25-chuyên-đề-Toán-Tiểu-học

["C\u00e1c ch\u1eef s\u1ed1 thu\u1ed9c l\u1edbp \u0111\u01a1n v\u1ecb l\u00e0 c\u00e1c ch\u1eef s\u1ed1 1 (gi\u1ed1ng nh\u01b0 c\u00e1c ch\u1eef s\u1ed1 0, kh\u00f4ng c\u1ea7n quan t\u00e2m). C\u00e1c ch\u1eef s\u1ed1 thu\u1ed9c l\u1edbp ngh\u00ecn l\u00e0 c\u00e1c ch\u1eef s\u1ed1 ch\u1eb5n kh\u00e1c nhau, kh\u00e1c 0 v\u00e0 nh\u1ecf h\u01a1n 8. l\u00e0 c\u00e1c s\u1ed1 2;4;6. C\u00f3 3 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m ngh\u00ecn, 2 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c ngh\u00ecn v\u00e0 1 c\u00e1ch ch\u1ecdn h\u00e0ng ngh\u00ecn. V\u1eady c\u00f3: 3 x 2 x 1 = 6 (s\u1ed1) \u0110\u00f3 l\u00e0: 246111 ; 264111 ; 426111 ; 462111 ; 624111 v\u00e0 642111 B\u00e0i 38: H\u00e3y cho bi\u1ebft c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 kh\u00e1c nhau, m\u00e0 ch\u1eef s\u1ed1 6 \u0111\u1ee9ng li\u1ec1n tr\u01b0\u1edbc ch\u1eef s\u1ed1 8. Gi\u1ea3i Ch\u1eef s\u1ed1 6 \u0111\u1ee9ng li\u1ec1n tr\u01b0\u1edbc ch\u1eef s\u1ed1 8 l\u00e0 68. Ta xem s\u1ed1 68 nh\u01b0 s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1, v\u1eady ch\u00fang ta c\u00f3 9 ch\u1eef s\u1ed1: 0, 1, 2, 3, 4, 5, \u201c68\u201d, 7, 9 Ta l\u1eadp s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 trong \u0111\u00f3 c\u00f3 ch\u1eef s\u1ed1 \u201c68\u201d. -Ch\u1eef s\u1ed1 \u201c68\u201d h\u00e0ng tr\u0103m: c\u00f3 8 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c v\u00e0 7 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb. C\u00f3 8 x 7 = 56 (s\u1ed1) - Ch\u1eef s\u1ed1 \u201c68\u201d h\u00e0ng ch\u1ee5c: c\u00f3 7 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m (kh\u00e1c 0) v\u00e0 7 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb. C\u00f3 7 x 7 = 49 (s\u1ed1) - Ch\u1eef s\u1ed1 \u201c68\u201d h\u00e0ng \u0111\u01a1n v\u1ecb: c\u00f3 7 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m (kh\u00e1c 0) v\u00e0 7 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c. C\u00f3 7 x 7 = 49 (s\u1ed1) V\u1eady c\u00f3 t\u1ea5t c\u1ea3: 56 + 49 + 49 = 154 (s\u1ed1) \u0110\u00e1p s\u1ed1: 154 s\u1ed1 B\u00e0i 39: Cho s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1, n\u1ebfu l\u1ea5y t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ed9ng v\u1edbi t\u00edch c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00e3 cho th\u00ec b\u1eb1ng ch\u00ednh s\u1ed1 \u0111\u00f3. T\u00ecm ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a s\u1ed1 \u0111\u00e3 cho. Gi\u1ea3i G\u1ecdi s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 ab (a > 0, a, b < 10) Ta c\u00f3: ab= a + b + a x b a x 10 + b = a + b + a x b a x 10 = a + a x b a x 10 = a x (1 + b) 10 = 1 + b => b = 10 - 1 b=9 \u0110\u00e1p s\u1ed1: Ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 9 B\u00e0i 40: 1.H\u00e3y cho bi\u1ebft c\u00f3 bao nhi\u00eau s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf h\u01a1n 2012. 