["\u0110\u1ed5i 3 gi\u1edd 41 ph\u00fat = 221 (ph\u00fat) \u0110i 1km \u0111\u01b0\u1eddng n\u1eb1m ngang m\u1ea5t: 60 : 5 = 12 (ph\u00fat) \u0110i xu\u1ed1ng d\u1ed1c 1km m\u1ea5t: 60 : 6 = 10 (ph\u00fat) \u0110i l\u00ean d\u1ed1c 1km m\u1ea5t: 60 : 4 = 15 (ph\u00fat) V\u1eady th\u1eddi gian \u0111\u1ec3 \u0111i 1km \u0111\u01b0\u1eddng d\u1ed1c (c\u1ea3 \u0111i v\u00e0 v\u1ec1) l\u00e0: 10 + 15 = 25 (ph\u00fat) C\u00f2n th\u1eddi gian \u0111\u1ec3 \u0111i 1km \u0111\u01b0\u1eddng b\u1eb1ng (c\u1ea3 \u0111i v\u00e0 v\u1ec1) l\u00e0: 12 x 2 = 24 (ph\u00fat) Gi\u1ea3 s\u1eed to\u00e0n b\u1ed9 qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb A \u0111\u1ebfn B \u0111\u1ec1u l\u00e0 \u0111\u01b0\u1eddng d\u1ed1c th\u00ec th\u1eddi gian \u0111\u1ec3 \u0111i 9km (c\u1ea3 \u0111i l\u1eabn v\u1ec1) l\u00e0: 25 x 9 = 225 (ph\u00fat) So v\u1edbi th\u1eddi gian \u0111i trong th\u1ef1c t\u1ebf (3 gi\u1edd 41 ph\u00fat) th\u00ec th\u1eddi gian \u0111\u00f3 nhi\u1ec1u h\u01a1n: 225 \u2013 221 = 4 (ph\u00fat) \u0110i 1km \u0111\u01b0\u1eddng d\u1ed1c l\u00e2u h\u01a1n 1km \u0111\u01b0\u1eddng b\u1eb1ng l\u00e0: 25 \u2013 24 = 1 (ph\u00fat) V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng n\u1eb1m ngang d\u00e0i l\u00e0: 4 : 1 = 4 (km) \u0110\u00e1p s\u1ed1: 4 km. B\u00e0i 7: Bi\u1ebft r\u1eb1ng tu\u1ed5i Tu\u1ea5n c\u00f3 bao nhi\u00eau ng\u00e0y th\u00ec tu\u1ed5i b\u1ed1 b\u1ea5y nhi\u00eau tu\u1ea7n, tu\u1ed5i Tu\u1ea5n c\u00f3 bao nhi\u00eau th\u00e1ng th\u00ec tu\u1ed5i \u00f4ng g\u1ed3m b\u1ea5y nhi\u00eau n\u0103m. \u00d4ng h\u01a1n b\u1ed1 30 tu\u1ed5i. T\u00ednh s\u1ed1 tu\u1ed5i m\u1ed7i ng\u01b0\u1eddi. H\u01b0\u1edbng d\u1eabn gi\u1ea3i Theo \u0111\u1ea7u b\u00e0i th\u00ec: tu\u1ed5i b\u1ed1 g\u1ea5p 7 l\u1ea7n tu\u1ed5i Tu\u1ea5n. Tu\u1ed5i \u00f4ng g\u1ea5p 12 l\u1ea7n tu\u1ed5i Tu\u1ea5n Coi tu\u1ed5i Tu\u1ea5n l\u00e0 1 ph\u1ea7n th\u00ec tu\u1ed5i b\u1ed1 l\u00e0 7 ph\u1ea7n, tu\u1ed5i \u00f4ng l\u00e0 12 ph\u1ea7n. Gi\u00e1 tr\u1ecb 1 ph\u1ea7n b\u1eb1ng nhau hay tu\u1ed5i Tu\u1ea5n l\u00e0: 30 : (12 \u2013 7) x 1 = 6 (tu\u1ed5i) Tu\u1ed5i b\u1ed1 l\u00e0: 6 x 7 = 42 (tu\u1ed5i) Tu\u1ed5i \u00f4ng l\u00e0: 6 x 12 = 72 (tu\u1ed5i) \u0110\u00e1p s\u1ed1: Tu\u1ea5n: 6 tu\u1ed5i; b\u1ed1: 42 tu\u1ed5i; \u00f4ng: 72 tu\u1ed5i. B\u00e0i 8: T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean bi\u1ebft r\u1eb1ng