460 bai toan vui luyen tri thong minh

["B\u00e0i 11 C\u00e2u 1: \u0110\u00e1p s\u1ed1: 24 ph\u00fat 40 x 6\/10 = 24 (ph\u00fat) C\u00e2u 2: \u0110\u00e1p s\u1ed1: \u0110\u00fang. C\u00e2u 3: \u0110\u00e1p s\u1ed1: 48. S\u1ed1 c\u1ea7n t\u00ecm b\u1eb1ng s\u1ed1 \u0111\u00e3 cho \u0111\u1ee9ng li\u1ec1n tr\u01b0\u1edbc n\u00f3 nh\u00e2n 3 16 x 3 = 48 C\u00e2u 4: \u0110\u00e1p s\u1ed1: C \u0110\u00e2y l\u00e0 h\u00ecnh duy nh\u1ea5t c\u00f3 3 c\u1ea1nh. C\u00e2u 5: \u0110\u00e1p s\u1ed1: 15 trang. C\u00e1ch 1: \u0110\u1ec3 \u0111\u00e1nh m\u00e1y b\u1ea3n th\u1ea3o xong trong 1 ng\u00e0y th\u00ec m\u1ed7i ng\u00e0y ph\u1ea3i \u0111\u00e1nh: 10 x 12 = 120 (trang)","C\u00e1ch 2: S\u1ed1 trang b\u1ea3n th\u1ea3o \u0111\u00f3 l\u00e0: 10 x 12 = 120 (trang) Mu\u1ed1n \u0111\u00e1nh 8 ng\u00e0y xong th\u00ec s\u1ed1 trang m\u1ed9t ng\u00e0y ph\u1ea3i \u0111\u00e1nh l\u00e0: 120 : 8 = 15 (trang) C\u00e2u 6: \u0110\u00e1p s\u1ed1: Ch\u1eef E kh\u00e1c v\u1edbi c\u00e1c ch\u1eef c\u00f2n l\u1ea1i v\u00ec \u0111\u00e2y l\u00e0 ch\u1eef c\u00e1i duy nh\u1ea5t \u0111\u01b0\u1ee3c t\u1ea1o n\u00ean b\u1edfi 4 \u0111o\u1ea1n th\u1eb3ng. C\u00e2u 7: \u0110\u00e1p s\u1ed1: Th\u1ee9 5 V\u00ec m\u1ed9t tu\u1ea7n c\u00f3 7 ng\u00e0y n\u00ean kh\u00f4ng th\u1ec3 c\u00f3 hai ng\u00e0y ch\u1ee7 nh\u1eadt l\u00e0 ng\u00e0y ch\u1eb5n li\u1ec1n nhau. V\u1eady gi\u1eefa hai ng\u00e0y ch\u1ee7 nh\u1eadt l\u00e0 ng\u00e0y ch\u1eb5n ph\u1ea3i c\u00f3 m\u1ed9t ng\u00e0y ch\u1ee7 nh\u1eadt l\u00e0 ng\u00e0y l\u1ebb. Trong th\u00e1ng \u0111\u00e3 cho c\u00f3 t\u1edbi 3 ng\u00e0y ch\u1ee7 nh\u1eadt l\u00e0 ng\u00e0y ch\u1eb5n. V\u1eady th\u00e1ng \u1ea5y ph\u1ea3i c\u00f3 xen k\u1ebd 2 ng\u00e0y ch\u1ee7 nh\u1eadt l\u00e0 ng\u00e0y l\u1ebb n\u1eefa. Suy ra trong th\u00e1ng n\u00e0y c\u00f3 5 ng\u00e0y ch\u1ee7 nh\u1eadt v\u00e0 ng\u00e0y ch\u1ee7 nh\u1eadt \u0111\u1ea7u ti\u00ean l\u00e0 ng\u00e0y ch\u1eb5n. V\u00ec t\u1eeb ng\u00e0y ch\u1ee7 nh\u1eadt th\u1ee9 nh\u1ea5t \u0111\u1ebfn ng\u00e0y ch\u1ee7 nh\u1eadt th\u1ee9 5 c\u00f3: 7 x (5 - 1) = 28 ng\u00e0y. M\u1ed9t th\u00e1ng ch\u1ec9 c\u00f3 t\u1ed1i \u0111a l\u00e0 31 ng\u00e0y n\u00ean ng\u00e0y ch\u1ee7 nh\u1eadt th\u1ee9 nh\u1ea5t ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 m\u00f9ng 1, m\u00f9ng 2 ho\u1eb7c m\u00f9ng 3. Song \u0111\u00f3 l\u1ea1i ph\u1ea3i l\u00e0 ng\u00e0y ch\u1eb5n n\u00ean ng\u00e0y ch\u1ee7 nh\u1eadt th\u1ee9 nh\u1ea5t l\u00e0 m\u00f9ng 2. Suy ra ng\u00e0y m\u00f9ng 9 v\u00e0 ng\u00e0y 16 c\u0169ng l\u00e0 ng\u00e0y ch\u1ee7 nh\u1eadt.","Do \u0111\u00f3 ng\u00e0y th\u1ee9 20 c\u1ee7a th\u00e1ng \u1ea5y l\u00e0 th\u1ee9 5.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: 90oC C\u00e2u 2: \u0110\u00e1p s\u1ed1: B (l\u00e0 ng\u01b0\u1eddi h\u1ecfi nhanh nh\u1ea5t). C\u00e2u 3: C\u00e1ch 1: S\u1ed1 th\u1ee9 nh\u1ea5t chi\u1ebfm 3 ph\u1ea7n, s\u1ed1 th\u1ee9 hai chi\u1ebfm 2 ph\u1ea7n th\u00ec \u0111\u01b0\u1ee3c t\u1ed5ng. T\u1ed5ng chi\u1ebfm 5 ph\u1ea7n, Hi\u1ec7u chi\u1ebfm 1 ph\u1ea7n. S\u1ed1 th\u1ee9 nh\u1ea5t chi\u1ebfm 3 ph\u1ea7n, s\u1ed1 th\u1ee9 hai chi\u1ebfm 2 ph\u1ea7n. Hi\u1ec7u b\u1eb1ng 1\/3 s\u1ed1 th\u1ee9 nh\u1ea5t. V\u1eady hi\u1ec7u 2 s\u1ed1 \u0111\u00e3 cho l\u00e0: 150 : 3 = 50 T\u1ed5ng hai s\u1ed1 \u0111\u00e3 cho l\u00e0: 50 x 5 = 250 V\u1eady T\u1ed4NG: 250 \u21d2 HI\u1ec6U: 50 C\u00e1ch 2:","Hi\u1ec7u hai s\u1ed1 \u0111\u00e3 cho l\u00e0: 150 : 3 = 50 S\u1ed1 th\u1ee9 hai l\u00e0: 150 - 50 = 100 T\u1ed5ng hai s\u1ed1 \u0111\u00e3 cho l\u00e0: 150 + 100 = 250. V\u1eady T\u1ed4NG : 250 \u21d2 HI\u1ec6U : 50. C\u00e2u 4: \u0110\u00e1p s\u1ed1: 17h50'. L\u1ea7n th\u1ee9 nh\u1ea5t c\u00f4 \u1ea5y \u0111\u1ebfn mu\u1ed9n 30 ph\u00fat, l\u1ea7n th\u1ee9 hai l\u00e0 (30 + 50) ph\u00fat, l\u1ea7n th\u1ee9 ba l\u00e0 (30 + 50 + 70) ph\u00fat. L\u1ea7n th\u1ee9 t\u01b0 l\u00e0 (30 + 50 + 70 + 90) ph\u00fat. Ta th\u1ea5y: C\u00e1c l\u1ea7n sau bao gi\u1edd c\u0169ng mu\u1ed9n h\u01a1n l\u1ea7n tr\u01b0\u1edbc 20 ph\u00fat. V\u1eady: sau c\u00f9ng l\u00e0 (30 + 50 + 70 + 90 + 110) ph\u00fat. C\u00e2u 5: \u0110\u00e1p s\u1ed1: H\u00ecnh C. C\u00e2u 6: \u0110\u00e1p s\u1ed1: Route A ng\u1eafn h\u01a1n. C\u00e2u 7: \u0110\u00e1p s\u1ed1: Sai. C\u00e2u 8: \u0110\u00e1p s\u1ed1: Lan: 15 c\u00e1i Ph\u01b0\u1ee3ng: 9 c\u00e1i","C\u00e2u 9: \u0110\u00e1p s\u1ed1: anh: 50 tu\u1ed5i, em: 35 tu\u1ed5i. Coi tu\u1ed5i em tr\u01b0\u1edbc \u0111\u00e2y l\u00e0 1 ph\u1ea7n (1) Ta c\u00f3: Tu\u1ed5i em hi\u1ec7n nay l\u00e0 4 ph\u1ea7n (2) Tu\u1ed5i anh tr\u01b0\u1edbc \u0111\u00e2y c\u0169ng 4 ph\u1ea7n (3) T\u1eeb (1) v\u00e0 (2) suy ra kho\u1ea3ng th\u1eddi gian tr\u01b0\u1edbc \u0111\u00e2y \u0111\u1ebfn hi\u1ec7n nay