2.H\u00e3y cho bi\u1ebft c\u00f3 bao nhi\u00eau s\u1ed1 t\u1ef1 nhi\u00ean l\u1ebb nh\u1ecf h\u01a1n 2012. 3.H\u00e3y cho bi\u1ebft c\u00f3 bao nhi\u00eau s\u1ed1 s\u1ed1 ch\u1eb5n c\u00f3 4 ch\u1eef s\u1ed1 nh\u1ecf h\u01a1n 2012? Gi\u1ea3i C\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb b\u00e9 \u0111\u1ebfn l\u1edbn: 0, 1, 2, 3, \u2026\u2026.. 1)C\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf h\u01a1n 2012 l\u00e0: 0, 1, 2, 3, \u2026\u2026\u2026, 2010, 2011 C\u00f3: 2011+1= 2012 (s\u1ed1) 2)C\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean l\u1ebb nh\u1ecf h\u01a1n 2012 l\u00e0: 1, 3, 5, \u2026., 2009, 2011. C\u00f3: (2011 \u2013 1):2+1 = 1006 (s\u1ed1)","(Hay xen k\u1ebb m\u1ed9t s\u1ed1 ch\u1eb5n v\u00e0 m\u1ed9t s\u1ed1 l\u1ebb n\u00ean c\u00f3: 2012 : 2 = 1006 (s\u1ed1)) 3)C\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean ch\u1eb5n c\u00f3 4 ch\u1eef s\u1ed1 nh\u1ecf h\u01a1n 2012 l\u00e0: 1000, 1002, 1004, \u2026\u2026, 2008, 2010. C\u00f3: (2010 \u2013 1000):2+1 = 506 (s\u1ed1) B\u00e0i 41: Cho m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh b\u1eb1ng c\u00e1ch gh\u00e9p c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 1999: 123456789101112\u2026\u202619951996199719981999 Gi\u1ea3i Ta chia c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 1999 th\u00e0nh 2 nh\u00f3m: t\u1eeb 000 \u0111\u1ebfn 999 v\u00e0 1000 \u0111\u1ebfn 1999. Ta th\u1ea5y: t\u1eeb 000 \u0111\u1ebfn 999 c\u00f3: 999 + 1 = 1000 (s\u1ed1) v\u00e0 c\u00f3 3 x 1000 = 3000 (ch\u1eef s\u1ed1) \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9. S\u1ed1 l\u1ea7n xu\u1ea5t hi\u1ec7n c\u1ee7a m\u1ed7i ch\u1eef s\u1ed1 l\u00e0: 3000 : 10 = 300 (l\u1ea7n) T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 t\u1eeb 000 \u0111\u1ebfn 999 l\u00e0: (0+1+2+3+4+5+6+7+8+9) x 300 = 13500 T\u01b0\u01a1ng t\u1ef1 nh\u01b0 v\u1eady cho c\u00e1c s\u1ed1 t\u1eeb 1000 \u0111\u1ebfn 1999 l\u1ea1i c\u00f3 th\u00eam 1000 ch\u1eef s\u1ed1 1. T\u1ed5ng c\u1ee7a nh\u00f3m 2 l\u00e0: 1000 + 13500 = 14500 T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 t\u1eeb 1 \u0111\u1ebfn 1999 l\u00e0: 14500 + 13500 = 28000. B\u00e0i 42: Cho c\u00e1c s\u1ed1 abc v\u00e0 cab v\u1edbi a-b=1; b-c=2. S\u1ed1 abc h\u01a1n s\u1ed1 cab bao nhi\u00eau \u0111\u01a1n v\u1ecb? V\u1edbi a- b = 1 ; b \u2013 c = 2 => a-c=1+2=3 Gi\u1ea3i X\u00e9t s\u1ed1 abc so v\u1edbi cab *.H\u00e0ng tr\u0103m a l\u1edbn h\u01a1n c ax100 - cx100 = 300 (l\u1edbn h\u01a1n) *.H\u00e0ng ch\u1ee5c b b\u00e9 