 tr\u00e2u. H\u00e3y t\u00ednh s\u1ed1 con m\u1ed7i lo\u1ea1i? C\u00e2u 8. S\u1ed1 n\u00e0o ti\u1ebfp theo d\u00e3y s\u1ed1 sau: 101,2 | 102,4 | 104,8 | 109,6 | ? C\u00e2u 9. S\u1ed1 3 ch\u1eef s\u1ed1 \u0110\u01a1n v\u1ecb gi\u1ed1ng ch\u1eef h\u00e0ng tr\u0103m \u0110\u1ed1 ai h\u1ecdc gi\u1ecfi, h\u1ecdc ch\u0103m \u0110\u00e1p bao nhi\u00eau s\u1ed1 \u0111\u00e2y kh\u00e2m ph\u1ee5c t\u00e0i","\u0110\u1ed1 em, \u0111\u1ed1 b\u1ea1n, \u0111\u1ed1 ai \u0110\u00e1p nhanh, xin th\u01b0\u1edfng tr\u00e0ng d\u00e0i v\u1ed7 tay. C\u00e2u 10. C\u1ea3 g\u00e0 v\u00e0 ch\u00f3 c\u00f3 100 ch\u00e2n. Bi\u1ebft s\u1ed1 g\u00e0 nhi\u1ec1u h\u01a1n s\u1ed1 ch\u00f3 8 con. H\u1ecfi c\u00f3 bao nhi\u00eau con g\u00e0, bao nhi\u00eau con ch\u00f3?","C\u00e2u 1. T\u00ecm s\u1ed1 c\u00f2n thi\u1ebfu? C\u00e2u 2. M\u1ec7nh \u0111\u1ec1 k\u1ebft trong \u0111o\u1ea1n lu\u1eadn sau \u0111\u00fang hay sai: T\u1ea5t c\u1ea3 b\u1ecfng ng\u00f4 l\u00e0 ng\u1ef1a. T\u1ea5t c\u1ea3 b\u1ecfng ng\u00f4 c\u00f3 n\u00fat \u0111\u1ecf v\u00e0 th\u1ec9nh tho\u1ea3ng ch\u01a1i b\u00e0i bingo. B\u00e1nh n\u01b0\u1edbng th\u1ec9nh tho\u1ea3ng ch\u01a1i b\u00e0i bingo. Kim c\u01b0\u01a1ng l\u00e0 b\u00e1nh n\u01b0\u1edbng, v\u00ec v\u1eady ng\u1ef1a v\u00e0 kim c\u01b0\u01a1ng th\u1ec9nh tho\u1ea3ng ch\u01a1i b\u00e0i bingo. C\u00e2u 3. C\u00f3 ba \u0111\u1ed9i thi\u1ebfu ni\u00ean A, B, C v\u1edbi t\u1ed5ng s\u1ed1 \u0111\u1ed9i vi\u00ean kho\u1ea3ng 40 \u0111\u1ebfn 50 em. \u0110\u1ec3 chu\u1ea9n b\u1ecb tham gia lao \u0111\u1ed9ng, nh\u00e0 tr\u01b0\u1eddng d\u1ef1 \u0111\u1ecbnh chia l\u1ea1i s\u1ed1 \u0111\u1ed9i vi\u00ean \u0111\u00f3 b\u1eb1ng c\u00e1ch chuy\u1ec3n t\u1eeb \u0111\u1ed9i A sang \u0111\u1ed9i B m\u1ed9t s\u1ed1 \u0111\u1ed9i vi\u00ean b\u1eb1ng s\u1ed1 \u0111\u1ed9i vi\u00ean c\u1ee7a \u0111\u1ed9i B, chuy\u1ec3n t\u1eeb \u0111\u1ed9i B sang \u0111\u1ed9i C m\u1ed9t s\u1ed1 \u0111\u1ed9i vi\u00ean b\u1eb1ng s\u1ed1 \u0111\u1ed9i vi\u00ean c\u1ee7a \u0111\u1ed9i C, chuy\u1ec3n t\u1eeb \u0111\u1ed9i C sang \u0111\u1ed9i A m\u1ed9t s\u1ed1 \u0111\u1ed9i vi\u00ean b\u1eb1ng s\u1ed1","\u0111\u1ed9i vi\u00ean c\u00f2n l\u1ea1i c\u1ee7a \u0111\u1ed9i A. Sau 3 l\u1ea7n chuy\u1ec3n nh\u01b0 v\u1eady th\u00ec s\u1ed1 \u0111\u1ed9i vi\u00ean c\u1ee7a ba \u0111\u1ed9i b\u1eb1ng nhau. H\u00e3y t\u00ednh s\u1ed1 \u0111\u1ed9i vi\u00ean \u1edf m\u1ed7i \u0111\u1ed9i thi\u1ebfu ni\u00ean l\u00fac ch\u01b0a chuy\u1ec3n. C\u00e2u 4. T\u00ed, Te \u0111\u00f4i b\u1ea1n th\u00edch \u0111\u00f9a Te khoe r\u1eb1ng: \\\"T\u1edb m\u1edbi mua \u0111\u1ed3ng h\u1ed3!\\\" T\u00ed c\u01b0\u1eddi: \\\"\u0110\u1eb9p qua sin - c\u00f4!\\\" \\\"M\u1ea5y ph\u00fat b\u1eb1ng ph\u1ea7n t\u01b0 ng\u00e0y?\\\" \\\"Gi\u1ecfi to\u00e1n nh\u01b0 c\u1eadu \u0111\u00e1p ngay kh\u00f3 g\u00ec!\\\" Gi\u00fap Te mau gi\u1ea3i li\u1ec1n \u0111i Nhi \u0111\u1ed3ng c\u00f9ng v\u1edbi thi\u1ebfu nhi \u0111ua t\u00e0i. C\u00e2u 5. T\u00ecm s\u1ed1 \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng: 10 4 5 9 3 13 10 5 846? C\u00e2u 6. 6 h\u00ecnh vu\u00f4ng b\u00ean tr\u00ean khi x\u1ebfp l\u1ea1i s\u1ebd t\u1ea1o th\u00e0nh h\u00ecnh h\u1ed9p n\u00e0o b\u00ean d\u01b0\u1edbi?","a. Ch\u1ec9 h\u00ecnh A b. Ch\u1ec9 h\u00ecnh B c. Ch\u1ec9 h\u00ecnh C d. Ch\u1ec9 h\u00ecnh D e. H\u00ecnh A v\u00e0 B f. H\u00ecnh A v\u00e0 C g. H\u00ecnh A v\u00e0 D h. H\u00ecnh B v\u00e0 C i. H\u00ecnh B v\u00e0 D j. H\u00ecnh C v\u00e0 D. C\u00e2u 7. T\u00ecm hai s\u1ed1 t\u1ef1 nhi\u00ean bi\u1ebft t\u1ed5ng c\u1ee7a ch\u00fang l\u00e0 9 v\u00e0 t\u00edch c\u1ee7a ch\u00fang l\u00e0 20.","C\u00e2u 8. \u0110\u1ed1 vui, vui \u0111\u1ed1 S\u1ed1 2 ch\u1eef s\u1ed1 Ch\u1ecdn sao h\u00e0ng ch\u1ee5c so ra H\u01a1n h\u00e0ng \u0111\u01a1n v\u1ecb, n\u00e0o ta thi t\u00e0i K\u1ec3 sao nhanh \u0111\u00fang, \u0111\u1eebng sai Nghe t\u00f4i h\u00f4 nh\u00e9 \\\"m\u1ed9t hai\\\" k\u1ec3 li\u1ec1n. C\u00e2u 9. C\u00f4 Lan c\u00f3 4 con g\u00e0 m\u00e1i. C\u00f4 nh\u1eadn th\u1ea5y r\u1eb1ng 1 con g\u00e0 c\u00e1ch 1 ng\u00e0y \u0111\u1ebb 1qu\u1ea3 tr\u1ee9ng, con th\u1ee9 2 c\u00e1ch 3 ng\u00e0y \u0111\u1ebb 1 qu\u1ea3 tr\u1ee9ng, con th\u1ee9 3 c\u00e1ch 4 ng\u00e0y \u0111\u1ebb 1 qu\u1ea3 tr\u1ee9ng v\u00e0 con th\u1ee9 4 c\u00e1ch 7 ng\u00e0y \u0111\u1ebb 1 qu\u1ea3 tr\u1ee9ng. M\u1ed9t l\u1ea7n c\u00f4 Lan l\u1ea5y trong chu\u1ed3ng \u0111\u01b0\u1ee3c 4 qu\u1ea3 tr\u1ee9ng v\u00e0 khoe v\u1edbi b\u00e0 h\u00e0ng x\u00f3m. B\u00e0 ta ch\u00fac m\u1eebng c\u00f4 v\u00e0 h\u1ecfi: S\u1ed1 ng\u00e0y ng\u1eafn nh\u1ea5t l\u00e0 m\u1ea5y ng\u00e0y (k\u1ec3 t\u1eeb b\u00e2y gi\u1edd) \u0111\u1ec3 