s\u1ed1 \u0111\u00f3 chia cho 4 hay chia cho 7 \u0111\u1ec1u d\u01b0 3, v\u00e0 hi\u1ec7u c\u1ee7a hai th\u01b0\u01a1ng l\u00e0 207. H\u01b0\u1edbng d\u1eabn gi\u1ea3i S\u1ed1 \u0111\u00f3 tr\u1eeb \u0111i 3 th\u00ec chia h\u1ebft cho 4 v\u00e0 7. V\u1eady hi\u1ec7u c\u1ee7a s\u1ed1 \u0111\u00f3 v\u00e0 3 chia h\u1ebft cho: 4 x 7 = 28. Gi\u1ea3 s\u1eed, hi\u1ec7u \u0111\u00f3 g\u1ed3m 28 ph\u1ea7n b\u1eb1ng nhau th\u00ec khi \u0111em hi\u1ec7u \u0111\u00f3 chia cho 7 ta \u0111\u01b0\u1ee3c th\u01b0\u01a1ng m\u1edbi l\u00e0 4 ph\u1ea7n v\u00e0 khi \u0111em hi\u1ec7u \u0111\u00f3 chia cho 4 ta \u0111\u01b0\u1ee3c th\u01b0\u01a1ng m\u1edbi l\u00e0 7 ph\u1ea7n.","V\u1eady hi\u1ec7u c\u1ee7a hai th\u01b0\u01a1ng l\u00fac n\u00e0y l\u00e0: 7 \u2013 4 = 3 (ph\u1ea7n) V\u00ec hi\u1ec7u c\u1ee7a hai th\u01b0\u01a1ng m\u1edbi c\u0169ng b\u1eb1ng hi\u1ec7u c\u1ee7a hai th\u01b0\u01a1ng trong \u0111\u1ea7u b\u00e0i n\u00ean 3 ph\u1ea7n n\u00f3i tr\u00ean b\u1eb1ng 207. Suy ra 1 ph\u1ea7n l\u00e0: 207 : 3 = 69. Hi\u1ec7u c\u1ee7a s\u1ed1 ph\u1ea3i t\u00ecm v\u00e0 3 l\u00e0: 69 x 28 = 1932. S\u1ed1 ph\u1ea3i t\u00ecm l\u00e0: 1932 + 3 = 1935. \u0110\u00e1p s\u1ed1: 1935 B\u00e0i 9: Tu\u1ed5i con hi\u1ec7n nay b\u1eb1ng 0,4 tu\u1ed5i m\u1eb9. C\u00e1ch \u0111\u00e2y 8 n\u0103m th\u00ec tu\u1ed5i con b\u1eb1ng 0,25 tu\u1ed5i m\u1eb9. T\u00ednh tu\u1ed5i con v\u00e0 tu\u1ed5i m\u1eb9 hi\u1ec7n nay. H\u01b0\u1edbng d\u1eabn gi\u1ea3i Ta th\u1ea5y tu\u1ed5i con hi\u1ec7n nay b\u1eb1ng: 0,4 : (1 \u2013 0,4) = 2 (hi\u1ec7u tu\u1ed5i m\u1eb9 v\u00e0 con) 3 Tu\u1ed5i con tr\u01b0\u1edbc \u0111\u00e2y 8 n\u0103m b\u1eb1ng: 0,25 : (1 \u2013 0,25) = 1 (hi\u1ec7u tu\u1ed5i m\u1eb9 v\u00e0 con) 3 V\u1eady 8 n\u0103m ch\u00ednh l\u00e0: 2 - 1 = 1 (hi\u1ec7u tu\u1ed5i m\u1eb9 v\u00e0 con) 3 3 3 Hi\u1ec7u tu\u1ed5i m\u1eb9 v\u00e0 con l\u00e0: 8: 1 = 24 (tu\u1ed5i) 3 Tu\u1ed5i con hi\u1ec7n nay l\u00e0: 24 x 2 = 16 (tu\u1ed5i) 3 Tu\u1ed5i m\u1eb9 hi\u1ec7n nay l\u00e0: 16 : 0,4 = 40 (tu\u1ed5i) \u0110\u00e1p s\u1ed1: m\u1eb9: 16 tu\u1ed5i; con: 40 tu\u1ed5i. B\u00e0i 10: S\u1ed1 s\u00e1ch ng\u0103n tr\u00ean b\u1eb1ng 1 s\u1ed1 s\u00e1ch \u1edf ng\u0103n d\u01b0\u1edbi. N\u1ebfu \u0111em 3 quy\u1ec3n s\u00e1ch \u1edf ng\u0103n 3 d\u01b0\u1edbi \u0111\u1ec3 l\u00ean ng\u0103n tr\u00ean th\u00ec s\u1ed1 s\u00e1ch \u1edf ng\u0103n tr\u00ean b\u1eb1ng 2 s\u1ed1 