l\u00e0: 4 - 1 = 3 (ph\u1ea7n) Do \u0111\u00f3: Tu\u1ed5i anh hi\u1ec7n nay l\u00e0: 4 + 3 = 7 (ph\u1ea7n) (4) Tu\u1ed5i em sau n\u00e0y c\u0169ng l\u00e0 7 ph\u1ea7n (5) T\u1eeb (1) v\u00e0 (3) ta th\u1ea5y s\u1ed1 tu\u1ed5i anh h\u01a1n em l\u00e0: 4 - 1 = 3 (ph\u1ea7n) V\u1eady t\u1eeb (5) ta th\u1ea5y tu\u1ed5i anh sau n\u00e0y l\u00e0: 7 + 3 = 10 (ph\u1ea7n) Do \u0111\u00f3 t\u1ed5ng s\u1ed1 tu\u1ed5i c\u1ee7a hai anh em l\u00e0: 10 + 7 = 17 (ph\u1ea7n) 1 ph\u1ea7n l\u00e0: 85 : 17 = 5 (tu\u1ed5i) Tu\u1ed5i anh sau n\u00e0y: 5 x 10 = 50 (tu\u1ed5i)","Tu\u1ed5i em sau n\u00e0y: 5 x 7 = 35 (tu\u1ed5i)","B\u00e0i 13 C\u00e2u 1: Nh\u1eadn x\u00e9t: 8 = 1 x 1 x 1 x 8 8=1x1x2x4 C\u00f3 t\u1ea5t c\u1ea3 16 s\u1ed1 c\u00f3 b\u1ed1n ch\u1eef s\u1ed1 m\u00e0 t\u00edch c\u00e1c ch\u1eef s\u1ed1 \u1ea5y b\u1eb1ng 8, \u0111\u00f3 l\u00e0 c\u00e1c s\u1ed1: 1181, 1124, 2114, 4112 1181, 1142, 2141, 4121 1811, 1214, 2411, 4211 8111, 1241 1412 1421 C\u00e2u 2: \u0110\u00e1p s\u1ed1: H\u00ecnh e v\u00ec hai h\u00ecnh tr\u00f2n b\u00ean trong d\u00ednh v\u00e0o nhau. C\u00e2u 3: \u0110\u00e1p s\u1ed1: M\u1eb9 T\u00fd c\u00f2n l\u1ea1i 2.000 \u0111\u1ed3ng. C\u00e2u 4: \u0110\u00e1p s\u1ed1: 2405","C\u00e2u 5: \u0110\u00e1p s\u1ed1: Mai: 170 b\u00f4ng hoa H\u1ed3ng: 450 b\u00f4ng hoa \u0110\u00e0o: 60 b\u00f4ng hoa. Vi\u1ebft t\u1eaft th\u1eddi gian (t\u00ednh theo ph\u00fat) m\u00e0 c\u00e1c b\u1ea1n Mai, H\u1ed3ng, \u0110\u00e0o d\u00f9ng \u0111\u1ec3 l\u00e0m hoa l\u00e0 mai, h\u1ed3ng, \u0111\u00e0o ta c\u00f3: mai + h\u1ed3ng + \u0111\u00e0o = 45 (1) 17 mai + 15 h\u1ed3ng + 12 \u0111\u00e0o = 680 (2) V\u00ec h\u1ed3ng = 3 mai n\u00ean thay v\u00e0o (1) v\u00e0 (2) ta c\u00f3: mai + 3 mai + \u0111\u00e0o = 45 17 mai + 45 mai + 12 \u0111\u00e0o = 680 Suy ra 4 mai + \u0111\u00e0o = 45 (3) 62 mai + 12 \u0111\u00e0o = 680 G\u1ea5p 12 l\u1ea7n c\u00e1c s\u1ed1 li\u1ec7u \u1edf (3) ta c\u00f3: 48 mai + 12 \u0111\u00e0o = 540(4) 62 mai + 12 \u0111\u00e0o = 680(5) So s\u00e1nh (5) v\u00e0 (4) b\u1eb1ng ph\u00e9p tr\u1eeb ta c\u00f3: 14 mai = 140 mai = 10 ph\u00fat V\u1eady: Th\u1eddi gian l\u00e0m hoa c\u1ee7a Mai l\u00e0 10 ph\u00fat","Th\u1eddi gian l\u00e0m hoa c\u1ee7a H\u1ed3ng l\u00e0: 10 x 3 = 30 (ph\u00fat) Th\u1eddi gian l\u00e0m hoa c\u1ee7a \u0110\u00e0o l\u00e0: 45 - (10 + 30) = 5 (ph\u00fat) Do \u0111\u00f3: Mai \u0111\u00e3 l\u00e0m: 17 x 10 = 170 (b\u00f4ng) H\u1ed3ng \u0111\u00e3 l\u00e0m: 15 x 30 = 450 (b\u00f4ng) \u0110\u00e0o \u0111\u00e3 l\u00e0m: 12 x 5 = 60 (b\u00f4ng) C\u00e2u 6: \u0110\u00e1p s\u1ed1: 119 S\u1ed1 \u0111\u00f3 c\u1ed9ng th\u00eam 1 th\u00ec chia h\u1ebft cho 2, 3, 5 t\u1ee9c l\u00e0 chia h\u1ebft cho 30. C\u00e1c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 30 v\u00e0 nh\u1ecf h\u01a1n 200 l\u00e0: 120, 150, 180. V\u1eady s\u1ed1 c\u1ea7n t\u00ecm ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 m\u1ed9t trong ba s\u1ed1: 119, 149 ho\u1eb7c 179. Trong ba s\u1ed1 n\u00e0y ch\u1ec9 c\u00f3 119 chia h\u1ebft cho 7 n\u00ean \u0111\u00f3 l\u00e0 s\u1ed1 c\u1ea7n t\u00ecm. C\u00e2u 7: \u0110\u00e1p s\u1ed1: 24 C\u1ed9ng l\u1ea7n l\u01b0\u1ee3t c\u00e1c s\u1ed1 3, 5, 7, 9 ta \u0111\u01b0\u1ee3c s\u1ed1 li\u1ec1n sau \u0111\u00f3. C\u00e2u 8: T\u1ef1 gi\u1ea3i.","C\u00e2u 9: Ta c\u00f3 m\u1ed9t s\u1ed1 c\u00e1ch tham kh\u1ea3o sau: 444 : 2 = 222 666 : 3 = 222 777 : 7 = 111 888 : 2 = 444 999 : 3 = 333 (C\u00f3 th\u1ec3 t\u00ecm th\u00eam nhi\u1ec1u v\u00ed d\u1ee5 kh\u00e1c) C\u00e2u 10: \u0110\u00e1p s\u1ed1: H\u00ecnh a.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: a C\u00e2u 2: B\u00e0i gi\u1ea3i: Coi s\u1ed1 m\u00e9t \u0111\u01b0\u1eddng s\u1eeda ch\u1eefa trong ng\u00e0y th\u1ee9 ba l\u00e0 1 ph\u1ea7n th\u00ec s\u1ed1 m\u00e9t \u0111\u01b0\u1eddng s\u1eeda ch\u1eefa \u0111\u01b0\u1ee3c trong hai ng\u00e0y \u0111\u1ea7u l\u00e0 2 ph\u1ea7n. V\u1eady 405m s\u1eeda \u0111\u01b0\u1ee3c trong c\u1ea3 ba ng\u00e0y l\u00e0: 2 + 1 = 3 (ph\u1ea7n) Do \u0111\u00f3 s\u1ed1 m\u00e9t \u0111\u01b0\u1eddng s\u1eeda ch\u1eefa \u0111\u01b0\u1ee3c trong ng\u00e0y th\u1ee9 ba l\u00e0: 405 : 3 = 135 (m) S\u1ed1 m\u00e9t \u0111\u01b0\u1eddng s\u1eeda \u0111\u01b0\u1ee3c trong hai ng\u00e0y \u0111\u1ea7u l\u00e0: 135 x 2 = 270 (m) S\u1ed1 m\u00e9t \u0111\u01b0\u1eddng s\u1eeda \u0111\u01b0\u1ee3c trong ng\u00e0y th\u1ee9 nh\u1ea5t l\u00e0: (270 - 30) : 2 = 120 (m) S\u1ed1 m\u00e9t \u0111\u01b0\u1eddng s\u1eeda trong ng\u00e0y th\u1ee9 hai l\u00e0: 120 + 30 = 150 (m) \u0110\u00e1p s\u1ed1: 120m, 150m v\u00e0 135m. C\u00e2u 3:","\u0110\u00e1p s\u1ed1: T\u1ed5ng s\u1ed1 ch\u00e2n: (3 x 2) + (4 x 4) = 22 ch\u00e2n C\u00e2u 4: \u0110\u00e1p s\u1ed1: H\u00ecnh c. C\u00e2u 5: \u0110\u00e1p s\u1ed1: S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 22, thay b\u1eb1ng s\u1ed1 40. D\u00e3y s\u1ed1 n\u00e0y theo quy lu\u1eadt l\u1ea7n l\u01b0\u1ee3t nh\u00e2n 2 r\u1ed3i c\u1ed9ng 4. 