c\u00f4 c\u00f3 th\u1ec3 l\u1ea5y \u0111\u01b0\u1ee3c 4 qu\u1ea3 tr\u1ee9ng n\u1eefa? B\u1ea1n h\u00e3y gi\u00fap c\u00f4 Lan nh\u00e9! C\u00e2u 10. M\u1ed9t s\u1ed1 ti\u1ec1n g\u1ed3m 20 t\u1edd b\u1ea1c v\u1eeba lo\u1ea1i 5.000 v\u1eeba lo\u1ea1i 10.000. S\u1ed1 ti\u1ec1n lo\u1ea1i 10.000 nhi\u1ec1u h\u01a1n s\u1ed1 ti\u1ec1n lo\u1ea1i 5.000 l\u00e0 125.000. T\u00ednh s\u1ed1 t\u1edd b\u1ea1c m\u1ed7i lo\u1ea1i?","B\u00e0i 40 C\u00e2u 1. K\u1ef7 v\u00e0 T\u1ef5 \u0111em g\u00e0 ra ch\u1ee3 \u0111\u1ec3 \u0111\u1ed5i l\u1ea5y ng\u1ef1a v\u00e0 b\u00f2. H\u1ecd t\u00ednh r\u1eb1ng: c\u1ee9 85 con g\u00e0 th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 con ng\u1ef1a v\u00e0 1 con b\u00f2, c\u1ee9 5 con ng\u1ef1a th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 12 con b\u00f2. Sau khi \u0111\u00e3 \u0111\u1ed5i \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 b\u00f2 v\u00e0 ng\u1ef1a, h\u1ecd b\u00e0n v\u1edbi nhau: K\u1ef7 n\u00f3i: \\\"N\u1ebfu ta \u0111\u1ed5i th\u00eam m\u1ed9t s\u1ed1 ng\u1ef1a n\u1eefa b\u1eb1ng \u0111\u00fang s\u1ed1 ng\u1ef1a m\u00e0 ta v\u1eeba \u0111\u1ed5i th\u00ec ta s\u1ebd \u0111\u01b0\u1ee3c 17 con c\u1ea3 ng\u1ef1a l\u1eabn b\u00f2, nh\u01b0ng nh\u01b0 th\u1ebf s\u1ed1 g\u00e0 kh\u00f4ng \u0111\u1ee7 \u0111\u1ec3 \u0111\u1ed5i\\\". T\u1ef5 n\u00f3i: \\\"N\u1ebfu ta \u0111\u1ed5i th\u00eam m\u1ed9t s\u1ed1 b\u00f2 n\u1eefa b\u1eb1ng \u0111\u00fang s\u1ed1 b\u00f2 m\u00e0 ta v\u1eeba \u0111\u1ed5i th\u00ec ch\u1eb3ng nh\u1eefng ta s\u1ebd \u0111\u01b0\u1ee3c 19 con c\u1ea3 b\u00f2 l\u1eabn ng\u1ef1a m\u00e0 s\u1ed1 g\u00e0 \u0111em \u0111i \u0111\u1ed5i c\u0169ng v\u1eeba v\u1eb7n h\u1ebft\\\". \u00dd h\u1ecd \u0111\u1ec1u \u0111\u00fang, b\u1ea1n h\u00e3y t\u00ednh xem K\u1ef7 v\u00e0 T\u1ef5 \u0111em bao nhi\u00eau con g\u00e0 ra ch\u1ee3\\\"? C\u00e2u 2. \u0110\u1ec3 gi\u00fap \u0111\u1ee1 c\u00e1c b\u1ea1n h\u1ecdc sinh ngh\u00e8o h\u1ecdc l\u1edbp ph\u1ed5 c\u1eadp ban \u0111\u00eam, l\u1edbp em \u0111\u00e3 mua 12 c\u00e2y b\u00fat c\u00e1c lo\u1ea1i, gi\u00e1 t\u1ed5ng c\u1ed9ng 36.000 \u0111\u1ed3ng. Bi\u1ebft m\u1ed7i c\u00e2y b\u00fat m\u00e1y gi\u00e1 6.000 \u0111\u1ed3ng, hai c\u00e2y b\u00fat bic gi\u00e1 3.000 \u0111\u1ed3ng, 4 c\u00e2y b\u00fat ch\u00ec gi\u00e1 3.000 \u0111\u1ed3ng. H\u1ecfi l\u1edbp em \u0111\u00e3 mua bao nhi\u00eau c\u00e2y b\u00fat m\u1ed7i lo\u1ea1i? C\u00e2u 3. Tr\u01b0\u1eddng em v\u1eeba m\u1edbi xong Tr\u0103m b\u00e0n \u0111em x\u1ebfp 9 ph\u00f2ng \u0111\u1ee7 y 5 ph\u00f2ng \u0111\u00f4ng h\u1ecdc sinh th\u00ec 12 b\u00e0n m\u1ed7i l\u1edbp, nh\u1edb ghi k\u1ebbo nh\u1ea7m","S\u1ed1 ph\u00f2ng c\u00f2n l\u1ea1i \u0111\u1ec1u ph\u00e2n S\u1ed1 b\u00e0n m\u1ed7i l\u1edbp xa g\u1ea7n t\u00ednh xem? \u0110\u1ed1 ai, \u0111\u1ed1 b\u1ea1n, \u0111\u1ed1 em \u0110\u00e1p nhanh \u0111\u00fang, th\u01b0\u1edfng qu\u00e0 k\u00e8m ph\u00e1o tay. C\u00e2u 4. Kho\u1ea3ng th\u1eddi gian t\u1eeb \u0111\u1ea7u th\u00e1ng cho t\u1edbi ng\u00e0y \u0111\u1ea1i h\u1ed9i Chi \u0111\u1ed9i g\u1ea5p 3 l\u1ea7n kho\u1ea3ng th\u1eddi gian t\u1eeb sau ng\u00e0y \u0111\u1ea1i h\u1ed9i Chi \u0111\u1ed9i \u0111\u1ebfn cu\u1ed1i th\u00e1ng. H\u1ecfi Chi \u0111\u1ed9i ti\u1ebfn h\u00e0nh \u0111\u1ea1i h\u1ed9i v\u00e0o ng\u00e0y n\u00e0o, th\u00e1ng n\u00e0o? C\u00e2u 5. Ng\u01b0\u1eddi ta \u0111\u1eb7t trong kho 6 th\u00f9ng r\u01b0\u1ee3u. T\u1eeb th\u00f9ng th\u1ee9 nh\u1ea5t \u0111\u1ebfn th\u00f9ng th\u1ee9 6 t\u01b0\u01a1ng \u1ee9ng ch\u1ee9a: 310 l\u00edt, 200 l\u00edt, 190 l\u00edt, 180 l\u00edt, 160 l\u00edt v\u00e0 150 l\u00edt. Ng\u00e0y th\u1ee9 nh\u1ea5t 2 ng\u01b0\u1eddi mang r\u01b0\u1ee3u \u0111i b\u00e1n, ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t b\u00e1n \u0111\u01b0\u1ee3c 2 th\u00f9ng, ng\u01b0\u1eddi th\u1ee9 hai b\u00e1n \u0111\u01b0\u1ee3c 3 th\u00f9ng, h\u01a1n n\u1eefa ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t b\u00e1n \u0111\u01b0\u1ee3c s\u1ed1 r\u01b0\u1ee3u b\u1eb1ng m\u1ed9t n\u1eeda s\u1ed1 r\u01b0\u1ee3u ng\u01b0\u1eddi th\u1ee9 hai \u0111\u00e3 b\u00e1n. H\u1ecfi th\u00f9ng r\u01b0\u1ee3u n\u00e0o c\u00f2n trong kho? C\u00e2u 6. C\u00f3 bao nhi\u00eau h\u00ecnh tam gi\u00e1c trong h\u00ecnh d\u01b0\u1edbi \u0111\u00e2y?","C\u00e2u 7. \u0110i\u1ec1n s\u1ed1 c\u00f2n thi\u1ebfu v\u00e0o ch\u1ed7 tr\u1ed1ng sao cho th\u00edch h\u1ee3p 432 420 96? C\u00e2u 8. H\u00ecnh n\u00e0o c\u00f3 t\u00ednh ch\u1ea5t kh\u00e1c v\u1edbi c\u00e1c h\u00ecnh c\u00f2n l\u1ea1i? C\u00e2u 9. \u0110i\u1ec1n s\u1ed1 v\u00e0o d\u1ea5u (?). 