s\u00e1ch \u1edf ng\u0103n d\u01b0\u1edbi. T\u00ednh s\u1ed1 s\u00e1ch \u1edf 5 m\u1ed7i ng\u0103n l\u00fac \u0111\u1ea7u. H\u01b0\u1edbng d\u1eabn gi\u1ea3i Khi \u0111em 3 cu\u1ed1n \u1edf ng\u0103n d\u01b0\u1edbi \u0111\u1ec3 l\u00ean ng\u0103n tr\u00ean th\u00ec t\u1ed5ng s\u1ed1 s\u00e1ch \u1edf hai ng\u0103n v\u1eabn kh\u00f4ng thay \u0111\u1ed5i. L\u00fac \u0111\u1ea7u s\u1ed1 s\u00e1ch \u1edf ng\u0103n tr\u00ean b\u1eb1ng: 1 : (1 + 3) = 1 (t\u1ed5ng s\u1ed1 s\u00e1ch) 4 V\u1ec1 sau s\u1ed1 s\u00e1ch \u1edf ng\u0103n tr\u00ean b\u1eb1ng: 2 : (2 + 5) = 2 (t\u1ed5ng s\u1ed1 s\u00e1ch) 7 V\u1eady 3 quy\u1ec3n s\u00e1ch ch\u00ednh l\u00e0: 2 - 1 = 1 (t\u1ed5ng s\u1ed1 s\u00e1ch) 7 4 28 T\u1ed5ng s\u1ed1 s\u00e1ch l\u00e0: 3 : 1 = 84 (quy\u1ec3n) 28 S\u1ed1 s\u00e1ch \u1edf ng\u0103n tr\u00ean l\u00fac \u0111\u1ea7u l\u00e0: 84 : 4 = 21 (quy\u1ec3n)","S\u1ed1 s\u00e1ch \u1edf ng\u0103n d\u01b0\u1edbi l\u00fac \u0111\u1ea7u l\u00e0: 21 x 3 = 63 (quy\u1ec3n) \u0110\u00e1p s\u1ed1: ng\u0103n tr\u00ean: 21 quy\u1ec3n; ng\u0103n d\u01b0\u1edbi: 63 quy\u1ec3n. B\u00e0i 11: Con trai h\u1ecfi m\u1eb9: \u201cM\u1eb9 \u01a1i, m\u1eb9 bao nhi\u00eau tu\u1ed5i?\u201d. M\u1eb9 tr\u1ea3 l\u1eddi: \u201cTu\u1ed5i con b\u00e2y gi\u1edd b\u1eb1ng 1 tu\u1ed5i m\u1eb9 tr\u01b0\u1edbc \u0111\u00e2y 8 n\u0103m. Sau 8 n\u0103m n\u1eefa th\u00ec tu\u1ed5i con l\u00fac \u0111\u00f3 s\u1ebd b\u1eb1ng 2 tu\u1ed5i m\u1eb9 b\u00e2y gi\u1edd\u201d. 4 5 Em h\u00e3y cho bi\u1ebft tu\u1ed5i m\u1eb9 b\u00e2y gi\u1edd l\u00e0 bao nhi\u00eau? H\u01b0\u1edbng d\u1eabn gi\u1ea3i Ta c\u00f3: tu\u1ed5i m\u1eb9 tr\u01b0\u1edbc \u0111\u00e2y \u2013 tu\u1ed5i con b\u00e2y gi\u1edd = tu\u1ed5i m\u1eb9 b\u00e2y gi\u1edd - tu\u1ed5i con sau n\u00e0y. G\u1ecdi t\u1eaft hai hi\u1ec7u tr\u00ean l\u00e0 hi\u1ec7u ta c\u00f3: Tu\u1ed5i m\u1eb9 tr\u01b0\u1edbc \u0111\u00e2y l\u00e0: 4 = 4 (hi\u1ec7u) 4\u22121 3 Tu\u1ed5i m\u1eb9 sau n\u00e0y l\u00e0: 4 = 5 (hi\u1ec7u) 5\u22122 3 Suy ra 8 n\u0103m n\u1eefa ch\u00ednh l\u00e0: 5 - 4 = 1 (hi\u1ec7u) 3 3 3 V\u1eady hi\u1ec7u l\u00e0: 8 : 1 = 24 (n\u0103m) 3 Tu\u1ed5i m\u1eb9 b\u00e2y gi\u1edd l\u00e0: 24 : (5 \u2013 2) x 5 = 40 (tu\u1ed5i) \u0110\u00e1p s\u1ed1: 40 tu\u1ed5i. B\u00e0i 12: Tu\u1ed5i anh hi\u1ec7n nay g\u1ea5p 3 l\u1ea7n tu\u1ed5i em tr\u01b0\u1edbc kia, l\u00fac anh b\u1eb1ng tu\u1ed5i em hi\u1ec7n nay. Khi