h\u01a1n a ax10 - bx10 = 10 (b\u00e9 h\u01a1n) *.H\u00e0ng \u0111\u01a1n v\u1ecb c b\u00e9 h\u01a1n b b - c = 2 (b\u00e9 h\u01a1n) S\u1ed1 abc l\u1edbn h\u01a1n cab: 300 - (10 + 2) = 288 \u0110\u00e1p s\u1ed1: 288 B\u00e0i 43 M\u1ed9t b\u1ea1n t\u00ecm t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 c\u00f3 n\u0103m ch\u1eef s\u1ed1 bi\u1ebft t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 l\u00e0 41 v\u00e0 s\u1ed1 \u0111\u00f3 kh\u00f4ng thay \u0111\u1ed5i n\u1ebfu vi\u1ebft c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 theo th\u1ee9 t\u1ef1 ng\u01b0\u1ee3c l\u1ea1i. H\u1ecfi b\u1ea1n \u0111\u00f3 t\u00ecm \u0111\u01b0\u1ee3c nhi\u1ec1u nh\u1ea5t bao nhi\u00eau s\u1ed1 tho\u1ea3 m\u00e3n y\u00eau c\u1ea7u? Gi\u1ea3i S\u1ed1 c\u00f3 n\u0103m ch\u1eef s\u1ed1 bi\u1ebft t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 l\u00e0 41 v\u00e0 s\u1ed1 \u0111\u00f3 kh\u00f4ng thay \u0111\u1ed5i n\u1ebfu vi\u1ebft c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 theo th\u1ee9 t\u1ef1 ng\u01b0\u1ee3c l\u1ea1i. Cho ta bi\u1ebft ch\u1eef s\u1ed1 \u1edf gi\u1eefa kh\u00f4ng thay \u0111\u1ed5i v\u00e0 l\u00e0 s\u1ed1 l\u1ebb > ho\u1eb7c = 5 (v\u00ec 4 ch\u1eef s\u1ed1 c\u00f2n l\u1ea1i c\u00f3 t\u1ed5ng l\u1edbn nh\u1ea5t 9x4=36), ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c ngh\u00ecn v\u00e0 h\u00e0ng \u0111\u01a1n v\u1ecb gi\u1ed1ng nhau, h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng ch\u1ee5c gi\u1ed1ng nhau. *.S\u1ed1 \u1edf gi\u1eef l\u00e0 5, ta c\u00f3 99599 *.S\u1ed1 \u1edf gi\u1eefa l\u00e0 7 th\u00ec t\u1ed5ng 4 s\u1ed1 c\u00f2n l\u1ea1i ph\u1ea3i l\u00e0 41-7=34. Hai ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb v\u00e0 h\u00e0ng ch\u1ee5c c\u00f3 t\u1ed5ng b\u1eb1ng 34:2=17. Ta c\u00f3 8 v\u00e0 9. C\u00e1c s\u1ed1 \u0111\u00f3 l\u00e0: 89798; 98789. *.S\u1ed1 \u1edf gi\u1eefa l\u00e0 9 th\u00ec t\u1ed5ng 4 s\u1ed1 c\u00f2n l\u1ea1i ph\u1ea3i l\u00e0 41-9=32. Hai ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb v\u00e0 h\u00e0ng","ch\u1ee5c c\u00f3 t\u1ed5ng b\u1eb1ng 32:2=16. Ta c\u00f3 8 v\u00e0 8 ho\u1eb7c 9 v\u00e0 7. C\u00e1c s\u1ed1 \u0111\u00f3 l\u00e0: 88988; 79997; 97979 C\u00e1c s\u1ed1 \u0111\u00f3 l\u00e0: 99599; 89798; 98789; 88988; 79997 v\u00e0 97979 B\u00e0i 44: T\u00ecm s\u1ed1 4a8b bi\u1ebft s\u1ed1 \u0111\u00f3 chia cho 2 c\u00f2n chia cho 5 v\u00e0 9 c\u00f9ng d\u01b0 l\u00e0 1 Gi\u1ea3i Chia h\u1ebft cho 2 l\u00e0 s\u1ed1 ch\u1eb5n, chia 5 d\u01b0 1 n\u00ean s\u1ed1 \u0111\u00f3 c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 b=6. Ta \u0111\u01b0\u1ee3c: 4a86 \u0110\u1ec3 s\u1ed1 n\u00e0y chia 9 d\u01b0 1 khi t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 chia cho 