7 | 49 | 441 | ? C\u00e2u 10. Th\u1ea7y hi\u1ec7u tr\u01b0\u1edfng \u0111\u1ebfn m\u1ed9t v\u01b0\u1eddn \u01b0\u01a1m c\u00e2y \u0111\u1ec3 mua c\u00e2y non v\u1ec1 tr\u1ed3ng xung quanh tr\u01b0\u1eddng. L\u1ea7n th\u1ee9 nh\u1ea5t th\u1ea7y mua 10 c\u00e2y ph\u01b0\u1ee3ng v\u00e0 8 c\u00e2y \u0111i\u1ec7p h\u1ebft t\u1ea5t c\u1ea3 l\u00e0 64.000 \u0111\u1ed3ng. L\u1ea7n th\u1ee9 hai mua 7 c\u00e2y ph\u01b0\u1ee3ng v\u00e0 4 c\u00e2y \u0111i\u1ec7p h\u1ebft t\u1ea5t c\u1ea3 40.000 \u0111\u1ed3ng. T\u00ednh gi\u00e1 ti\u1ec1n 1 c\u00e2y ph\u01b0\u1ee3ng v\u00e0 1 c\u00e2y \u0111i\u1ec7p.","C\u00e2u 1. Trong cu\u1ed9c h\u1ecdp c\u00f3 12 ng\u01b0\u1eddi tham d\u1ef1. M\u1ed7i m\u1ed9t ng\u01b0\u1eddi l\u1ea1i b\u1eaft tay nh\u1eefng ng\u01b0\u1eddi c\u00f2n l\u1ea1i tr\u01b0\u1edbc v\u00e0 sau cu\u1ed9c h\u1ecdp. H\u1ecfi c\u00f3 t\u1ed5ng s\u1ed1 bao nhi\u00eau c\u00e1i b\u1eaft tay? C\u00e2u 2. C\u00f3 bao nhi\u00eau tu\u1ea7n m\u00e0 c\u00f3 156 gi\u1edd? C\u00e2u 3. H\u00ecnh n\u00e0o kh\u00e1c v\u1edbi c\u00e1c h\u00ecnh c\u00f2n l\u1ea1i? C\u00e2u 4. S\u1ed1 n\u00e0o thay cho X? 14 - (-7) - (-7) = X","C\u00e2u 5. Nam gi\u1ea3i 1 b\u00e0i to\u00e1n v\u00e0 l\u00e0m 4 ph\u00e9p t\u00ednh m\u1ea5t 38 ph\u00fat 16 gi\u00e2y. Nam gi\u1ea3i m\u1ed9t b\u00e0i to\u00e1n l\u00e2u g\u1ea5p 4 l\u1ea7n m\u1ed9t ph\u00e9p t\u00ednh. H\u1ecfi trung b\u00ecnh Nam l\u00e0m 1 ph\u00e9p t\u00ednh m\u1ea5t bao nhi\u00eau th\u1eddi gian? C\u00e2u 6. T\u1ed5ng 3 ch\u1eef s\u1ed1 b\u1eb1ng 10 S\u1ed1 3 ch\u1eef s\u1ed1, \u0111\u1ed1 ng\u01b0\u1eddi gi\u1ecfi ch\u0103m \u0110\u01a1n v\u1ecb g\u1ea5p 9 h\u00e0ng tr\u0103m S\u1ed1 g\u00ec \u0111\u00e1p \u0111\u00fang, \u0111\u00e2y kh\u00e2m ph\u1ee5c t\u00e0i Th\u1eed xem gi\u1ea3i nh\u1ea5t v\u1ec1 ai Nghe t\u00f4i h\u00f4 nh\u00e9, \\\"m\u1ed9t, hai\\\" \u0111\u00e1p li\u1ec1n. C\u00e2u 7. H\u00f4m nay l\u00e0 th\u1ee9 t\u01b0. Ng\u00e0y th\u1ee9 t\u01b0 k\u1ec3 t\u1eeb ng\u00e0y h\u00f4m qua l\u00e0 ng\u00e0y n\u00e0o? a. Ch\u1ee7 nh\u1eadt b. Th\u1ee9 2 c. Th\u1ee9 6 d. Th\u1ee9 5 e. Th\u1ee9 7 C\u00e2u 8. N\u1ebfu b\u1ea1n \u0111em nh\u00e2n b\u1ea5t k\u1ef3 hai trong c\u00e1c s\u1ed1 thu\u1ed9c d\u00e3y s\u1ed1 sau, b\u1ea1n s\u1ebd \u0111\u1ec1u c\u00f3 k\u1ebft qu\u1ea3 l\u00e0 m\u1ed9t s\u1ed1 l\u1ebb: 9, 11, 3, 7, 17.","\u0110i\u1ec1u n\u00e0y \u0111\u00fang hay sai? C\u00e2u 9. \u0110\u1ea7u n\u0103m h\u1ecdc, c\u00f4 gi\u00e1o d\u1ef1 \u0111\u1ecbnh s\u1eafp x\u1ebfp ch\u1ed7 ng\u1ed3i cho l\u1edbp em. C\u00f4 \u0111\u1ecbnh x\u1ebfp m\u1ed7i b\u00e0n 3 em, nh\u01b0 v\u1eady th\u00ec 4 em kh\u00f4ng c\u00f3 ch\u1ed7 ng\u1ed3i, c\u00f2n n\u1ebfu x\u1ebfp m\u1ed7i b\u00e0n 4 em th\u00ec c\u00f2n tr\u1ed1ng m\u1ed9t b\u00e0n. H\u1ecfi: - Trong l\u1edbp c\u00f3 bao nhi\u00eau b\u00e0n? - S\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp l\u00e0 bao nhi\u00eau?","PH\u1ea6N II: H\u01af\u1edaNG D\u1eaaN - B\u00c0I GI\u1ea2I B\u00e0i 1 C\u00e2u 1: \u0110\u00e1p s\u1ed1: 132 T\u01b0 duy theo c\u00e1ch n\u00e0y: Ng\u01b0\u1eddi \u0111\u1ea7u ti\u00ean b\u1eaft tay v\u1edbi 11 ng\u01b0\u1eddi, ng\u01b0\u1eddi th\u1ee9 hai c\u0169ng b\u1eaft tay v\u1edbi 11 ng\u01b0\u1eddi, nh\u01b0ng c\u00e1c em ch\u1ec9 t\u00ednh 10 ng\u01b0\u1eddi v\u00ec c\u00e1i b\u1eaft tay v\u1edbi ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t \u0111\u00e3 \u0111\u01b0\u1ee3c t\u00ednh. Sau \u0111\u00f3 ng\u01b0\u1eddi th\u1ee9 ba s\u1ebd t\u00ednh l\u00e0 9 ng\u01b0\u1eddi, ng\u01b0\u1eddi th\u1ee9 t\u01b0 t\u00ednh l\u00e0 8 ng\u01b0\u1eddi v\u00e0 ti\u1ebfp t\u1ee5c s\u1ebd c\u00f3 66 c\u00e1i b\u1eaft tay tr\u01b0\u1edbc v\u00e0 sau cu\u1ed9c h\u1ecdp, t\u1ed5ng s\u1ed1 l\u00e0 132 c\u00e1i b\u1eaft tay. C\u00e2u 2: \u0110\u00e1p s\u1ed1: T\u1ea5t c\u1ea3 c\u00e1c tu\u1ea7n. M\u1ed9t c\u00e2u h\u1ecfi \u0111\u00e1nh l\u1eeba! C\u00e2u 3: \u0110\u00e1p s\u1ed1: H\u00ecnh C v\u00ec \u0111\u00e2y l\u00e0 h\u00ecnh duy nh\u1ea5t c\u00f3 3 \u0111\u01b0\u1eddng k\u1ebb. C\u00e2u 4: \u0110\u00e1p s\u1ed1: 28 C\u00e2u 5: \u0110\u00e1p s\u1ed1: 4 ph\u00fat 47 gi\u00e2y.","Gi\u1ea3i m\u1ed9t b\u00e0i to\u00e1n l\u00e2u g\u1ea5p 4 l\u1ea7n l\u00e0m m\u1ed9t ph\u00e9p t\u00ednh n\u00ean trong 38 ph\u00fat 18 gi\u00e2y Nam gi\u1ea3i \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 ph\u00e9p t\u00ednh l\u00e0: 4 + 4 = 8 (ph\u00e9p t\u00ednh) Th\u1eddi gian gi\u1ea3i m\u1ed9t ph\u00e9p t\u00ednh l\u00e0: 38 ph\u00fat 16 gi\u00e2y : 8 = 4 ph\u00fat 47 gi\u00e2y. C\u00e2u 6: \u0110\u00e1p s\u1ed1: S\u1ed1 109 H\u00e0ng \u0111\u01a1n v\u1ecb g\u1ea5p 9 l\u1ea7n h\u00e0ng tr\u0103m \u21d2 H\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 9, h\u00e0ng tr\u0103m l\u00e0 