20 x 2 = 40, 44 - 4 = 40 C\u00e2u 6: \u0110\u00e1p s\u1ed1: 21 qu\u1ea3 L\u00fac \u0111\u1ea7u An c\u00f3 6 + 3 + 1 + (6 + 3 + 1) + 1 = 21 (qu\u1ea3) C\u00e2u 7: B\u00e0i gi\u1ea3i: Gi\u1ea3 s\u1eed \u0111\u1ed9i b\u00f3ng \u0111\u00f3 th\u1eafng c\u1ea3 15 tr\u1eadn th\u00ec \u0111\u01b0\u1ee3c th\u01b0\u1edfng: 12 x 15 = 180 (tri\u1ec7u) Nh\u01b0ng v\u00ec c\u00f3 nh\u1eefng tr\u1eadn kh\u00f4ng th\u1eafng \u0111\u01b0\u1ee3c n\u00ean b\u1ecb thi\u1ec7t m\u1ea5t: 180 - 36 = 144 (tri\u1ec7u) C\u1ee9 m\u1ed7i l\u1ea7n kh\u00f4ng \u0111\u01b0\u1ee3c th\u01b0\u1edfng th\u00ec \u0111\u1ed9i \u0111\u00f3 thi\u1ec7t m\u1ea5t: 12 + 12 = 24 (tri\u1ec7u) S\u1ed1 tr\u1eadn kh\u00f4ng \u0111\u01b0\u1ee3c th\u01b0\u1edfng l\u00e0:","144 : 24 = 6 (tr\u1eadn) S\u1ed1 tr\u1eadn \u0111\u00e3 th\u1eafng l\u00e0: 16 - 6 = 10 (tr\u1eadn) \u0110\u00e1p s\u1ed1: 10 tr\u1eadn th\u1eafng. C\u00e2u 8: \u0110\u00e1p s\u1ed1: \u0110\u00fang. C\u00e2u 9: \u0110\u00e1p s\u1ed1: a. Kh\u00f4ng th\u1ec3 n\u00f3i c hay b cao h\u01a1n.","B\u00e0i 15 C\u00e2u 1: \u0110\u00e1p s\u1ed1: b. Ch\u1ec9 c\u1ea7n bi\u1ebft 3 trong 4 c\u1ea1nh b\u1ea5t k\u1ef3. C\u00e2u 2: + C\u00e2n l\u1ea7n m\u1ed9t: B\u01b0\u1edbc 1: \u0110\u1eb7t qu\u1ea3 c\u00e2n 1kg l\u00ean 1 \u0111\u0129a. B\u01b0\u1edbc 2: L\u1ea7n l\u01b0\u1ee3t \u0111\u1ed5 g\u1ea1o v\u00e0o 2 \u0111\u0129a c\u00e2n cho \u0111\u1ebfn khi c\u00e2n th\u0103ng b\u1eb1ng. Nh\u01b0 v\u1eady: \u0110\u0129a c\u00e2n th\u1ee9 nh\u1ea5t c\u00f3 7kg g\u1ea1o. \u0110\u0129a c\u00e2n th\u1ee9 hai c\u00f3 6kg g\u1ea1o + 1kg qu\u1ea3 c\u00e2n. + C\u00e2n l\u1ea7n hai: B\u01b0\u1edbc 1: L\u1ea5y g\u1ea1o ra kh\u1ecfi \u0111\u0129a c\u00e2n Nh\u1edb \u0111\u1ec3 ri\u00eang 7kg g\u1ea1o, v\u1eabn \u0111\u1ec3 qu\u1ea3 c\u00e2n 1kg tr\u00ean 1 \u0111\u0129a c\u00e2n. B\u01b0\u1edbc 2: L\u1ea7n l\u01b0\u1ee3t \u0111\u1ed5 7kg v\u00e0o 2 \u0111\u0129a c\u1ea7n cho \u0111\u1ebfn khi c\u00e2n th\u0103ng b\u1eb1ng. Nh\u01b0 v\u1eady: \u0110\u0129a c\u00e2n th\u1ee9 nh\u1ea5t c\u00f3 4kg g\u1ea1o. \u0110\u0129a c\u00e2n th\u1ee9 hai c\u00f3 3kg g\u1ea1o + 1kg qu\u1ea3 c\u00e2n. Sau hai l\u1ea7n c\u00e2n ta \u0111\u00e3 l\u1ea5y ri\u00eang ra \u0111\u01b0\u1ee3c 4kg g\u1ea1o.","C\u00e2u 3: \u0110\u00e1p s\u1ed1: 333.062.500 \u0111\u00f4la. C\u00e2u 4: \u0110\u00e1p s\u1ed1: 22176 \u0111\u00f4la C\u00e2u 5: \u0110\u00e1p s\u1ed1: S\u1ed1 81 C\u1ed9t 2 = C\u1ed9t 1 - 4 C\u1ed9t 3 = C\u1ed9t 2 + 5 C\u1ed9t 4 = C\u1ed9t 3 - 6 C\u00e2u 6: S\u00e1ch: 12.000\u0111, v\u1edf: 4.000\u0111. G\u1ee3i \u00fd: 2 quy\u1ec3n s\u00e1ch v\u00e0 2 quy\u1ec3n v\u1edf gi\u00e1: 16.000 x 2 = 32.000 (\u0111) \u0110\u1ed5i 2 quy\u1ec3n s\u00e1ch l\u1ea5y 5 quy\u1ec3n v\u1edf th\u00ec b\u1edbt \u0111\u01b0\u1ee3c 4.000 (\u0111), l\u00fac \u0111\u00f3 ch\u1ec9 ph\u1ea3i tr\u1ea3: 32.000 - 4.000 = 28.000 (\u0111) \u0110\u00f3 l\u00e0 gi\u00e1 c\u1ee7a: 5 + 2 = 7 (quy\u1ec3n v\u1edf) T\u1ef1 gi\u1ea3i C\u00e2u 7: Anh: 18 tu\u1ed5i, em: 15 tu\u1ed5i. Coi tu\u1ed5i em tr\u01b0\u1edbc \u0111\u00e2y l\u00e0 1 ph\u1ea7n th\u00ec:","Tu\u1ed5i anh tr\u01b0\u1edbc \u0111\u00e2y l\u00e0 : 1 ph\u1ea7n + 3 (tu\u1ed5i) Tu\u1ed5i em hi\u1ec7n nay c\u0169ng l\u00e0: 1 ph\u1ea7n + 3 (tu\u1ed5i) Tu\u1ed5i anh hi\u1ec7n nay l\u00e0: 1 ph\u1ea7n + 3 + 3 = 1 ph\u1ea7n + 6 (tu\u1ed5i) V\u00ec (1 ph\u1ea7n + 6 tu\u1ed5i) n\u00e0y c\u0169ng ch\u00ednh l\u00e0 1,5 ph\u1ea7n, n\u00ean 0,5 ph\u1ea7n l\u00e0 6 tu\u1ed5i. Suy ra 1 ph\u1ea7n l\u00e0: 6 : 0,5 = 12 (tu\u1ed5i) V\u1eady tu\u1ed5i em hi\u1ec7n nay l\u00e0: 12 + 3 = 15 (tu\u1ed5i) Tu\u1ed5i anh hi\u1ec7n nay l\u00e0: 15 + 3 = 18 (tu\u1ed5i) C\u00e2u 8: \u0110\u00e1p s\u1ed1: 30 em. T\u00f3m t\u1eaft: 20 ng\u00e0y: 120 em (20 - 4 ) ng\u00e0y: (120 + ?) em S\u1ed1 ng\u00e0y \u0103n th\u1ef1c t\u1ebf l\u00e0: 20 - 4 = 16 (ng\u00e0y) S\u1ed1 em b\u00e9 \u0103n th\u1ef1c t\u1ebf l\u00e0: 120 x 20 : 16 = 150 (em) S\u1ed1 em b\u00e9 m\u1edbi \u0111\u1ebfn th\u00eam l\u00e0:","150 - 120 = 30 (em) C\u00e2u 9: \u0110\u00e1p s\u1ed1: 63.000\u0111 v\u00e0 112.000\u0111 L\u1ea7n th\u1ee9 hai b\u00e1n nhi\u1ec1u h\u01a1n l\u1ea7n th\u1ee9 nh\u1ea5t l\u00e0: 16 - 9 = 7 (l\u00edt) Gi\u00e1 m\u1ed7i l\u00edt n\u01b0\u1edbc m\u1eafm l\u00e0: 49.000 : 7 = 7.000 (\u0111) L\u1ea7n th\u1ee9 nh\u1ea5t thu \u0111\u01b0\u1ee3c: 7.000 x 9 = 63.000 (\u0111) L\u1ea7n th\u1ee9 hai thu \u0111\u01b0\u1ee3c: 63.000 + 49.000 = 112.000 (\u0111) C\u00e2u 10: \u0110\u00e1p s\u1ed1: 5.000\u0111 v\u00e0 