tu\u1ed5i em b\u1eb1ng tu\u1ed5i anh hi\u1ec7n nay th\u00ec t\u1ed5ng s\u1ed1 tu\u1ed5i c\u1ee7a hai ng\u01b0\u1eddi s\u1ebd l\u00e0 28. T\u00ednh tu\u1ed5i c\u1ee7a m\u1ed7i ng\u01b0\u1eddi hi\u1ec7n nay. H\u01b0\u1edbng d\u1eabn gi\u1ea3i N\u1ebfu tu\u1ed5i em tr\u01b0\u1edbc \u0111\u00e2y l\u00e0 1 ph\u1ea7n th\u00ec tu\u1ed5i anh hi\u1ec7n nay (c\u0169ng l\u00e0 tu\u1ed5i em sau n\u00e0y) l\u00e0 3 ph\u1ea7n nh\u01b0 th\u1ebf. Do anh lu\u00f4n h\u01a1n em m\u1ed9t s\u1ed1 tu\u1ed5i kh\u00f4ng \u0111\u1ed5i n\u00ean tu\u1ed5i em hi\u1ec7n nay (c\u0169ng l\u00e0 tu\u1ed5i anh tr\u01b0\u1edbc kia) l\u00e0 2 ph\u1ea7n v\u00e0 tu\u1ed5i anh sau n\u00e0y l\u00e0 4 ph\u1ea7n. Gi\u00e1 tr\u1ecb m\u1ed9t ph\u1ea7n b\u1eb1ng nhau l\u00e0: 28 : (3 + 4) = 4 (tu\u1ed5i) Tu\u1ed5i anh hi\u1ec7n nay l\u00e0: 4 x 3 = 12 (tu\u1ed5i) Tu\u1ed5i em hi\u1ec7n nay l\u00e0: 4 x 2 = 8 (tu\u1ed5i) \u0110\u00e1p s\u1ed1: anh: 12 tu\u1ed5i; em: 8 tu\u1ed5i. B\u00e0i 13: M\u1ed9t chi\u1ebfc thuy\u1ec1n \u0111i xu\u00f4i d\u00f2ng t\u1eeb A \u0111\u1ebfn B h\u1ebft 3 gi\u1edd, \u0111i ng\u01b0\u1ee3c d\u00f2ng t\u1eeb B l\u00ean A h\u1ebft 4 gi\u1edd. H\u1ecfi m\u1ed9t kh\u00f3m b\u00e8o tr\u00f4i t\u1eeb A \u0111\u1ebfn B h\u1ebft m\u1ea5y gi\u1edd? H\u01b0\u1edbng d\u1eabn gi\u1ea3i","T\u1ec9 s\u1ed1 th\u1eddi gian xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng l\u00e0: 3. C\u00f9ng m\u1ed9t qu\u00e3ng \u0111\u01b0\u1eddng n\u00ean t\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c 4 xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng l\u00e0: 4. Suy ra n\u1ebfu v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng g\u1ed3m 4 ph\u1ea7n th\u00ec v\u1eadn t\u1ed1c 3 ng\u01b0\u1ee3c d\u00f2ng l\u00e0 3 ph\u1ea7n. V\u1eady v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc (c\u0169ng ch\u00ednh l\u00e0 v\u1eadn t\u1ed1c tr\u00f4i xu\u00f4i c\u1ee7a kh\u00f3m b\u00e8o) l\u00e0: 4\u22123 = 0,5 (ph\u1ea7n) 2 T\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc v\u00e0 v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng c\u1ee7a thuy\u1ec1n l\u00e0: 0,5 : 4 = 1. 8 V\u1eady th\u1eddi gian kh\u00f3m b\u00e8o tr\u00f4i t\u1eeb A \u0111\u1ebfn B g\u1ea5p 8 l\u1ea7n th\u1eddi gian xu\u00f4i d\u00f2ng v\u00e0 l\u00e0: 3 x 8 = 24 (gi\u1edd) \u0110\u00e1p s\u1ed1: 24 gi\u1edd."]
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