1 T\u1ed5ng ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb v\u00e0 h\u00e0ng tr\u0103m l\u00e0: \u21d2 9 + 1 = 10 Theo \u0111\u1ec1, t\u1ed5ng 3 ch\u1eef s\u1ed1 l\u00e0 10. V\u1eady ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c ph\u1ea3i l\u00e0 0 V\u1eady s\u1ed1 3 ch\u1eef s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 109 C\u00e2u 7: \u0110\u00e1p s\u1ed1: Th\u1ee9 7. C\u00e2u 8: \u0110\u00e1p s\u1ed1: \u0110\u00fang. C\u00e2u 9: 8 c\u00e1i b\u00e0n v\u00e0 28 h\u1ecdc sinh.","V\u00ec x\u1ea3y ra tr\u01b0\u1eddng h\u1ee3p tr\u1ed1ng m\u1ed9t c\u00e1i b\u00e0n n\u00ean s\u1ed1 b\u00e0n kh\u00f4ng th\u1ec3 l\u00e0 1. Ta b\u1eaft \u0111\u1ea7u th\u1eed t\u1eeb s\u1ed1 b\u00e0n l\u00e0 2. L\u00fac \u0111\u00f3: Theo c\u00e1ch s\u1eafp x\u1ebfp m\u1ed7i b\u00e0n 3 em th\u00ec s\u1ed1 h\u1ecdc sinh l\u00e0: 2 x 3 + 4 = 10 (HS) Theo c\u00e1ch s\u1eafp x\u1ebfp m\u1ed7i b\u00e0n 4 em th\u00ec s\u1ed1 h\u1ecdc sinh l\u00e0: (2 - 1) x 4 = 4 (HS) V\u00ec 10 HS kh\u00e1c 4 HS n\u00ean tr\u01b0\u1eddng h\u1ee3p n\u00e0y b\u1ecb lo\u1ea1i. C\u1ee9 nh\u01b0 v\u1eady, ta th\u1eed v\u1edbi s\u1ed1 b\u00e0n l\u00e0 3, 4, 5, ... v\u00e0 ghi k\u1ebft qu\u1ea3 v\u00e0o b\u1ea3ng nh\u01b0 sau: S\u1ed1 b\u00e0n S\u1ed1 HS (n\u1ebfu x\u1ebfp 3 em 1 b\u00e0n) S\u1ed1 HS (n\u1ebfu x\u1ebfp 4 em 1 b\u00e0n) Nh\u1eadn x\u00e9t K\u1ebft lu\u1eadn 3 3 x 3 + 4 = 13 (HS) (3 - 1) x 4 = 8 (HS) 13 kh\u00e1c 8 Lo\u1ea1i 4 4 x 3 + 4 = 16(HS) (4 - 1) x 4 = 12(HS) 16 kh\u00e1c 12 Lo\u1ea1i 5 5 x 3 + 4 = 19 (HS) (5 - 1) x 4 = 16(HS) 19 kh\u00e1c 16 Lo\u1ea1i 6 6 x 3 + 4 = 22(HS) (6 - 1) x 4= 20( HS ) 22 kh\u00e1c 20 Lo\u1ea1i 7 7 x 3 + 4 = 25(HS) (7 - 1) x 4=24 (HS) 25 kh\u00e1c 24 Lo\u1ea1i 8 8 x 3 +4 = 28(HS) ( 8 - 1 ) x 4 = 28 (HS) 28 b\u1eb1ng 28 Ch\u1ecdn V\u1eady l\u1edbp em c\u00f3 8 c\u00e1i b\u00e0n v\u00e0 28 h\u1ecdc sinh.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: H\u00ecnh A. C\u00e2u 2: \u0110\u00e1p s\u1ed1: 88 (S\u1ed1 k\u1ebf sau b\u1eb1ng s\u1ed1 li\u1ec1n tr\u01b0\u1edbc nh\u00e2n 3 r\u1ed3i tr\u1eeb cho 14). C\u00e2u 3: \u0110\u00e1p s\u1ed1: 16. S\u1ed1 tu\u1ed5i c\u1ee7a ch\u1ecb Mai l\u00fac n\u00e0o c\u0169ng h\u01a1n tu\u1ed5i Mai l\u00e0 8 tu\u1ed5i. Do v\u1eady khi ch\u1ecb Mai 16 tu\u1ed5i th\u00ec s\u1ebd g\u1ea5p \u0111\u00f4i tu\u1ed5i Mai. C\u00e2u 4: \u0110\u00e1p s\u1ed1: d. C\u00e2u 5: \u0110\u00e1p s\u1ed1: T\u1ed5ng s\u1ed1 l\u00e0 15 l\u1ea7n. C\u00e2u 6: \u0110\u00e1p s\u1ed1: H\u00ecnh D. C\u00e2u 7: \u0110\u00e1p s\u1ed1: 22 con g\u00e0 v\u00e0 14 con ch\u00f3.","Gi\u1ea3 s\u1eed s\u1ed1 ch\u00f3 l\u00e0 10, l\u00fac \u0111\u00f3 s\u1ed1 g\u00e0 s\u1ebd l\u00e0: 10 + 8 = 18 (con g\u00e0) V\u1eady hi\u1ec7u s\u1ed1 ch\u00e2n ch\u00f3 v\u00e0 ch\u00e2n g\u00e0 s\u1ebd l\u00e0: 10 x 4 - 18 x 2 = 4 (ch\u00e2n) N\u1ebfu ta th\u00eam m\u1ed9t con g\u00e0 v\u00e0 th\u00eam m\u1ed9t con ch\u00f3 th\u00ec hi\u1ec7u s\u1ed1 g\u00e0 v\u00e0 ch\u00f3 kh\u00f4ng thay \u0111\u1ed5i (v\u1eabn l\u00e0 8), nh\u01b0ng hi\u1ec7u s\u1ed1 ch\u00e2n s\u1ebd t\u0103ng th\u00eam: 4 - 2 = 2 (ch\u00e2n) V\u1eady s\u1ed1 g\u00e0 (c\u0169ng l\u00e0 s\u1ed1 ch\u00f3) ph\u1ea3i th\u00eam v\u00e0o l\u00e0: 8 : 2 = 4 (con) Suy ra s\u1ed1 g\u00e0 l\u00e0: 18 + 4 = 22 (con) C\u00f2n s\u1ed1 ch\u00f3 l\u00e0: 36 - 22 = 14 (con) C\u00e2u 8: \u0110\u00e1p s\u1ed1: 30 xe. S\u1ed1 l\u1ed1p xe c\u00f3 t\u1ea5t c\u1ea3 l\u00e0: 40 + 50 = 90 (chi\u1ebfc) M\u1ed7i xe x\u00edch l\u00f4 c\u00f3 3 b\u00e1nh, c\u1ea7n 3 chi\u1ebfc l\u1ed1p S\u1ed1 xe \u0111\u01b0\u1ee3c l\u1eafp l\u00e0: 90 : 3 = 30 (xe) C\u00e2u 9: \u0110\u00e1p s\u1ed1: L\u00e0 s\u1ed1 15 v\u00ec \u0111\u00e2y l\u00e0 s\u1ed1 duy nh\u1ea5t l\u00e0 s\u1ed1 l\u1ebb.","C\u00e2u 10: \u0110\u00e1p s\u1ed1: 27 c\u00f4ng nh\u00e2n. G\u1ee3i \u00fd: T\u00ednh s\u1ed1 c\u00f4ng nh\u00e2n c\u1ea7n c\u00f3 \u0111\u1ec3 d\u1ec7t 180 t\u00e0 \u00e1o trong 2 ng\u00e0y b\u1eb1ng c\u00e1ch gi\u1ea3i hai b\u00e0i to\u00e1n tam su\u1ea5t nh\u01b0 sau: B\u00e0i to\u00e1n th\u1ee9 nh\u1ea5t: tam su\u1ea5t ngh\u1ecbch (120 t\u00e1) 3 ng\u00e0y \u21d2 12 ng\u01b0\u1eddi (120 t\u00e1) 2 ng\u00e0y \u21d2 ? ng\u01b0\u1eddi Sau khi t\u00ecm ra k\u1ebft qu\u1ea3 18 ng\u01b0\u1eddi, th\u00ec gi\u1ea3i ti\u1ebfp. B\u00e0i to\u00e1n th\u1ee9 hai: tam su\u1ea5t thu\u1eadn (2 ng\u00e0y) 120 t\u00e1 \u21d2 18 ng\u01b0\u1eddi (2 ng\u00e0y) 180 t\u00e1 \u21d2 ? ng\u01b0\u1eddi","B\u00e0i 3 C\u00e2u 1: \u0110\u00e1p s\u1ed1: Sai. C\u00e2u \u0111\u00fang ph\u1ea3i l\u00e0 m\u1ed9t s\u1ed1 x\u00fac x\u00edch ng\u00e1y. C\u00e2u 2: \u0110\u00e1p s\u1ed1: 64 D\u00e3y s\u1ed1 n\u00e0y