2.000\u0111. Gi\u00e1 12 quy\u1ec3n s\u00e1ch nhi\u1ec1u h\u01a1n gi\u00e1 9 quy\u1ec3n v\u1edf 42.000\u0111. Gi\u1ea3m t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 li\u1ec7u \u0111i 3 l\u1ea7n ta th\u1ea5y: Gi\u00e1 4 quy\u1ec3n s\u00e1ch nhi\u1ec1u h\u01a1n gi\u00e1 3 quy\u1ec3n v\u1edf 14.000\u0111. M\u1eb7t kh\u00e1c ta c\u00f3: Gi\u00e1 4 quy\u1ec3n s\u00e1ch nhi\u1ec1u h\u01a1n gi\u00e1 8 quy\u1ec3n v\u1edf 4.000\u0111. V\u1eady gi\u00e1 ti\u1ec1n c\u1ee7a 8 - 3 = 5 (quy\u1ec3n v\u1edf) l\u00e0: 14.000 - 4.000 = 10.000(\u0111) Gi\u00e1 1 quy\u1ec3n v\u1edf l\u00e0: 10.000 : 5 = 2.000(\u0111)","Gi\u00e1 4 quy\u1ec3n s\u00e1ch l\u00e0: 8 x 2.000 + 4.000 = 20.000(\u0111) Gi\u00e1 1 quy\u1ec3n s\u00e1ch l\u00e0: 20.000 : 4 = 5.000(\u0111)","C\u00e2u 1: \u0110\u00e1p s\u1ed1: T\u00e8o c\u00f2n l\u1ea1i 3 vi\u00ean bi xanh, 3 vi\u00ean bi h\u1ed3ng v\u00e0 10 vi\u00ean bi v\u00e0ng. C\u00e2u 2: \u0110\u00e1p s\u1ed1: 54, 81 C\u00e1c s\u1ed1 c\u1ea7n t\u00ecm l\u1ea7n l\u01b0\u1ee3t b\u1eb1ng s\u1ed1 \u0111\u1ee9ng tr\u01b0\u1edbc nh\u00e2n v\u1edbi 2, s\u1ed1 \u0111\u1ee9ng sau c\u1ed9ng v\u1edbi s\u1ed1 \u0111\u1ee9ng tr\u01b0\u1edbc. C\u00e2u 3: \u0110\u00e1p s\u1ed1: 115, 151, 511 5=5x1x1 Do v\u1eady s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 m\u00e0 t\u00edch 3 s\u1ed1 \u0111\u00f3 b\u1eb1ng 5 l\u00e0: 115, 151, 511 C\u00e2u 4: \u0110\u00e1p s\u1ed1: H\u00ecnh 1 v\u00e0 h\u00ecnh 4 C\u00e2u 5: \u0110\u00e1p s\u1ed1: 9 s\u1ed1. C\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 m\u00e0 3 ch\u1eef s\u1ed1 \u0111\u00f3 gi\u1ed1ng nhau l\u00e0: 111, 222, 333, 444, 555, 666, 777, 888, 999 V\u1eady c\u00f3 t\u1ea5t c\u1ea3 9 s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 m\u00e0 3 ch\u1eef s\u00f3 \u0111\u00f3 gi\u1ed1ng nhau.","C\u00e2u 6: Th\u1eadt ra \u00f4ng T\u00e2m kh\u00f4ng m\u1ea5t ti\u1ec1n cho \u00f4ng Tu\u1ea5n, l\u00e0 ng\u01b0\u1eddi \u0111\u00e3 \u0111\u1ed5i n\u0103m t\u1edd 10.000\u0111 \u0111\u1ec3 l\u1ea5y t\u1edd 50.000\u0111 ti\u1ec1n m\u1eb7t. \u00d4ng T\u00e2m m\u1ea5t \u0111\u00f4i gi\u00e0y tr\u1ecb gi\u00e1 40.000\u0111 v\u00e0 t\u1edd 10.000\u0111 \u0111\u1ec3 tr\u1ea3 l\u1ea1i cho kh\u00e1ch h\u00e0ng t\u1ed5ng c\u1ed9ng l\u00e0 50.000\u0111. C\u00e2u 7: 19 - 7 = 12 12 : 3 = 4 V\u1eady : 19 - 7 = 3 x 4 C\u00e2u 8: \u0110\u00e1p s\u1ed1: M\u00e0u lam. M\u00e0u \u00e1o tr\u00f9ng v\u1edbi m\u00e0u c\u1ea7u v\u1ed3ng. Lam l\u00e0 m\u00e0u th\u1ee9 n\u0103m trong c\u1ea7u v\u1ed3ng. C\u00e2u 9: \u0110\u00e1p s\u1ed1: 41 C\u1ed9t A g\u1ed3m c\u00e1c s\u1ed1 l\u1ebb C\u1ed9t B g\u1ed3m c\u00e1c s\u1ed1 ch\u1eb5n C\u1ed9t C g\u1ed3m tr\u00ecnh t\u1ef1 c\u1ee7a c\u00e1c s\u1ed1 b\u00ecnh ph\u01b0\u01a1ng C\u1ed9t D g\u1ed3m tr\u00ecnh t\u1ef1 c\u1ee7a c\u00e1c s\u1ed1 nguy\u00ean t\u1ed1. C\u00e2u 10: \u0110\u00e1p s\u1ed1: \u0110\u00fang.","B\u00e0i 17 C\u00e2u 1: \u0110\u00e1p s\u1ed1: 21 v\u00e0 27 C\u00e1c s\u1ed1 c\u1ea7n t\u00ecm l\u1ea7n l\u01b0\u1ee3t b\u1eb1ng s\u1ed1 \u0111\u1ee9ng tr\u01b0\u1edbc c\u1ed9ng v\u1edbi 6, s\u1ed1 \u0111\u1ee9ng sau tr\u1eeb \u0111i 3. 18 + 6 = 24 24 \u2013 3 = 21 21 + 6 = 27 C\u00e2u 2: Gi\u00e1 khung h\u00ecnh: 10.000\u0111 G\u1ea1t t\u00e0n: 2.000\u0111 Gh\u1ebf: 60.000 \u0111 B\u00e0n: 240.000\u0111 C\u00e2u 3: B\u00e0i gi\u1ea3i: C\u00e1ch 1: Trong ba ng\u00e0y ng\u01b0\u1eddi ta \u0111\u00e3 \u0111\u1eafp \u0111\u01b0\u1ee3c: 715 + 815 + 528 = 2058 (m) Trung b\u00ecnh m\u1ed7i ng\u00e0y ng\u01b0\u1eddi ta \u0111\u1eafp \u0111\u01b0\u1ee3c:","2058 : 3 = 686 (m) C\u00e1ch 2: Trung b\u00ecnh m\u1ed7i ng\u00e0y ng\u01b0\u1eddi ta \u0111\u1eafp \u0111\u01b0\u1ee3c: (715 + 815 + 528) : 3 = 686 (m) \u0110\u00e1p s\u1ed1: 686m C\u00e2u 4: \u0110\u00e1p s\u1ed1: H\u00ecnh a. C\u00e2u 5: \u0110\u00e1p s\u1ed1: 3 S\u1ed1 sau b\u1eafng s\u1ed1 k\u1ec1 tr\u01b0\u1edbc c\u1ed9ng 2, r\u1ed3i chia cho 2. C\u00e2u 6: B\u00e0i gi\u1ea3i: S\u1ed1 ch\u1eb5n k\u1ebf ti\u1ebfp sau 10 l\u00e0 12 S\u1ed1 l\u1ebb li\u1ec1n tr\u01b0\u1edbc 20 l\u00e0 19 T\u1ed5ng 2 s\u1ed1 tr\u00ean l\u00e0: 12 + 19 = 31 \u0110\u00e1p s\u1ed1: 31 C\u00e2u 7: B\u00e0i gi\u1ea3i:","Chu vi h\u1ed3 sen h\u00ecnh tr\u00f2n l\u00e0: 42 x 15 = 630 (m) B\u00e1n k\u00ednh h\u1ed3 sen l\u00e0: 630 : 3,14 : 2 = 100 (m) \u0110\u00e1p s\u1ed1: 100 m. C\u00e2u 8: \u0110\u00e1p s\u1ed1: 49,5 S\u1ed1 th\u1ee9 nh\u1ea5t c\u1ed9ng 3, s\u1ed1 th\u1ee9 hai nh\u00e2n 3.