9 c\u0169ng d\u01b0 1. M\u00e0 4+8+6=18 chia h\u1ebft cho 9 v\u1eady a=1. S\u1ed1 \u0111\u00f3 l\u00e0 4186 B\u00e0i 45: C\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau m\u00e0 c\u00e1c s\u1ed1 \u0111\u00f3 \u0111\u1ec1u kh\u00f4ng chia h\u1ebft cho 5 ? Gi\u1ea3i C\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0: 9x9x8= 648 (s\u1ed1) C\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau chia h\u1ebft cho 5 l\u00e0: -C\u00f3 ch\u1eef s\u1ed1 5 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb: c\u00f3 8 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m (s\u1ed1 0 kh\u00f4ng \u1edf h\u00e0ng tr\u0103m), 8 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c. N\u00ean c\u00f3 8x8=64 (s\u1ed1) -C\u00f3 ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb: c\u00f3 9 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m, 8 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c. N\u00ean c\u00f3 9x8=72 (s\u1ed1) C\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau m\u00e0 kh\u00f4ng chia h\u1ebft cho 5 c\u00f3: 648 - (64 + 72) = 512 (s\u1ed1) B\u00e0i 46: H\u00e3y vi\u1ebft th\u00eam 3 ch\u1eef s\u1ed1 v\u00e0o b\u00ean ph\u1ea3i s\u1ed1 567 \u0111\u1ec3 \u0111\u01b0\u1ee3c s\u1ed1 l\u1ebb c\u00f3 6 ch\u1eef s\u1ed1 kh\u00e1c nhau, khi chia s\u1ed1 \u0111\u00f3 cho 5 v\u00e0 9 \u0111\u1ec1u d\u01b0 1. Gi\u1ea3i S\u1ed1 l\u1ebb m\u00e0 chia cho 5 d\u01b0 1 ph\u1ea3i c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 1. Ta \u0111\u01b0\u1ee3c 567**1 \u0110\u1ec3 chia cho 9 d\u01b0 1 th\u00ec t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u0169ng chia 9 d\u01b0 1. T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0: 5+6+7+*+*+1 = 19 + *+* 19 \u0111\u00e3 chia cho 9 d\u01b0 1 n\u00ean *+* ph\u1ea3i chia h\u1ebft cho 9 Hai s\u1ed1 c\u00f3 m\u1ed9t ch\u1eef s\u1ed1 kh\u00e1c nhau v\u00e0 kh\u00e1c 5;6;7;1c\u00f3 t\u1ed5ng chia h\u1ebft cho 9 ph\u1ea3i l\u00e0 0 v\u00e0 9. S\u1ed1 \u0111\u00f3 l\u00e0: 567091 ho\u1eb7c 567901 B\u00e0i 43: T\u1eeb c\u00e1c ch\u1eef s\u1ed1 : 0 , 2 , 3 , 5 , 6 , 7 , c\u00f3 th\u1ec3 l\u1eadp \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ed1 ch\u1eb5n g\u1ed3m 4 ch\u1eef s\u1ed1 kh\u00e1c nhau? Gi\u1ea3i H\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 ch\u1eef s\u1ed1 0: 5 c\u00e1ch l\u1ef1a ch\u1ecdn h\u00e0ng ngh\u00ecn, 4 c\u00e1ch l\u1ef1a ch\u1ecdn h\u00e0ng tr\u0103m, 3 c\u00e1ch l\u1ef1a ch\u1ecdn h\u00e0ng ch\u1ee5c. C\u00f3 5 x 4 x 3 = 60 (s\u1ed1) H\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 2 ho\u1eb7c 6: 4 x 4 x 3 = 48 (s\u1ed1) S\u1ed1 s\u1ed1 ch\u1eb5n c\u00f3 4 ch\u1eef s\u1ed1 kh\u00e1c nhau: 60 + 48 x 2 = 156 (s\u1ed1) B\u00e0i 44: T\u1eeb