theo quy lu\u1eadt, s\u1ed1 sau b\u1eb1ng s\u1ed1 tr\u01b0\u1edbc nh\u00e2n 2. Do v\u1eady s\u1ed1 ti\u1ebfp theo d\u00e3y s\u1ed1 n\u00e0y l\u00e0 s\u1ed1 64. C\u00e2u 3: \u0110\u00e1p s\u1ed1: C nh\u1eadn \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n B 3.150.000 \u0111\u1ed3ng. C\u00e2u 4: \u0110\u00e1p s\u1ed1:An c\u00f2n l\u1ea1i 4 c\u00e1i. C\u00e2u 5: \u0110\u00e1p s\u1ed1: H\u00ecnh b C\u00e2u 6: B\u00e0i gi\u1ea3i:","N\u1ebfu coi s\u1ed1 qu\u00fdt l\u00e0 m\u1ed9t ph\u1ea7n th\u00ec 36 qu\u1ea3 g\u1ed3m: 4 - 1 = 3 (ph\u1ea7n) S\u1ed1 qu\u1ea3 qu\u00fdt l\u00e0: 36 : 3 = 12 (qu\u1ea3) S\u1ed1 qu\u1ea3 cam l\u00e0: 12 x 4 = 48 (qu\u1ea3) \u0110\u00e1p s\u1ed1: 48 qu\u1ea3 cam v\u00e0 12 qu\u1ea3 qu\u00fdt. C\u00e2u 7: B\u00e0i gi\u1ea3i: Chu vi khung \u1ea3nh l\u00e0: (80 + 60) x 2 = 280 (cm) S\u1ed1 hoa c\u1ea7n d\u00f9ng l\u00e0: 280 : 10 = 28 (b\u00f4ng hoa) \u0110\u00e1p s\u1ed1: 28 b\u00f4ng hoa. C\u00e2u 8: B\u00e0i gi\u1ea3i: Gi\u00e1 ti\u1ec1n 1 c\u00e1i b\u00e0n h\u01a1n gi\u00e1 ti\u1ec1n 1 c\u00e1i gh\u1ebf l\u00e0: 160.000 - 100.000 = 60.000 (\u0111) N\u1ebfu nh\u01b0 s\u1ed1 b\u00e0n b\u1eb1ng s\u1ed1 gh\u1ebf th\u00ec vi\u1ec7c thay \u0111\u1ed5i s\u1ed1 b\u00e0n v\u00e0 s\u1ed1 gh\u1ebf cho nhau s\u1ebd kh\u00f4ng l\u00e0m gi\u00e1 ti\u1ec1n thay \u0111\u1ed5i, song v\u00ec ph\u1ea3i tr\u1ea3 th\u00eam 120.000\u0111 n\u1eefa n\u00ean th\u1ef1c t\u1ebf s\u1ed1 gh\u1ebf ph\u1ea3i nhi\u1ec1u h\u01a1n s\u1ed1 b\u00e0n l\u00e0: 120.000 : 60.000 = 2 (c\u00e1i) S\u1ed1 ti\u1ec1n mua hai gh\u1ebf l\u00e0:","100.000 x 2 = 200.000 (\u0111) N\u1ebfu b\u1edbt \u0111i 2 c\u00e1i gh\u1ebf th\u00ec s\u1ed1 b\u00e0n b\u1eb1ng s\u1ed1 gh\u1ebf, do \u0111\u00f3 s\u1ed1 ti\u1ec1n mua c\u1ea3 b\u00e0n l\u1eabn gh\u1ebf l\u00fac \u0111\u00f3 s\u1ebd l\u00e0: 1.240.000 - 200.000 = 1.040.000 (\u0111) S\u1ed1 ti\u1ec1n mua 1 b\u00e0n v\u00e0 1 gh\u1ebf l\u00e0: 160.000 + 100.000 = 260.000 (\u0111) S\u1ed1 b\u00e0n \u0111\u00e3 mua l\u00e0: 1.040.000 : 260.000 = 4 (c\u00e1i) S\u1ed1 gh\u1ebf \u0111\u00e3 mua l\u00e0: 4 + 2 = 6 (c\u00e1i) \u0110\u00e1p s\u1ed1: 4 c\u00e1i b\u00e0n v\u00e0 6 c\u00e1i gh\u1ebf.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: T\u1ed5ng c\u1ed9ng b\u1ea1n \u0103n 14 qu\u1ea3. C\u00e2u 2: \u0110\u00e1p s\u1ed1: C\u00f3 12 c\u00e1i. C\u00e2u 3: \u0110\u00e1p s\u1ed1: M\u1ed9t gi\u1edd c\u00f3 60 ph\u00fat. C\u00e2u 4: \u0110\u00e1p s\u1ed1: 495.000\u0111. C\u00e2u 5: \u0110\u00e1p s\u1ed1: M\u1ea5t 1 ph\u00fat c\u1ea3 \u0111o\u00e0n t\u00e0u s\u1ebd qua h\u1ebft \u0111\u01b0\u1eddng h\u1ea7m (0,5 + 0,25) x 80\/60 = 1 (ph\u00fat). C\u00e2u 6: \u0110\u00e1p s\u1ed1: b. 32\/25 C\u00e2u 7: \u0110\u00e1p s\u1ed1: 14 = 2 x 7","14 = 1 x 2 x 7. C\u00e2u 8: \u0110\u00e1p s\u1ed1: \u1ede \u0111\u00e1y c\u00e1i l\u1ed3ng n\u01a1i m\u00e0 l\u00e0 m\u1ed9t h\u00ecnh \u0111a gi\u00e1c (Polly \u0111\u00e3 \u0111i). \u0110\u00e1p \u00e1n e. C\u00e2u 9: \u0110\u00e1p s\u1ed1: 56 ph\u00fat. 12m = 120dm S\u1ed1 \u0111o\u1ea1n g\u1ed7 l\u00e0: 120 : 15 = 8 (\u0111o\u1ea1n) S\u1ed1 l\u1ea7n c\u01b0a l\u00e0: 8 - 1 = 7 (l\u1ea7n) Th\u1eddi gian c\u1ee7a m\u1ed7i l\u1ea7n c\u01b0a v\u00e0 ngh\u1ec9 l\u00e0: 6 + 2 = 8 (ph\u00fat) Th\u1eddi gian \u0111\u1ec3 c\u01b0a xong c\u00e2y g\u1ed7 l\u00e0: 7 x 8 = 56 (ph\u00fat) C\u00e2u 10: \u0110\u00e1p s\u1ed1: 480 S\u1ed1 \u1edf gi\u1eefa b\u1eb1ng hai l\u1ea7n t\u00edch 2 s\u1ed1 \u1edf ngo\u00e0i.","B\u00e0i 5 C\u00e2u 1: \u0110\u00e1p s\u1ed1: H\u00ecnh C. C\u00e2u 2: \u0110\u00e1p s\u1ed1: C\u00f3 30 gi\u00e2y. C\u00e2u 3: \u0110\u00e1p s\u1ed1: M\u1ed7i b\u1ea1n \u0111\u01b0\u1ee3c 3 qu\u1ea3. C\u00e2u 4: \u0110\u00e1p s\u1ed1: 22 (S\u1ed1 li\u1ec1n sau b\u1eb1ng t\u1ed5ng 2 s\u1ed1 li\u1ec1n tr\u01b0\u1edbc tr\u1eeb \u0111i 1). C\u00e2u 5: \u0110\u00e1p s\u1ed1: 25 C\u00e2u 6: \u0110\u00e1p s\u1ed1: 52 x 51 x 50 x 49 = 6.497.400 c\u00e1ch. C\u00e2u 7: \u0110\u00e1p s\u1ed1: H = 29, T = 16, K = 40 H = 8 + 5 + 5 + 3 + 8 = 29","T = 3 + 8 + 5 = 16 K = 5 + 3 +3 + 5 + 8 + 8 + 3 +5 = 40 C\u00e2u 8: \u0110\u00e1p s\u1ed1: H\u00ecnh C. C\u00e2u 9: 1kg t\u00e1o: 5.000\u0111 1kg m\u1eadn: 2.5000\u0111 V\u00ec 21 : 14 = 1,5 n\u00ean th\u1ec3 t\u00edch t\u00e1o b\u1eb1ng th\u1ec3 t\u00edch 1,5kg m\u1eadn. N\u1ebfu c\u1ea3 s\u1ecdt \u0111\u1ef1ng t\u00e1o th\u00ec ch\u1ec9 n\u1eb7ng 14kg. B\u00e2y gi\u1edd n\u1ebfu thay th\u1ebf 1kg t\u00e1o b\u1eb1ng 1,5kg m\u1eadn th\u00ec s\u1ecdt s\u1ebd n\u1eb7ng th\u00eam: 1,5 - 1 = 0,5(kg) T\u1eeb 14kg tr\u1edf l\u00ean 18kg, s\u1ecdt \u0111\u00e3 n\u1eb7ng th\u00eam 18 - 14 = 4(kg) V\u1eady ta \u0111\u00e3 thay th\u1ebf: 