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: 7 Xu\u1ea5t ph\u00e1t t\u1eeb 12 di chuy\u1ec3n theo chi\u1ec1u kim \u0111\u1ed3ng h\u1ed3. C\u00e1c \u00f4 \u0111\u1ed1i di\u1ec7n trong s\u01a1 \u0111\u1ed3 s\u1ebd l\u00e0 c\u1ed9ng 1, r\u1ed3i c\u1ed9ng 2\u2026 C\u00e2u 2: \u0110\u00e1p s\u1ed1: 11 Theo h\u00e0ng ngang s\u1ed1 \u0111\u1ea7u ti\u00ean chia 2, c\u1ed9ng v\u1edbi s\u1ed1 th\u1ee9 hai nh\u00e2n 2, b\u1eb1ng s\u1ed1 th\u1ee9 3 V\u00ed d\u1ee5: (4 : 2) + (9 x 2) = 20 C\u00e2u 3: B\u00e0i gi\u1ea3i: C\u00e1ch 1: Trong m\u1ed9t ng\u00e0y 8 ng\u01b0\u1eddi s\u1eeda \u0111\u01b0\u1ee3c: 64 : 2 = 32 (m) Trong 5 ng\u00e0y 8 ng\u01b0\u1eddi s\u1eeda \u0111\u01b0\u1ee3c: 32 x 5 = 160 (m) Trong 5 ng\u00e0y 1 ng\u01b0\u1eddi s\u1eeda \u0111\u01b0\u1ee3c; 160 : 8 = 20 (m)","Trong 5 ng\u00e0y 9 ng\u01b0\u1eddi s\u1eeda \u0111\u01b0\u1ee3c: 20 x 9 = 180 (m) C\u00e1ch 2: N\u1ebfu ta coi 1 ng\u01b0\u1eddi l\u00e0m trong 1 ng\u00e0y \u0111\u01b0\u1ee3c 1 c\u00f4ng th\u00ec s\u1ed1 c\u00f4ng \u0111\u1ec3 s\u1eeda 64m \u0111\u01b0\u1eddng l\u00e0: 8 x 2 = 16 (c\u00f4ng) 5 ng\u01b0\u1eddi l\u00e0m trong 9 ng\u00e0y \u0111\u01b0\u1ee3c: 5 x 9 = 45 (c\u00f4ng) V\u1edbi 45 c\u00f4ng ta s\u1eeda \u0111\u01b0\u1ee3c: (64 x 45) : 16 = 180 (m) \u0110\u00e1p s\u1ed1: 180 m C\u00e2u 4: \u0110\u00e1p s\u1ed1: A = 16, B = 27, C = 39 C\u00e2u 5: \u0110\u00e1p s\u1ed1: H\u00ecnh C C\u00e2u 6: \u0110\u00e1p s\u1ed1: 10 Theo h\u00e0ng ngang, s\u1ed1 sau c\u00f9ng b\u1eb1ng t\u1ed5ng 2 s\u1ed1 \u0111\u1ea7u tr\u1eeb cho s\u1ed1 th\u1ee9 ba. C\u00e2u 7: \u0110\u00e1p s\u1ed1: 204","82 + 72 + 62 + 52 + 42 + 32 + 22 + 12 = 204 C\u00e2u 8: \u0110\u00e1p s\u1ed1: 16 S\u1ed1 ngo\u00e0i c\u1ed9t tr\u00e1i chia s\u1ed1 ngo\u00e0i c\u1ed9t ph\u1ea3i, r\u1ed3i nh\u00e2n 2. C\u00e2u 9: \u0110\u00e1p s\u1ed1: 101,5km. H\u01b0\u1edbng d\u1eabn: - T\u00ecm hi\u1ec7u v\u1eadn t\u1ed1c c\u1ee7a hai xe. - T\u00ecm kho\u1ea3ng c\u00e1ch gi\u1eefa hai xe du l\u1ecbch b\u1eaft \u0111\u1ea7u \u0111i. - T\u00ecm th\u1eddi gian \u0111\u1ec3 xe du l\u1ecbch \u0111u\u1ed5i k\u1ecbp xe kh\u00e1ch, t\u1eeb \u0111\u00f3 t\u00ednh \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb A \u0111\u1ebfn B.","B\u00e0i 19 C\u00e2u 1: \u0110\u00e1p s\u1ed1: Chris l\u00e0m v\u1ee1 l\u1ecd hoa. C\u00e2u 2: \u0110\u00e1p s\u1ed1: Kho th\u1ee9 nh\u1ea5t 402 bao. Kho th\u1ee9 hai 378 bao. C\u00e2u 3: \u0110\u00e1p s\u1ed1: 1, 2, 3 1+2+3=6 1x2x3=6 V\u1eady 3 s\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 1, 2, 3 C\u00e2u 4: \u0110\u00e1p s\u1ed1: 69 S\u1ed1 tr\u01b0\u1edbc tr\u1eeb \u0111i s\u1ed1 \u0111\u1ee9ng k\u1ec1 ngay sau n\u00f3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u1ea7n l\u01b0\u1ee3t l\u00e0: 1, 2, 4, 8, 16 C\u00e2u 5: B\u00e0i gi\u1ea3i: Tr\u01b0\u1edbc h\u1ebft ta t\u00ednh s\u1ed1 tr\u1eadn \u0111\u1ea5u:","C\u00f3 4 \u0111\u1ed9i, m\u1ed7i \u0111\u1ed9i \u0111\u1ea5u 3 tr\u1eadn, v\u1eady c\u00f3: 4 x 3 = 12 (tr\u1eadn) Nh\u01b0ng n\u1ebfu t\u00ednh nh\u01b0 v\u1eady th\u00ec m\u1ed7i tr\u1eadn \u0111\u01b0\u1ee3c t\u00ednh 2 l\u1ea7n v\u1eady th\u1ef1c ra ch\u1ec9 c\u00f3: 12 : 2 = 6 (tr\u1eadn) N\u1ebfu c\u1ea3 6 tr\u1eadn n\u00e0y \u0111\u1ec1u c\u00f3 ph\u00e2n th\u1eafng, thua th\u00ec t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u1ea3 4 \u0111\u1ed9i s\u1ebd l\u00e0: 6 x 3 = 18 (\u0111i\u1ec3m) C\u1ee9 m\u1ed7i tr\u1eadn h\u00f2a th\u00ec t\u1ed5ng s\u1ed1 tr\u00ean s\u1ebd b\u1ecb b\u1edbt \u0111i: 3 - (1 + 1) = 1 (\u0111i\u1ec3m) S\u1ed1 \u0111i\u1ec3m b\u1ecb b\u1edbt \u0111i l\u00e0: 18 - 16 = 2 (\u0111i\u1ec3m) S\u1ed1 tr\u1eadn h\u00f2a l\u00e0: 2 : 1 = 2 (tr\u1eadn) S\u1ed1 tr\u1eadn th\u1eafng l\u00e0: 6 - 2 = 4 (tr\u1eadn) \u0110\u00e1p s\u1ed1: 4 tr\u1eadn th\u1eafng v\u00e0 2 tr\u1eadn h\u00f2a. C\u00e2u 6: \u0110\u00e1p s\u1ed1: \u0110\u00fang. C\u00e2u 7: B\u00e0i gi\u1ea3i:","\u0110\u1ed5i \u0111\u01a1n v\u1ecb: 3km = 3.000m S\u1ed1 c\u00e2y \u1edf m\u1ed9t b\u00ean \u0111\u01b0\u1eddng l\u00e0: 3.000 : 20 + 1 = 151 (c\u00e2y) Theo th\u1ee9 t\u1ef1 th\u00ec 2 c\u00e2y d\u01b0\u01a1ng r\u1ed3i \u0111\u1ebfn 1 c\u00e2y b\u1ea1ch \u0111\u00e0n r\u1ed3i \u0111\u1ebfn 1 c\u00e2y tr\u00e0m n\u00ean n\u1ebfu ta coi 4 c\u00e2y l\u1eadp th\u00e0nh 1 nh\u00f3m th\u00ec s\u1ed1 nh\u00f3m l\u00e0: 151 : 4 = 37 (nh\u00f3m) d\u01b0 3 c\u00e2y, 3 c\u00e2y \u0111\u00f3 ch\u00ednh l\u00e0 2 c\u00e2y d\u01b0\u01a1ng, 1 c\u00e2y b\u1ea1ch \u0111\u00e0n. V\u1eady s\u1ed1 c\u00e2y d\u01b0\u01a1ng \u1edf m\u1ed7i b\u00ean \u0111\u01b0\u1eddng l\u00e0: 37 x 2 + 2 = 76 (c\u00e2y) S\u1ed1 c\u00e2y d\u01b0\u01a1ng \u1edf hai b\u00ean \u0111\u01b0\u1eddng l\u00e0: 76 x 2 = 152 (c\u00e2y) S\u1ed1 c\u00e2y b\u1ea1ch \u0111\u00e0n \u1edf hai b\u00ean \u0111\u01b0\u1eddng l\u00e0: (37 + 1) x 2 = 76 (c\u00e2y) S\u1ed1 c\u00e2y tr\u00e0m \u1edf hai b\u00ean \u0111\u01b0\u1eddng l\u00e0: 37 x 2 = 74 (c\u00e2y) \u0110\u00e1p s\u1ed1: 152 c\u00e2y d\u01b0\u01a1ng, 76 c\u00e2y b\u1ea1ch \u0111\u00e0n