c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean : 2,3,7,9,a,b; b\u1ea1n B\u00ecnh \u0111\u00e3 gh\u00e9p ch\u00fang th\u00e0nh t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 c\u00f3 6 ch\u1eef s\u1ed1 kh\u00e1c nhau . B\u00ecnh cho bi\u1ebft t\u1ed5ng c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 c\u00f3 6 ch\u1eef s\u1ed1 kh\u00e1c nhau n\u00e0y l\u00e0 ch\u1eef s\u1ed1 6 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb B\u00ecnh nh\u1edd c\u00e1c b\u1ea1n t\u00ecm gi\u00fap hai s\u1ed1 t\u1ef1 nhi\u00ean a, b ?","Gi\u1ea3i C\u00f3 6 ch\u1eef s\u1ed1 kh\u00e1c nhau n\u00ean c\u00f3 s\u1ed1 c\u00e1ch l\u1ef1a ch\u1ecdn nh\u01b0 sau:. Tr\u01b0\u1eddng h\u1ee3p 1: N\u1ebfu a v\u00e0 b kh\u00e1c 0. H\u00e0ng tr\u0103m ngh\u00ecn: c\u00f3 6 l\u1ef1a ch\u1ecdn; Ch\u1ee5c ngh\u00ecn: 5; Ngh\u00ecn: 4; Tr\u0103m: 3; Ch\u1ee5c: 2; \u0110\u01a1n v\u1ecb: 1 V\u1eady c\u00f3: 6x5x4x3x2x1 = 720 (s\u1ed1) M\u1ed7i h\u00e0ng , m\u1ed7i ch\u1eef s\u1ed1 xu\u1ea5t hi\u1ec7n: 720 : 6 = 120 (l\u1ea7n) T\u1ed5ng c\u1ee7a h\u00e0ng \u0111\u01a1n v\u1ecb: (2+3+7+9+a+b) x 120 l\u00e0 s\u1ed1 tr\u00f2n ch\u1ee5c (120 c\u00f3 t\u1eadn c\u00f9ng l\u00e0 0_lo\u1ea1i). Tr\u01b0\u1eddng h\u1ee3p 2: N\u1ebfu a v\u00e0 b c\u00f3 1 l\u00e0 ch\u1eef s\u1ed1 0 (gi\u1ea3 s\u1eed a=0).. Tr\u0103m ngh\u00ecn: 5; Ch\u1ee5c ngh\u00ecn: 5; Ngh\u00ecn: 4; Tr\u0103m: 3; Ch\u1ee5c: 2; \u0110\u01a1n v\u1ecb:1 C\u00f3: 5x5x4x3x2x1 = 600 (s\u1ed1) Gi\u1ea3m \u0111i 720-600= 120 (s\u1ed1) do kh\u00f4ng c\u00f3 ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng cao nh\u1ea5t n\u00ean s\u1ed1 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb \u0111\u01b0\u1ee3c chia ra 120 ch\u1eef s\u1ed1 0 v\u00e0 (600 -120) : 5 = 96 m\u1ed7i ch\u1eef s\u1ed1 c\u00f2n l\u1ea1i. (kh\u00f4ng c\u00f3 ch\u1eef s\u1ed1 0 \u1edf 120 s\u1ed1 n\u00e0y n\u00ean 120 s\u1ed1 0 \u1edf c\u00e1c s\u1ed1 c\u00f3 h\u00e0ng cao nh\u1ea5t kh\u00e1c 0) T\u1ed5ng c\u1ee7a h\u00e0ng \u0111\u01a1n v\u1ecb: (2+3+7+9+b) x 96 = (21+b) x 96 c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 6 Suy ra: (21+b) c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 1 ho\u1eb7c l\u00e0 6. N\u1ebfu l\u00e0 1 th\u00ec b=0 (21+0=21_lo\u1ea1i v\u00ec a=0). V\u1eady 21+b c\u00f3 t\u1eadn c\u00f9ng l\u00e0 6 n\u00ean b=5 (21+5=26) \u0110\u00e1p s\u1ed1: a=0 v\u00e0 b=5 (ho\u1eb7c ng\u01b0\u1ee3c l\u1ea1i) B\u00e0i 45: N\u1ebfu abc l\u00e0 s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 th\u1ecfa m\u00e3n: 1: 0,abc = a + b + c th\u00ec abc l\u00e0 bao nhi\u00eau? Gi\u1ea3i Ta c\u00f3: 1: 0,abc = a + b + c hay (a+b+c) x abc = 1000 Hay 1000 : abc = a+b+c 1000 chia h\u1ebft cho s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 