4 : 0,5 = 8(kg) t\u00e1o b\u1eb1ng m\u1eadn. Suy ra s\u1ed1 t\u00e1o trong r\u1ed5 l\u00e0: 14 - 8 = 6(kg) C\u00f2n s\u1ed1 m\u1eadn trong r\u1ed5 l\u00e0: 18 - 6 = 12(kg) Gi\u00e1 ti\u1ec1n t\u00e1o v\u00e0 m\u1eadn trong r\u1ed5 \u0111\u1ec1u l\u00e0: 60.000 : 2 = 30.000(\u0111) Gi\u00e1 1kg t\u00e1o l\u00e0:","30.000 : 6 = 5.000(\u0111) Gi\u00e1 1kg m\u1eadn l\u00e0: 30.000 : 12 = 2.5000(\u0111) C\u00e2u 10: \u0110\u00e1p s\u1ed1: 83 v\u00e0 38 24 = 6 x 4 24 = 8 x 3 V\u1eady c\u00e1c s\u1ed1 c\u1ea7n t\u00ecm c\u00f3 th\u1ec3 l\u00e0: 64, 46, 86, 38 Th\u1eed l\u1ea1i t\u1ed5ng hai ch\u1eef s\u1ed1: 6 + 4 = 10 V\u1eady ta lo\u1ea1i 2 s\u1ed1 64 v\u00e0 46 8 + 3 = 11 V\u1eady ta t\u00ecm \u0111\u01b0\u1ee3c 2 s\u1ed1 theo \u0111\u1ec1 b\u00e0i l\u00e0 83 v\u00e0 38","C\u00e2u 1: \u0110\u00e1p s\u1ed1: \u0110\u00fang. C\u00e2u 2: \u0110\u00e1p s\u1ed1: 8 C\u00e2u 3: Cha: 40 tu\u1ed5i, con: 13 tu\u1ed5i. Tu\u1ed5i c\u1ee7a cha l\u00e0: (53 + 27) : 2 = 40 (tu\u1ed5i) Tu\u1ed5i c\u1ee7a con l\u00e0: 53 - 40 = 13 (tu\u1ed5i) C\u00e2u 4: \u0110\u00e1p s\u1ed1: 15 \u0111i\u1ec3m. C\u00e2u 5: \u0110\u00e1p s\u1ed1: H\u00ecnh b C\u00e2u 6: \u0110\u00e1p s\u1ed1: C\u00e2u C","C\u00e2u 7: \u0110\u00e1p s\u1ed1: 78 S\u1ed1 gi\u1eefa b\u1eb1ng t\u1ed5ng hai s\u1ed1 ngo\u00e0i nh\u00e2n 3 C\u00e2u 8: T\u1ef1 gi\u1ea3i. C\u00e2u 9: 7 xe 4 b\u00e1nh ch\u1edf 5 t\u1ea5n 5 xe 6 b\u00e1nh ch\u1edf 6 t\u1ea5n 6 xe 8 b\u00e1nh ch\u1edf 6 t\u1ea5n. Gi\u1ea3 s\u1eed c\u1ea3 18 xe \u0111\u1ec1u ch\u1edf 6 t\u1ea5n th\u00ec s\u1ed1 t\u1ea5n ch\u1edf \u0111\u01b0\u1ee3c l\u00e0: 6 x 18 = 108 (t\u1ea5n) S\u1ed1 t\u1ea5n d\u01b0 ra l\u00e0: 108 - 101 = 7 (t\u1ea5n) S\u1ed1 t\u1ea5n d\u01b0 ra n\u00e0y l\u00e0 do ta \u0111\u00e3 thay xe ch\u1edf 5 t\u1ea5n b\u1eb1ng xe ch\u1edf 6 t\u1ea5n. V\u1eady s\u1ed1 xe 5 t\u1ea5n l\u00e0: 7 : (6 - 5) = 7 (xe) S\u1ed1 h\u00e0ng do 7 xe 5 t\u1ea5n ch\u1edf l\u00e0: 7 x 5 = 35 (t\u1ea5n) S\u1ed1 h\u00e0ng do c\u00e1c lo\u1ea1i xe 6 t\u1ea5n ch\u1edf l\u00e0: 101 - 35 = 66 (t\u1ea5n)","S\u1ed1 b\u00e1nh xe c\u1ee7a hai lo\u1ea1i xe ch\u1edf 6 t\u1ea5n l\u00e0: 106 - 4 x 7 = 78 (b\u00e1nh) S\u1ed1 xe 6 t\u1ea5n l\u00e0: 18 - 7 = 11 (xe) Gi\u1ea3 s\u1eed 11 xe \u0111\u1ec1u l\u00e0 lo\u1ea1i xe 6 b\u00e1nh th\u00ec s\u1ed1 b\u00e1nh xe l\u00e0: 6 x 11 = 66 (b\u00e1nh) S\u1ed1 b\u00e1nh xe b\u1ecb h\u1ee5t \u0111i l\u00e0: 78 - 66 = 12 (b\u00e1nh) 12 b\u00e1nh b\u1ecb h\u1ee5t \u0111i n\u00e0y l\u00e0 do ta \u0111\u00e3 thay c\u00e1c xe 8 b\u00e1nh b\u1eb1ng xe 6 b\u00e1nh. V\u1eady s\u1ed1 xe 8 b\u00e1nh l\u00e0: 12 : (8 - 6) = 6 (xe) S\u1ed1 xe 6 b\u00e1nh l\u00e0: 11 - 6 = 5 (xe) C\u00e2u 10: \u0110\u00e1p s\u1ed1: 42 c\u00e2y. 1km = 1.000m S\u1ed1 kho\u1ea3ng c\u00e1ch 50m trong 1.000m l\u00e0: 1.000 : 50 = 20 (kho\u1ea3ng c\u00e1ch) S\u1ed1 c\u00e2y \u1edf m\u1ed7i b\u00ean \u0111\u01b0\u1eddng l\u00e0: 20 + 1 = 21 (c\u00e2y) S\u1ed1 c\u00e2y \u1edf c\u1ea3 hai b\u00ean \u0111\u01b0\u1eddng l\u00e0:","21 x 2 = 42 (c\u00e2y) V\u1eady l\u00e0 c\u00f3: 42 c\u00e2y. Ghi nh\u1edb: T\u1eeb b\u01b0\u1edbc t\u00ednh th\u1ee9 hai ta th\u1ea5y: N\u1ebfu c\u00f3 tr\u1ed3ng c\u00e2y \u1edf c\u1ea3 hai \u0111\u1ea7u \u0111\u01b0\u1eddng th\u00ec: S\u1ed1 c\u00e2y = S\u1ed1 kho\u1ea3ng c\u00e1ch + 1.","B\u00e0i 7 C\u00e2u 1: \u0110\u00e1p s\u1ed1: B\u00e0i ki\u1ec3m tra c\u00f3 25 c\u00e2u. T\u00e8o tr\u1ea3 l\u1eddi sai 10 c\u00e2u v\u00e0 \u0111\u1ea1t 60% t\u1ed5ng \u0111i\u1ec3m, suy ra T\u00e8o \u0111\u00e3 m\u1ea5t 40% t\u1ed5ng \u0111i\u1ec3m trong 10 c\u00e2u sai \u0111\u00f3. V\u1eady 1 c\u00e2u s\u1ebd c\u00f3 4% t\u1ed5ng \u0111i\u1ec3m. S\u1ed1 c\u00e2u T\u00e8o l\u00e0m \u0111\u00fang l\u00e0 60% : 4% = 15 (c\u00e2u) V\u1eady b\u00e0i ki\u1ec3m tra s\u1ebd c\u00f3: 15 + 10 = 25 (c\u00e2u) C\u00e2u 2: \u0110\u00e1p s\u1ed1: 15 + 1 + 15 = 31 (ng\u01b0\u1eddi). C\u00e2u 3: \u0110\u00e1p s\u1ed1: 3 ng\u00e0y S\u1ed1 ng\u00e0y c\u00f2n l\u1ea1i ph\u1ea3i l\u00e0m theo d\u1ef1 \u0111\u1ecbnh l\u00e0: 7 - 3 = 4 (ng\u00e0y) S\u1ed1 ng\u01b0\u1eddi sau khi \u0111\u01b0\u1ee3c b\u1ed5 sung l\u00e0: 240 + 80 = 320 (ng\u01b0\u1eddi) V\u1edbi 1 ng\u01b0\u1eddi, \u0111\u1ec3 l\u00e0m xong qu\u00e3ng \u0111\u00ea c\u00f2n l\u1ea1i, c\u1ea7n: 4 x 240 = 960 (ng\u00e0y) V\u1edbi 320 ng\u00e0y, \u0111\u1ec3 l\u00e0m xong qu\u00e3ng \u0111\u00ea c\u00f2n l\u1ea1i, c\u1ea7n: 960 : 320 = 3 (ng\u00e0y)","C\u00e2u 4: 45 x 3 = 135 60 x 3 = 180 80 x 3 = 240 V\u1eady s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 105 x 3 = 315 C\u00e2u 5: \u0110\u00e1p s\u1ed1: 15 \u0111\u00f4la C\u00e2u 6: \u0110\u00e1p s\u1ed1: 10 