v\u00e0 74 c\u00e2y tr\u00e0m. C\u00e2u 8: B\u00e0i gi\u1ea3i: 3 l\u1ea7n chi\u1ec1u d\u00e0i m\u1ed7i c\u00e2y v\u1ea3i sau khi c\u1eaft l\u00e0: 1000 - (10 + 20 + 25) = 945 (m) Chi\u1ec1u d\u00e0i c\u1ee7a m\u1ed7i c\u00e2y v\u1ea3i sau khi c\u1eaft l\u00e0: 945 : 3 = 315 (m)","Chi\u1ec1u d\u00e0i c\u00e2y v\u1ea3i \u0111\u1ecf l\u00fac ban \u0111\u1ea7u l\u00e0: 315 + 10 = 325 (m) Chi\u1ec1u d\u00e0i c\u1ee7a c\u00e2y v\u1ea3i xanh l\u00fac ban \u0111\u1ea7u l\u00e0: 315 + 20 = 335 (m) Chi\u1ec1u d\u00e0i c\u1ee7a c\u00e2y v\u1ea3i v\u00e0ng l\u00fac ban \u0111\u1ea7u l\u00e0: 315 + 25 = 340 (m) \u0110\u00e1p s\u1ed1: \u0110\u1ecf: 325m, Xanh: 335m, V\u00e0ng: 340m","C\u00e2u 1: \u0110\u00e1p s\u1ed1: 88 S\u1ed1 \u1edf gi\u1eefa b\u1eb1ng hi\u1ec7u c\u1ee7a 2 s\u1ed1 ngo\u00e0i nh\u00e2n 4. C\u00e2u 2: \u0110\u00e1p s\u1ed1: 9 km. Ta vi\u1ebft t\u1eaft: qu\u00e3ng \u0111\u01b0\u1eddng \u0111i b\u1ed9 trong 6 gi\u1edd l\u00e0 \\\"6 b\u1ed9\\\", qu\u00e3ng \u0111\u01b0\u1eddng \u0111i ng\u1ef1a trong 11 gi\u1edd l\u00e0 \\\"11 ng\u1ef1a\\\". Theo \u0111\u1ea7u b\u00e0i ta c\u00f3: L\u01b0\u1ee3t \u0111i th\u1ee9 nh\u1ea5t: 5 ng\u1ef1a + 6 b\u1ed9 = 80km L\u01b0\u1ee3t \u0111i th\u1ee9 hai: 11 ng\u1ef1a - 6 b\u1ed9 = 64km G\u1ed9p c\u1ea3 hai l\u1ea1i ta c\u00f3: 5 ng\u1ef1a + 6 b\u1ed9 + 11 ng\u1ef1a - 6 b\u1ed9 = 80 km + 64km Hay: 16 ng\u1ef1a = 144km Suy ra: 1 ng\u1ef1a = 144 : 18 = 9(km) V\u1eady m\u1ed7i gi\u1edd ng\u1ef1a \u0111i \u0111\u01b0\u1ee3c 9km.","C\u00e2u 3: M\u1ed9t \u0111\u00f4i t\u1ea5t ngo\u00e0i gi\u00e1 3.500\u0111 M\u1ed9t \u0111\u00f4i t\u1ea5t trong h\u1ed9p gi\u00e1: 19.500 : 6 = 3.250 (\u0111) Ti\u1ebft ki\u1ec7m \u0111\u01b0\u1ee3c: 3.500 - 3.250 = 250 (\u0111) 250 : 3.500 x 100% = 7,1% C\u00e2u 4: \u0110\u00e1p s\u1ed1: 2 116 x 3 = 348 x 2 = 696 C\u00e2u 5: \u0110\u00e1p s\u1ed1: T = 15, W = 38, Z = 40 T = 5 + 7 + 3 = 15 W = 3 + 7 + 5 + 7 + 3 + 5 + 3 + 5 = 38 Z = 7 + 7 + 5 + 3 + 5 + 7 + 3 + 3 = 40 C\u00e2u 6: \u0110\u00e1p s\u1ed1: B\u1ed1: 36 tu\u1ed5i, m\u1eb9: 33 tu\u1ed5i Con l\u1edbn: 12 tu\u1ed5i, con b\u00e9: 6 tu\u1ed5i. V\u00ec s\u1ed1 tu\u1ed5i c\u1ee7a b\u1ed1 l\u00e0 t\u00edch c\u1ee7a hai s\u1ed1 t\u1ef1 nhi\u00ean gi\u1ed1ng nhau n\u00ean s\u1ed1 \u0111\u00f3 ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 m\u1ed9t trong c\u00e1c s\u1ed1 25, 36, 49, 64, ...","X\u00e9t t\u1eebng tr\u01b0\u1eddng h\u1ee3p V\u00ec tu\u1ed5i b\u1ed1 m\u1edbi l\u00e0 49 m\u00e0 t\u1ed5ng s\u1ed1 tu\u1ed5i c\u1ee7a b\u1ed1n ng\u01b0\u1eddi \u0111\u00e3 l\u00e0 116 > 87 n\u00ean ta kh\u00f4ng x\u00e9t ti\u1ebfp c\u00e1c tr\u01b0\u1eddng h\u1ee3p tu\u1ed5i b\u1ed1 l\u00e0 64, 81... C\u00e2u 7: \u0110\u00e1p s\u1ed1: 18 8 x 8 = 64 ho\u00e1n v\u1ecb = 46 5 x 5 = 25 ho\u00e1n v\u1ecb = 52 4 x 4 = 16 ho\u00e1n v\u1ecb = 61 7 x 7 = 49 ho\u00e1n v\u1ecb = 94","6 x 6 = 36 ho\u00e1n v\u1ecb = 63 9 x 9 = 81 ho\u00e1n v\u1ecb = 18 C\u00e2u 8: a. 14m 5dm = ? dm 14m 5dm = 145dm b. 1km 50m = ? m 1km 50m = 1050m c. 2500m = ? km ? m 2500m = 2km 500m d. 968 mm = ? dm ? mm 968 mm = 9dm 68mm","B\u00e0i 21 C\u00e2u 1: \u0110\u00e1p s\u1ed1: H\u00ecnh D. C\u00e2u 2: \u0110\u00e1p s\u1ed1: 24 s\u1ed1. C\u00e1ch 1: V\u1edbi 4 ch\u1eef s\u1ed1 1, 2, 3, 4 ta vi\u1ebft \u0111\u01b0\u1ee3c 24 s\u1ed1 sau: C\u00e1ch 2: Suy lu\u1eadn nh\u01b0 sau: V\u1edbi 4 ch\u1eef s\u1ed1 1, 2, 3, 4 C\u00f3 4 c\u00e1ch \u0111\u1eb7t ch\u1eef h\u00e0ng ngh\u00ecn l\u00e0 1, ho\u1eb7c 2, ho\u1eb7c 3, ho\u1eb7c 4. V\u1edbi m\u1ed7i c\u00e1ch \u0111\u1eb7t ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn, c\u00f3 3 c\u00e1ch \u0111\u1eb7t ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m l\u00e0 m\u1ed9t trong 3 s\u1ed1 c\u00f2n l\u1ea1i. V\u1edbi m\u1ed7i c\u00e1ch \u0111\u1eb7t ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m th\u00ec c\u00f3 2 c\u00e1ch \u0111\u1eb7t ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c l\u00e0 m\u1ed9t trong hai ch\u1eef s\u1ed1 c\u00f2n l\u1ea1i. V\u1edbi m\u1ed7i c\u00e1ch \u0111\u1eb7t ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c th\u00ec c\u00f3 m\u1ed9t c\u00e1ch \u0111\u1eb7t ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 1 ch\u1eef s\u1ed1 c\u00f2n l\u1ea1i.","V\u1eady c\u00f3 t\u1ea5t c\u1ea3 : 4 x 3 x 2 x 1 = 24 (s\u1ed1) C\u00e2u 3: \u0110\u00e1p s\u1ed1: C\u00f3 t\u1ed5ng c\u1ed9ng: 10 + 1 + 10 = 21 tham gia. C\u00e2u 4: B\u00e0i gi\u1ea3i: S\u1ed1 v\u1edf m\u1ed7i b\u1ea1n nh\u1eadn \u0111\u01b0\u1ee3c l\u00e0: 1000 : 50 = 20 (quy\u1ec3n v\u1edf) S\u1ed1 v\u1edf kh\u1ed1i Ba nh\u1eadn \u0111\u01b0\u1ee3c l\u00e0: 20 x 12 = 240 (quy\u1ec3n v\u1edf) \u0110\u00e1p s\u1ed1: 240 quy\u1ec3n v\u1edf. C\u00e2u 5: B\u00e0i