c\u00f3 c\u00e1c tr\u01b0\u1eddng h\u1ee3p 125 x 8 = 1000 => a=1; b=2; c=5 250 x 4 = 1000 (lo\u1ea1i) 500 x 2 = 1000 (lo\u1ea1i) V\u1eady: abc = 125 B\u00e0i 46: T\u00ecm s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1. Bi\u1ebft r\u1eb1ng s\u1ed1 \u0111\u00f3 g\u1ea5p 21 l\u1ea7n hi\u1ec7u c\u1ee7a ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c v\u00e0 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb. Gi\u1ea3i G\u1ecdi s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 ab. Ta c\u00f3: ab = 21 x (a-b) 10.a+b = 21.a - 21.b 11.a = 22.b Suy ra: a = b x 2 Ta c\u00f3 c\u00e1c s\u1ed1 sau: 21; 42; 63; 84 B\u00e0i 47: T\u1ed5ng c\u1ee7a hai s\u1ed1 l\u00e0 2009, gi\u1eefa hai s\u1ed1 tr\u00ean c\u00f3 5 s\u1ed5 l\u1ebb. T\u00ecm hai s\u1ed1. Gi\u1ea3i T\u1ed5ng 2 s\u1ed1 t\u1ef1 nhi\u00ean l\u00e0 l\u1ebb, t\u1ee9c ph\u1ea3i c\u00f3 1 ch\u1eb5n v\u00e0 1 l\u1ebb. Gi\u1eefa ch\u00fang c\u00f3 5 s\u1ed1 l\u1ebb, ph\u1ea3i c\u00f3 5 s\u1ed1 ch\u1eb5n. V\u1eady gi\u1eefa ch\u00fang c\u00f3 10 s\u1ed1. Hi\u1ec7u ch\u00fang l\u00e0: 10 + 1 = 11 S\u1ed1 b\u00e9: (2009 \u2013 11) : 2 = 999","S\u01a1 l\u1edbn: 2009 \u2013 999 = 1010 \u0110\u00e1p s\u00f4: 999 v\u00e0 1010 B\u00e0i 48: S\u1ed1 n\u00e0y n\u1eb1m trong ph\u1ea1m vi c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 58. Khi vi\u1ebft \\\"n\u00f3\\\" kh\u00f4ng s\u1eed d\u1ee5ng c\u00e1c ch\u1eef s\u1ed1 1 ; 2 ; 3. Ngo\u00e0i ra \\\"n\u00f3\\\" l\u00e0 s\u1ed1 l\u1ebb v\u00e0 kh\u00f4ng chia h\u1ebft cho c\u00e1c s\u1ed1 3 ; 5 ; 7. V\u1eady \\\"n\u00f3\\\" l\u00e0 s\u1ed1 n\u00e0o ? Gi\u1ea3i N\u00f3 l\u00e0 s\u1ed1 l\u1ebb n\u1eb1m trong ph\u1ea1m vi c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 58, khi vi\u1ebft n\u00f3 kh\u00f4ng s\u1eed d\u1ee5ng c\u00e1c ch\u1eef s\u1ed1 1 ; 2 ; 3 v\u00e0 kh\u00f4ng chia h\u1ebft cho 5 n\u00ean ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng c\u1ee7a n\u00f3 l\u00e0 7 ho\u1eb7c 9. C\u00e1c s\u1ed1 c\u00f3 th\u1ec3 l\u00e0: 7 ; 9 ; 47 ; 49 v\u00e0 57. Kh\u00f4ng chia h\u1ebft cho 3; 5; 7 n\u00ean trong c\u00e1c s\u1ed1 tr\u00ean ch\u1ec9 c\u00f3 s\u1ed1 47 l\u00e0 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. V\u1eady n\u00f3 l\u00e0 s\u1ed1 47. B\u00e0i 49: ch\u00f9m b\u00e0i l\u1eadp s\u1ed1 1\/.T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf nh\u1ea5t \u0111\u01b0\u1ee3c vi\u1ebft b\u1edfi c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau v\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 b\u1eb1ng 25. Gi\u1ea3i S\u1ed1 nh\u1ecf nh\u1ea5t khi c\u00f3 \u00edt ch\u1eef s\u1ed1 nh\u1ea5t, gi\u00e1 tr\u1ecb t\u1eebng ch\u1eef s\u1ed1 l\u1edbn nh\u1ea5t c\u00f3 th\u1ec3. H\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 