khay. 3 t\u00e1 ly c\u00f3 l\u00e0: 12 x 3 = 36 (ly) S\u1ed1 ly c\u00f3 t\u1ea5t c\u1ea3 l\u00e0: 14 + 36 = 50 (ly) S\u1ed1 khay c\u1ea7n d\u00f9ng l\u00e0: 50 : 5 = 10 (khay) C\u00e2u 7: \u0110\u00e1p s\u1ed1: B. Th\u00eam hai \u0111\u01b0\u1eddng v\u00e0o t\u00f2a nh\u00e0 th\u1ee9 hai t\u1eeb tr\u00e1i sang \u0111\u1ec3 t\u1ea1o th\u00e0nh ch\u1eef \\\"X\\\" trong chu\u1ed7i ch\u1eef c\u00e1i l\u1ed9n ng\u01b0\u1ee3c l\u1ea1i: U, V, W, X, Y h\u00ecnh th\u00e0nh t\u1eeb c\u00e1c m\u00e1i c\u1ee7a t\u00f2a th\u00e1p. C\u00e2u 8:","\u0110\u00e1p s\u1ed1: Ch\u1ec9 c\u00f2n l\u1ea1i m\u1ed9t con b\u1ecb b\u1eafn ch\u1ebft v\u00ec c\u00e1c con kh\u00e1c nghe th\u1ea5y ti\u1ebfng s\u00fang n\u1ed5 li\u1ec1n c\u1ea5t c\u00e1nh bay \u0111i. C\u00e2u 9: \u0110\u00e1p s\u1ed1: H\u00ecnh c. C\u00e2u 10: S\u00e1u s\u1ed1 l\u00e0: 6, 8, 10, 12, 14, 16 Trong 6 s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp th\u00ec - S\u1ed1 th\u1ee9 hai h\u01a1n s\u1ed1 th\u1ee9 nh\u1ea5t 2 \u0111\u01a1n v\u1ecb - S\u1ed1 th\u1ee9 ba h\u01a1n s\u1ed1 th\u1ee9 nh\u1ea5t 4 \u0111\u01a1n v\u1ecb - S\u1ed1 th\u1ee9 t\u01b0 h\u01a1n s\u1ed1 th\u1ee9 nh\u1ea5t 6 \u0111\u01a1n v\u1ecb - S\u1ed1 th\u1ee9 n\u0103m h\u01a1n s\u1ed1 th\u1ee9 nh\u1ea5t 8 \u0111\u01a1n v\u1ecb - S\u1ed1 th\u1ee9 s\u00e1u h\u01a1n s\u1ed1 th\u1ee9 nh\u1ea5t 10 \u0111\u01a1n v\u1ecb N\u1ebfu thay t\u1ea5t c\u1ea3 s\u00e1u s\u1ed1 \u0111\u1ec1u b\u1eb1ng s\u1ed1 th\u1ee9 nh\u1ea5t th\u00ec t\u1ed5ng c\u1ee7a c\u1ea3 s\u00e1u s\u1ed1 \u0111\u1ec1u gi\u1ea3m \u0111i: 2 + 4 + 6 + 8 + 10 = 30 (\u0111\u01a1n v\u1ecb). V\u1eady s\u00e1u l\u1ea7n s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0: 66 - 30 = 36 S\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0: 36 : 6 = 6 S\u00e1u s\u1ed1 \u0111\u00f3 l\u00e0: 6, 8, 10, 12, 14, 16","C\u00e2u 1: \u0110\u00e1p s\u1ed1: 9 tu\u1ed5i. C\u00e2u 2. \u0110\u00e1p s\u1ed1: Sai. M\u1ec7nh \u0111\u1ec1 \u0111\u00fang ph\u1ea3i l\u00e0 M\u1ed9t s\u1ed1 chi\u1ebfc \u0111\u1ed3ng h\u1ed3 t\u1ea1o ra m\u00e1y vidio. C\u00e2u 3: 18 ng\u00e0y Ng\u01b0\u1eddi th\u1ee3 l\u00e0m trong 18 ng\u00e0y th\u00ec \u0111\u01b0\u1ee3c: 18 x 20.000 = 360.000(\u0111) Nh\u01b0ng ng\u01b0\u1eddi th\u1ee3 kh\u00f4ng l\u00e0m trong 12 ng\u00e0y th\u00ec b\u1ecb tr\u1eeb: 12 x 30.000 = 360.000(\u0111) C\u00e2u 4: \u0110\u00e1p s\u1ed1: a. C\u00e2u 5: \u0110\u00e1p s\u1ed1: d. C\u00e2u 6:","\u0110\u00e1p s\u1ed1: H\u00ecnh C C\u00e2u 7: \u0110\u00e1p s\u1ed1: 51 ch\u1eef s\u1ed1 0 T\u1eeb 1 \u0111\u1ebfn 99 ta c\u00f3 c\u00e1c s\u1ed1 mang ch\u1eef s\u1ed1 0 l\u00e0: 10, 20, 30, 40, 50, 60, 70, 80, 90 V\u1eady ta c\u1ea7n 9 ch\u1eef s\u1ed1 0 \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 t\u1eeb 101 \u0111\u1ebfn 109 ta c\u1ea7n 9 ch\u1eef s\u1ed1 0 \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 t\u1eeb 201 \u0111\u1ebfn 209 ta c\u1ea7n 9 ch\u1eef s\u1ed1 0 \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 tr\u00f2n ch\u1ee5c 110, 120, ...190 ta c\u1ea7n 9 ch\u1eef s\u1ed1 0 \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 tr\u00f2n ch\u1ee5c 210, 220...290 ta c\u1ea7n 9 ch\u1eef s\u1ed1 0 \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 100, 200, 300 ta c\u1ea7n 6 ch\u1eef s\u1ed1 0 V\u1eady \u0111\u1ec3 vi\u1ebft t\u1eeb 1 \u0111\u1ebfn 300, ta c\u1ea7n s\u1ed1 ch\u1eef s\u1ed1 0 l\u00e0: 9 x 5 + 6 = 51 (ch\u1eef s\u1ed1 0) C\u00e2u 8: S\u1ed1 th\u1ee9 hai l\u00e0 8 G\u1ee3i \u00fd: Coi s\u1ed1 th\u1ee9 hai l\u00e0 1 ph\u1ea7n th\u00ec: - S\u1ed1 th\u1ee9 nh\u1ea5t g\u1ed3m 10 ph\u1ea7n, s\u1ed1 th\u1ee9 t\u01b0 c\u0169ng v\u1eady - S\u1ed1 th\u1ee9 ba g\u1ed3m 4 ph\u1ea7n T\u1ed5ng c\u1ee7a 4 s\u1ed1 l\u00e0:","50 x 4 = 200 K\u1ebft lu\u1eadn 4 s\u1ed1 l\u1ea7n l\u01b0\u1ee3t l\u00e0: 80, 8, 32, 80 C\u00e2u 9: \u0110\u00e1p s\u1ed1: \u0110\u00fang. C\u00e2u 10: \u0110\u00e1p s\u1ed1: 77 (S\u1ed1 gi\u1eefa b\u1eb1ng 1\/2 t\u00edch 2 s\u1ed1 ngo\u00e0i).","B\u00e0i 9 C\u00e2u 1: \u0110\u00e1p s\u1ed1: 12 \u0111o\u1ea1n \u0111\u01b0\u1eddng 1h = 60 ph\u00fat 5 ph\u00fat \u0111i \u0111\u01b0\u1ee3c 1 \u0111o\u1ea1n \u0111\u01b0\u1eddng. V\u1eady 60 ph\u00fat \u0111i \u0111\u01b0\u1ee3c: 60 x 1 : 5 = 12 (\u0111o\u1ea1n \u0111\u01b0\u1eddng) C\u00e2u 2: \u0110\u00e1p s\u1ed1: 17 tu\u1ed5i. C\u00e2u 3: \u0110\u00e1p s\u1ed1: b C\u00e2u 4: \u0110\u00e1p s\u1ed1: 0 C\u00e2u 5: \u0110\u00e1p s\u1ed1: 412 Tr\u00ecnh t\u1ef1 l\u00e0 1, 2, 4, 8, 16, 32, 64, 128, 256; s\u1eafp x\u1ebfp th\u00e0nh c\u00e1c nh\u00f3m ba: 124, 816, 326, 412, 825","C\u00e2u 6: \u0110\u00e1p s\u1ed1: 8 ng\u01b0\u1eddi. 75 + 65 + 85 + 80 = 305 