gi\u1ea3i: S\u1ed1 h\u1ecdc sinh c\u1ee7a hai l\u1edbp l\u00e0: 47 + 43 = 90 (h\u1ecdc sinh) S\u1ed1 v\u1edf c\u1ee7a m\u1ed7i h\u1ecdc sinh l\u00e0: 450 : 90 = 5 (quy\u1ec3n) S\u1ed1 v\u1edf c\u1ee7a l\u1edbp A l\u00e0: 5 x 47 = 235 (quy\u1ec3n) S\u1ed1 v\u1edf c\u1ee7a l\u1edbp B l\u00e0: 5 x 43 = 215 (quy\u1ec3n)","\u0110\u00e1p s\u1ed1: L\u1edbp A: 235 quy\u1ec3n v\u1edf, l\u1edbp B: 215 quy\u1ec3n v\u1edf C\u00e2u 6: L\u1ea5y th\u01b0\u1edbc \u0111o s\u1ebd ch\u00ednh x\u00e1c v\u00ec n\u1ebfu \u0111\u1ebfm c\u00e1c vi\u00ean g\u1ea1ch ch\u00fang ta s\u1ebd kh\u00f4ng t\u00ednh \u0111\u01b0\u1ee3c c\u1ea3 nh\u1eefng ch\u1ed7 gi\u1eefa 2 vi\u00ean g\u1ea1ch. C\u00e2u 7: \u0110\u00e1p s\u1ed1: 35 T\u1ed5ng c\u1ee7a 3 s\u1ed1 ch\u1eb5n l\u1edbn nh\u1ea5t: 16 + 22 + 24 = 62 T\u00edch c\u1ee7a 2 s\u1ed1 l\u1ebb nh\u1ecf nh\u1ea5t: 3 x 9 = 27 V\u1eady t\u1ed5ng tr\u1eeb t\u00edch s\u1ebd l\u00e0: 62 - 27 = 35 C\u00e2u 8: \u0110\u00e1p s\u1ed1: 9 546 + 241 = 787 664 + 227 = 891 527 + 127 = 654 C\u00e2u 9: \u0110\u00e1p s\u1ed1: 1 gi\u1edd. 40 km t\u1ee9c 40 km\/h = 1 gi\u1edd.","60 km t\u1ee9c 60 km\/h = 1 gi\u1edd.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: 12 S\u1ed1 tr\u01b0\u1edbc h\u01a1n s\u1ed1 sau 6 \u0111\u01a1n v\u1ecb. C\u00e2u 2: B\u00e0i gi\u1ea3i: V\u00ec c\u00f3 5 \u0111\u1ed9i n\u00ean m\u1ed7i \u0111\u1ed9i ph\u1ea3i \u0111\u1ea5u 4 tr\u1eadn a. Ta c\u00f3 7 = 3 + 3 + 1 = 3 + 1 + 1 + 1 + 1 v\u1eady ch\u1ec9 c\u00f3 hai kh\u1ea3 n\u0103ng: V\u1edbi 7 = 3 + 1 + 1 + 1 + 1 th\u00ec \u0111\u1ed9i \u0111\u00f3 th\u1eafng 1 tr\u1eadn v\u00e0 h\u00f2a 4 tr\u1eadn. V\u1eady \u0111\u1ed9i \u0111\u00f3 \u0111\u1ea5u t\u1edbi 5 tr\u1eadn (v\u00f4 l\u00fd): lo\u1ea1i V\u1edbi 7 = 3 + 3 + 1 th\u00ec \u0111\u1ed9i \u0111\u00f3 th\u1eafng 2 tr\u1eadn v\u00e0 h\u00f2a 1 tr\u1eadn. Kh\u1ea3 n\u0103ng n\u00e0y h\u1ee3p l\u00fd v\u00ec khi \u0111\u00f3 \u0111\u1ed9i b\u1ecb thua: 4 - (2 + 1) = 1 (tr\u1eadn), t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a \u0111\u1ed9i l\u00e0: 3 + 3 + 1 + 0 = 7 (\u0111i\u1ec3m) Tr\u1ea3 l\u1eddi: a. 2 th\u1eafng, 1 h\u00f2a, 1 thua. b. 2 th\u1eafng, 2 h\u00f2a. c. Kh\u00f4ng th\u1ec3 x\u1ea3y ra vi\u1ec7c \u0111\u1ed9i \u0111\u00f3 \u0111\u01b0\u1ee3c 11 \u0111i\u1ec3m. C\u00e2u 3: \u0110\u00e1p s\u1ed1: A = 15, B = 6, C = 3, D = 1","C\u00e2u 4: \u0110\u00e1p s\u1ed1: H\u00ecnh 4 C\u00e2u 5: \u0110\u00e1p s\u1ed1: 70 qu\u1ea3. T\u1ed5ng s\u1ed1 na b\u00e0 t\u1eb7ng cho 3 ch\u00e1u l\u00e0 (t\u1ee9c s\u1ed1 na m\u1ed9t ph\u1ea7n) 5 + 3 + 6 = 14 (qu\u1ea3) S\u1ed1 na b\u00e0 c\u00f3 l\u00e0: 14 x 5 = 70 (qu\u1ea3) Ch\u00fa \u00fd: Ta c\u00f3 th\u1ec3 gi\u1ea3i gh\u00e9p nh\u01b0 sau S\u1ed1 na b\u00e0 c\u00f3 l\u00e0: (5 + 3 + 6) x 5 = 70 (qu\u1ea3) C\u00e2u 6: \u0110\u00e1p s\u1ed1: An: 46, B\u00ecnh: 42, Chi: 34 C\u00e2u 7: C\u00e2u 8:","B\u00e0i gi\u1ea3i: T\u1ed5ng s\u1ed1 c\u00f3 t\u1ea5t c\u1ea3 s\u1ed1 ch\u00e2n l\u00e0: (7 x 4) + (3 x 2) + (5 x 2) + (6 x 4) = 28 + 6 + 10 + 24 = 68 (ch\u00e2n)","B\u00e0i 23 C\u00e2u 1: \u0110\u00e1p s\u1ed1: Nh\u00e0 \u0111\u00f3 \u0111u\u1ed5i l\u00e0 \u0111\u00fang v\u00ec anh ta ng\u1ee7 th\u00ec l\u00e0m sao tr\u00f4ng \u0111\u01b0\u1ee3c nh\u00e0. C\u00e2u 2: a. N\u1ebfu 2 \u0111\u1ee9ng \u1edf h\u00e0ng ng\u00e0n ta c\u00f3 6 s\u1ed1: 2047, 2074, 2704, 2740, 2407, 2470. N\u1ebfu 4 \u0111\u1ee9ng \u1edf h\u00e0ng ng\u00e0n ta c\u00f3 6 s\u1ed1: 4027, 4072, 4207, 4270, 4702, 4720. N\u1ebfu 7 \u0111\u1ee9ng \u1edf h\u00e0ng ng\u00e0n ta c\u00f3 6 s\u1ed1: 7024, 7042, 7204, 7240, 7402, 7420 V\u1eady c\u00f3 t\u1ea5t c\u1ea3 18 s\u1ed1. b. 2.047 = 2.000 + 40 + 7; 2.074 = 2000 + 70 + 4 2704 = 2000 + 700 + 4; 2740 = 2000 + 700 + 40 c. T\u1ed5ng c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 n\u00f3i tr\u00ean c\u00f3: (2 + 4 + 7) x 6 = 78 (ng\u00e0n) T\u01b0\u01a1ng t\u1ef1 t\u1ed5ng \u1ea5y c\u0169ng c\u00f3: (2 + 4 + 7) x 4 = 52 tr\u0103m, 52 ch\u1ee5c, 52 \u0111\u01a1n v\u1ecb. V\u1eady t\u1ed5ng ph\u1ea3i t\u00ecm l\u00e0: 78.000 + 52 x (100 + 10 + 1) = 83.772","C\u00e2u 3: \u0110\u00e1p s\u1ed1: 9 S\u1ed1 \u1edf v\u1ecb tr\u00ed th\u1ee9 3, theo chi\u1ec1u ngang, b\u1eb1ng t\u1ed5ng 2 s\u1ed1 \u0111\u1ea7u tr\u1eeb cho s\u1ed1 cu\u1ed1i. C\u00e2u 4: B\u00e0i gi\u1ea3i: 1 th\u00f9ng \u0111\u1ef1ng n\u1eeda s\u1ed1 d\u1ea7u th\u00ec n\u1eb7ng 24kg 2 th\u00f9ng, m\u1ed7i th\u00f9ng \u0111\u1ef1ng n\u1eeda s\u1ed1 d\u1ea7u, n\u1eb7ng l\u00e0: 24 x 2 = 48 (kg) 48kg ch\u00ednh l\u00e0 t\u1ed5ng tr\u1ecdng l\u01b0\u1ee3ng 2 v\u1ecf th\u00f9ng c\u1ed9ng v\u1edbi s\u1ed1 d\u1ea7u \u0111\u1ef1ng \u0111\u1ea7y 1 th\u00f9ng. Ri\u00eang 1 v\u1ecf th\u00f9ng v\u00e0 s\u1ed1 d\u1ea7u \u0111\u1ef1ng \u0111\u1ea7y 1 th\u00f9ng n\u1eb7ng 44 kg. V\u1eady 1 v\u1ecf th\u00f9ng n\u1eb7ng l\u00e0: 48 - 44 = 4 (kg) \u0110\u00e1p s\u1ed1: 4kg. C\u00e2u 5: \u0110\u00e1p s\u1ed1: Lan nhi\u1ec1u tu\u1ed5i nh\u1ea5t. C\u00e2u 6: B\u00e0i gi\u1ea3i: N\u1ebfu mua 3 c\u00e2y b\u00fat m\u00e1y v\u00e0 3 c\u00e2y b\u00fat bi th\u00ec 3 c\u00e2y b\u00fat m\u00e1y \u0111\u1eaft h\u01a1n 3 c\u00e2y b\u00fat bi:","10.000 x 3 = 30.000 (\u0111) So v\u1edbi l\u1ea7n mua tr\u01b0\u1edbc th\u00ec s\u1ed1 b\u00fat m\u00e1y gi\u1ea3m \u0111i: 5 - 3 = 2 (b\u00fat m\u00e1y) Gi\u00e1 ti\u1ec1n 2 c\u00e2y b\u00fat \u0111\u00f3 l\u00e0: 54.000 - 30.000 = 24.000 (\u0111) Gi\u00e1 m\u1ed9t c\u00e2y b\u00fat m\u00e1y l\u00e0: 24.000 : 2 = 12.000 (\u0111) Gi\u00e1 m\u1ed9t c\u00e2y b\u00fat bi l\u00e0: 12.000 - 10.000 = 2.000 (\u0111) \u0110\u00e1p s\u1ed1: B\u00fat m\u00e1y: 12.000\u0111, b\u00fat bi: 2.000\u0111. C\u00e2u 7: \u0110\u00e1p s\u1ed1: 48 D\u00e3y s\u1ed1 l\u1ea7n l\u01b0\u1ee3t c\u1ed9ng 2, 4, 8, 16.","C\u00e2u 1: \u0110\u00e1p s\u1ed1: b. 70 C\u00e2u 2: \u0110\u00e1p s\u1ed1: Ng\u01b0\u1eddi n\u00f4ng d\u00e2n c\u00f2n l\u1ea1i: 17 - 9 = 8 (con) C\u00e2u 3: \u0110\u00e1p s\u1ed1: \u0110o\u00e0n 1 c\u00f3 9 \u00f4t\u00f4, \u0111o\u00e0n 2 c\u00f3 15 \u00f4t\u00f4. C\u00e2u 4: \u0110\u00e1p s\u1ed1: Con kh\u1ec9 b\u1ecb r\u01a1i xu\u1ed1ng n\u01b0\u1edbc s\u1ebd c\u00f3 b\u1ed9 l\u00f4ng b\u1ecb \u01b0\u1edbt s\u0169ng. C\u00e2u 5: B\u00e0i gi\u1ea3i: T\u1ed5ng 5 s\u1ed1 ch\u1eb5n \u0111\u1ea7u ti\u00ean l\u00e0: 0 + 2 + 4 + 6 + 8 = 20 T\u1ed5ng 5 s\u1ed1 l\u1ebb \u0111\u1ea7u ti\u00ean l\u00e0: 1 + 3 + 5 + 7 + 9 = 25 T\u1ed5ng 5 s\u1ed1 l\u1ebb \u0111\u1ea7u ti\u00ean h\u01a1n t\u1ed5ng 5 s\u1ed1 ch\u1eb5n \u0111\u1ea7u ti\u00ean l\u00e0:","25 - 20 = 5 C\u00e2u 6: \u0110\u00e1p s\u1ed1: H\u00ecnh A. C\u00e2u 7: B\u00e0i gi\u1ea3i: M\u1ed7i ng\u00e0y 38 ng\u01b0\u1eddi s\u1eeda \u0111\u01b0\u1ee3c: 1330 : 5 = 266 (m) M\u1ed7i ng\u00e0y 1 ng\u01b0\u1eddi s\u1eeda \u0111\u01b0\u1ee3c: 266 : 38 = 7 (m) Mu\u1ed1n s\u1eeda 1.470m \u0111\u01b0\u1eddng trong 2 ng\u00e0y th\u00ec m\u1ed7i ng\u00e0y ph\u1ea3i s\u1eeda: 1.470 : 2 = 735 (m) S\u1ed1 ng\u01b0\u1eddi c\u1ea7n c\u00f3 \u0111\u1ec3 s\u1eeda 1.470m \u0111\u01b0\u1eddng trong 2 ng\u00e0y: 735 : 7 = 105 (ng\u01b0\u1eddi) \u0110\u00e1p s\u1ed1: 105 ng\u01b0\u1eddi.","B\u00e0i 25 C\u00e2u 1: B\u00e0i gi\u1ea3i: Hai l\u1ea7n t\u1ed5ng c\u1ed9ng s\u1ed1 ti\u1ec1n th\u01b0\u1edfng c\u1ee7a ba ch\u1ecb l\u00e0: 200.000 + 150.000 + 220.000 = 570.000 (\u0111\u1ed3ng) V\u1eady t\u1ed5ng s\u1ed1 ti\u1ec1n th\u01b0\u1edfng c\u1ee7a ba ch\u1ecb l\u00e0: 570.000 : 2 = 285.000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n ch\u1ecb An \u0111\u01b0\u1ee3c th\u01b0\u1edfng l\u00e0: 285.000 - 150.000 = 135.000 (\u0111) S\u1ed1 ti\u1ec1n ch\u1ecb Ba \u0111\u01b0\u1ee3c th\u01b0\u1edfng l\u00e0: 285.000 - 220.000 = 65.000 (\u0111) S\u1ed1 ti\u1ec1n ch\u1ecb C\u00fac \u0111\u01b0\u1ee3c th\u01b0\u1edfng l\u00e0: 285.000 - 200.000 = 85.000 (\u0111) \u0110\u00e1p s\u1ed1: Ch\u1ecb An: 135.000\u0111 Ch\u1ecb Ba: 65.000\u0111 Ch\u1ecb C\u00fac: 85.000\u0111 C\u00e2u 2: \u0110\u00e1p s\u1ed1: 25 \u00f4.","C\u00e2u 3: \u0110\u00e1p s\u1ed1: Tr\u00ean l\u00e0 8, d\u01b0\u1edbi l\u00e0 7. C\u00e1c s\u1ed1 c\u1ea7n t\u00ecm l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 \u1edf c\u1ed9t 2 + c\u1ed9t 3 \u2013 c\u1ed9t 1 C\u00e2u 4: T\u1ed5ng s\u1ed1 15 l\u1ea7n. C\u00e2u 5: \u0110\u00e1p s\u1ed1: X = 9, Y = 17, Z = 6. X=3+2+1+2+1=9 Y = 2 + 1 + 3 + 3 + 2 + 2 + 1 + 3 = 17 Z=2+1+3=6 C\u00e2u 6: \u0110\u00e1p s\u1ed1: 54 s\u1ed1. T\u1eeb 10 \u0111\u1ebfn 99 c\u00f3 c\u00e1c s\u1ed1: 18, 27, 36,..., 81, 90 t\u1ee9c l\u00e0 c\u00f3 9 s\u1ed1. T\u1eeb 100 \u0111\u1ebfn 199 c\u00f3 c\u00e1c s\u1ed1: 108, 117, 126,..., 180 t\u1ee9c l\u00e0 c\u00f3 9 s\u1ed1. T\u1eeb 200 \u0111\u1ebfn 299 c\u00f3 c\u00e1c s\u1ed1: 207, 216, 225,..., 270 t\u1ee9c l\u00e0 c\u00f3 8 s\u1ed1. T\u1eeb 300 \u0111\u1ebfn 399 c\u00f3 c\u00e1c s\u1ed1: 306, 315, 324,..., 360 t\u1ee9c l\u00e0 c\u00f3 7 s\u1ed1. ... T\u1eeb 800 \u0111\u1ebfn 899 c\u00f3 c\u00e1c s\u1ed1: 801, 810 t\u1ee9c l\u00e0 c\u00f3 2 s\u1ed1. T\u1eeb 900 \u0111\u1ebfn 999 c\u00f3 1 s\u1ed1: 900 V\u1eady t\u1ea5t c\u1ea3 c\u00f3:","9 + (9 + 8 + 7 + ... + 2 + 1) = 9 + 45 = 54 (s\u1ed1 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 b\u1eb1ng 9). C\u00e2u 7: \u0110\u00e1p s\u1ed1: e. 4 l\u1ea7n. C\u00e2u 8: \u0110\u00e1p s\u1ed1: 15 v\u00e0 13 C\u00e1c s\u1ed1 c\u1ea7n t\u00ecm b\u1eb1ng s\u1ed1 \u0111\u1ee9ng c\u00e1ch n\u00f3 1 s\u1ed1 l\u1ea7n l\u01b0\u1ee3t tr\u1eeb \u0111i 2 r\u1ed3i c\u1ed9ng th\u00eam 5 \u0111\u01a1n v\u1ecb."]