9; h\u00e0ng ch\u1ee5c l\u00e0 8; h\u00e0ng tr\u0103m l\u00e0 7. V\u1eady h\u00e0ng ngh\u00ecn l\u00e0 1 \u0111\u1ec3 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 b\u1eb1ng 25. S\u1ed1 \u0111\u00f3 l\u00e0: 1 789 2\/.T\u00ecm s\u1ed1 l\u1edbn nh\u1ea5t \u0111\u01b0\u1ee3c vi\u1ebft b\u1edfi c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau v\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 b\u1eb1ng 23. Gi\u1ea3i S\u1ed1 l\u1edbn nh\u1ea5t khi c\u00f3 nhi\u1ec1u ch\u1eef s\u1ed1 nh\u1ea5t, gi\u00e1 tr\u1ecb t\u1eebng ch\u1eef s\u1ed1 nh\u1ecf nh\u1ea5t c\u00f3 th\u1ec3. Ta ch\u1ecdn c\u00e1c ch\u1eef s\u1ed1 nh\u1ecf nh\u1ea5t l\u00e0: 0; 1; 2; 3; 4; 5 v\u00e0 8 \u0111\u1ec3 c\u00f3 0+1+2+3+4+5+8=23. S\u1ed1 l\u1edbn nh\u1ea5t \u0111\u00f3 l\u00e0: 8 543 210 3\/.T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t kh\u00e1c 0 v\u00e0 chia h\u1ebft cho 2; 3; 4; 5 v\u00e0 6. Gi\u1ea3i S\u1ed1 chia h\u1ebft cho 6 th\u00ec chia h\u1ebft cho 2 v\u00e0 cho 3. S\u1ed1 b\u00e9 nh\u1ea5t v\u1eeba chia h\u1ebft cho 4, v\u1eeba chia h\u1ebft cho 6 l\u00e0: 2x2x3=12 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 12 x 5 = 60 4\/.T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t kh\u00e1c 1 v\u00e0 khi chia s\u1ed1 \u0111\u00f3 cho 2; 3; 4; 5 v\u00e0 6 th\u00ec c\u00f9ng c\u00f3 s\u1ed1 d\u01b0 b\u1eb1ng 1. Gi\u1ea3i Nh\u01b0 b\u00e0i 3, \u0111\u1ec3 \u0111\u1ec1u d\u01b0 1 ta th\u00eam v\u00e0o s\u1ed1 b\u1ecb chia 1 \u0111\u01a1n v\u1ecb. 60 + 1 = 61 5\/.T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t sao cho khi chia s\u1ed1 \u0111\u00f3 cho 2; 3; 4; 5 v\u00e0 6 th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 d\u01b0 l\u1ea7n l\u01b0\u1ee3t l\u00e0 1; 2; 3; 4 v\u00e0 5. Gi\u1ea3i Nh\u01b0 b\u00e0i 3, \u0111\u1ec3 \u0111\u1ec1u c\u00f3 s\u1ed1 d\u01b0 b\u00e9 h\u01a1n s\u1ed1 chia 1 \u0111\u01a1n v\u1ecb th\u00ec ta b\u1edbt \u1edf s\u1ed1 b\u1ecb chia 1 \u0111\u01a1n v\u1ecb. 60 \u2013 1 = 59 B\u00e0i 50: Hai s\u1ed1 c\u00f3 t\u1ed5ng b\u1eb1ng 839, bi\u1ebft n\u1ebfu x\u00f3a ch\u1eef s\u1ed1 3 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a s\u1ed1 l\u1edbn th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 b\u00e9. T\u00ecm s\u1ed1 b\u00e9.","S\u1ed1 l\u1edbn g\u1ea5p 10 l\u1ea7n s\u1ed1 b\u00e9 c\u1ed9ng th\u00eam 3 \u0111\u01a1n v\u1ecb. Nh\u01b0 v\u1eady b\u1edbt 3 \u0111\u01a1n v\u1ecb \u1edf s\u1ed1 l\u1edbn th\u00ec s\u1ed1 l\u1edbn s\u1ebd g\u1ea5p 10 l\u1ea7n s\u1ed1 b\u00e9. L\u00fac n\u00e0y t\u1ed5ng s\u1ebd c\u00f2n 839 \u2013 3 = 836 T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 10 + 1 = 11 (ph\u1ea7n) S\u1ed1 b\u00e9: 836 : 11 = 76 S\u1ed1 l\u01a1n: 839 \u2013 76 = 763 