Con s\u1ed1 n\u00e0y cho bi\u1ebft c\u00f3 ba lo\u1ea1i m\u00e1y cho t\u1ea5t c\u1ea3 100 ng\u01b0\u1eddi, b\u1ed1n lo\u1ea1i m\u00e1y cho 5 ng\u01b0\u1eddi trong s\u1ed1 h\u1ecd. C\u00e2u 7: \u0110\u00e1p s\u1ed1: \u0110\u00fang C\u00e2u 8: 3+2=5 5 x 2 = 10 10 + 2 = 12 C\u1ee9 theo quy lu\u1eadt \u0111\u00f3, s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 52. C\u00e2u 9: \u0110\u00e1p s\u1ed1: 10 \u0111\u01a1n v\u1ecb. V\u00ec ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c ph\u1ea3i kh\u00e1c 0 n\u00ean ch\u1ec9 c\u00f3 9 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 \u1edf h\u00e0ng ch\u1ee5c. V\u1edbi m\u1ed7i c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 \u1edf h\u00e0ng ch\u1ee5c th\u00ec c\u00f3 10 - 1 = 9 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb. S\u1ebd c\u00f3 t\u1ea5t c\u1ea3 9 x 9 = 81 (s\u1ed1) V\u1eady c\u00e1c s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 kh\u00e1c nhau \u0111\u01b0\u1ee3c vi\u1ebft t\u1eeb 10 ch\u1eef s\u1ed1 0, 1, 2,..., 8, 9 l\u00e0 81 s\u1ed1. + X\u00e9t d\u00e3y s\u1ed1 10, 11, 12,..., 98, 99 D\u00e3y s\u1ed1 tr\u00ean g\u1ed3m 45 s\u1ed1 ch\u1eb5n v\u00e0 45 s\u1ed5 l\u1ebb nh\u01b0 sau:","10 12 14 16 .......... 94 96 98 (A) 11 13 15 17 .......... 95 97 99 (B) M\u1ed7i s\u1ed1 ch\u1eb5n \u1edf tr\u00ean \u0111\u1ec1u b\u1eb1ng s\u1ed1 l\u1ebb \u1edf d\u01b0\u1edbi tr\u1eeb \u0111i 1, do \u0111\u00f3 t\u1ed5ng A c\u00e1c s\u1ed1 ch\u1eb5n \u1edf d\u00e3y tr\u00ean b\u1eb1ng t\u1ed5ng B c\u00e1c s\u1ed1 l\u1ebb \u1edf d\u00e3y d\u01b0\u1edbi tr\u1eeb \u0111i 45: A - B = 45 Trong d\u00e3y s\u1ed1 ch\u1eb5n \u1edf tr\u00ean, c\u00f3 4 s\u1ed1 g\u1ed3m 2 ch\u1eef s\u1ed1 gi\u1ed1ng nhau l\u00e0: 22, 44, 66, 88. V\u1eady t\u1ed5ng c\u00e1c s\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0: A - (22 + 44 + 66 + 88) = A - 220 = B - 45 - 220 = B - 265 Trong d\u00e3y s\u1ed1 l\u1ebb \u1edf d\u01b0\u1edbi, c\u00f3 5 s\u1ed1 g\u1ed3m 2 ch\u1eef s\u1ed1 gi\u1ed1ng nhau l\u00e0: 11, 33, 55, 77, 99. V\u1eady t\u1ed5ng c\u00e1c s\u1ed1 l\u1ebb c\u00f3 2 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0: B - (11 + 33 + 55 + 77 + 99) = B - 275 = B - 265 - 10 So s\u00e1nh (B - 265) v\u00e0 (B - 265 - 10) ta th\u1ea5y t\u1ed5ng c\u00e1c s\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u1edbn h\u01a1n t\u1ed5ng c\u00e1c s\u1ed1 l\u1ebb c\u00f3 2 ch\u1eef s\u1ed1 kh\u00e1c nhau 10 \u0111\u01a1n v\u1ecb.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: b C\u00e2u 2: \u0110\u00e1p s\u1ed1: 100 c\u00e1i b\u00e1nh. C\u00e2u 3: \u0110\u00e1p s\u1ed1: 30 gh\u1ebf 5 gh\u1ebf c\u00f3 20 ch\u00e2n X gh\u1ebf c\u00f3 120 ch\u00e2n X = 120 x 5 : 20 = 30 (gh\u1ebf) C\u00e2u 4: \u0110\u00e1p s\u1ed1: 13 ch\u1ed7 37 ng\u01b0\u1eddi, m\u1ed7i ng\u01b0\u1eddi tr\u1ea3 51.000\u0111. C\u00e2u 5: \u0110\u00e1p s\u1ed1: S\u1ed1 35, l\u1ea5y s\u1ed1 \u0111\u1ee9ng tr\u01b0\u1edbc tr\u1eeb \u0111i 5 \u0111\u01a1n v\u1ecb. C\u00e2u 6: B\u00e0i gi\u1ea3i:","M\u1ed7i b\u1ed9, s\u1ed1 gh\u1ebf c\u1ea7n b\u1ecf b\u1edbt ra ngo\u00e0i l\u00e0: 6 - 4 = 2 (gh\u1ebf) T\u1ed5ng s\u1ed1 gh\u1ebf c\u1ea7n b\u1ecf b\u1edbt ra ngo\u00e0i l\u00e0: 2 x 10 = 20 (gh\u1ebf) \u0110\u00e1p s\u1ed1: 20 gh\u1ebf. Ch\u00fa \u00fd: Ta c\u00f3 th\u1ec3 gi\u1ea3i gh\u00e9p nh\u01b0 sau: S\u1ed1 gh\u1ebf c\u1ea7n b\u1ecf b\u1edbt ra ngo\u00e0i l\u00e0: (6 - 4) x 10 = 20 (gh\u1ebf) C\u00e2u 7: 15 con tr\u00e2u, 11 con b\u00f2 v\u00e0 19 con ng\u1ef1a. T\u1eeb hai \u0111i\u1ec1u ki\u1ec7n \u0111\u1ea7u ta th\u1ea5y: S\u1ed1 ng\u1ef1a h\u01a1n s\u1ed1 tr\u00e2u l\u00e0: 30 - 26 = 4 (con) K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n cu\u1ed1i ta gi\u1ea3i b\u00e0i to\u00e1n t\u00ecm hai s\u1ed1 bi\u1ebft t\u1ed5ng l\u00e0 34 v\u00e0 hi\u1ec7u l\u00e0 4 \u0111\u1ec3 c\u00f3: S\u1ed1 ng\u1ef1a l\u00e0: (34 + 4) : 2 = 19 (con) S\u1ed1 tr\u00e2u l\u00e0: 19 - 4 = 15 (con) V\u1eady s\u1ed1 b\u00f2 l\u00e0: 26 - 15 = 11 (con) C\u00e2u 8: \u0110\u00e1p s\u1ed1: \u0110\u00fang.","C\u00e2u 9: \u0110\u00e1p s\u1ed1: 98 c\u00e2y. S\u1ed1 \\\"kho\u1ea3ng c\u00e1ch\\\" l\u00e0: 750 : 15 = 50 (kho\u1ea3ng c\u00e1ch) S\u1ed1 c\u00e2y ph\u1ea3i tr\u1ed3ng \u1edf m\u1ed9t b\u00ean \u0111\u01b0\u1eddng l\u00e0: 50 - 1 = 49 (c\u00e2y) S\u1ed1 c\u00e2y ph\u1ea3i tr\u1ed3ng \u1edf c\u1ea3 hai b\u00ean \u0111\u01b0\u1eddng l\u00e0: 49 x 2 = 98 (c\u00e2y) Ghi ch\u00fa: B\u00e0i to\u00e1n n\u00e0y thu\u1ed9c lo\u1ea1i \\\"Kh\u00f4ng c\u00f3 c\u00e2y \u1edf hai \u0111\u1ea7u \u0111\u01b0\u1eddng\\\". \u1ede \u0111\u00e2y ta th\u1ea5y: S\u1ed1 c\u00e2y = s\u1ed1 kho\u1ea3ng c\u00e1ch - 1"]