B\u00e0i 51: C\u00f3 bao nhi\u00eau s\u1ed1 chia h\u1ebft cho 5 \u0111\u01b0\u1ee3c l\u1eadp t\u1eeb c\u00e1c ch\u1eef s\u1ed1: 4, 7, 8, 6, 5 Gi\u1edbi h\u1ea1n th\u00eam cho \u0111\u1ec1 b\u00e0i m\u1ed9t ch\u00fat l\u00e0 c\u00f3 c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau. -S\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1 l\u00e0 1 (s\u1ed1 5) -S\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1: C\u00f3 4 s\u1ed1. -S\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 c\u00f3: 4 x 3 = 12 (s\u1ed1) -S\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 c\u00f3: 4 x 3 x 2 = 24 (s\u1ed1) -S\u1ed1 c\u00f3 5 ch\u1eef s\u1ed1 c\u00f3: 4 x 3 x 2 x 1 = 24 (s\u1ed1) B\u00e0i 52: Hai s\u1ed1 c\u00f3 t\u1ed5ng l\u00e0 344. n\u1ebfu g\u1ea5p 3 l\u1ea7n s\u1ed1 th\u1ee9 nh\u1ea5t v\u00e0 g\u1ea5p 4 l\u1ea7n s\u1ed1 th\u1ee9 hai th\u00ec \u0111\u01b0\u1ee3c t\u1ed5ng l\u00e0 1914. T\u00ecm s\u1ed1 th\u1ee9 nh\u1ea5t. N\u1ebfu g\u1ea5p c\u1ea3 2 s\u1ed1 l\u00ean 3 l\u1ea7n th\u00ec t\u1ed5ng m\u1edbi s\u1ebd l\u00e0: 344 x 3 = 1032 S\u1ed1 th\u1ee9 hai: 1914 \u2013 1032 = 882 S\u1ed1 th\u1ee9 nh\u1ea5t: 344 \u2013 882 = 1376 B\u00e0i 53: C\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau s\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1? Gi\u1ea3i T\u1eeb 10 \u0111\u1ebfn 99 c\u00f3 99-10+1=90 (s\u1ed1). Trong \u0111\u00f3 c\u00f3 90 : 2 = 45 (s\u1ed1) ch\u1eb5n v\u00e0 45 s\u1ed1 l\u1ebb. C\u00e1ch kh\u00e1c: S\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1 th\u00ec h\u00e0ng \u0111\u01a1n v\u1ecb c\u00f3 th\u1ec3 l\u00e0: 0;2;4;6;8. M\u1ed7i ch\u1eef s\u1ed1 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb ta c\u00f3 9 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c. S\u1ed1 s\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1 l\u00e0: 5 x 9 = 45 (s\u1ed1) \u0110\u00e1p s\u1ed1: 45 s\u1ed1 B\u00e0i 54: C\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau s\u1ed1 l\u1ebb c\u00f3 2 ch\u1eef s\u1ed1 kh\u00e1c nhau? Gi\u1ea3i S\u1ed1 l\u1ebb c\u00f3 2 ch\u1eef s\u1ed1 th\u00ec h\u00e0ng \u0111\u01a1n v\u1ecb c\u00f3 th\u1ec3 l\u00e0: 1;3;5;7;9 M\u1ed7i ch\u1eef s\u1ed1 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb ta c\u00f3 8 c\u00e1ch ch\u1ecdn ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c (kh\u00f4ng ch\u1ecdn ch\u1eef s\u1ed1 0 v\u00e0 ch\u1eef s\u1ed1 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb). S\u1ed1 s\u1ed1 l\u1ebb c\u00f3 2 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0: 5 x 8 = 40 (s\u1ed1) \u0110\u00e1p s\u1ed1: 40 s\u1ed1 B\u00e0i 55: Mu\u1ed1n vi\u1ebft c\u00e1c s\u1ed1 t\u1eeb 1000 \u0111\u1ebfn 2013 c\u1ea7n bao nhi\u00eau ch\u1eef s\u1ed1 8 Gi\u1ea3i T\u1eeb 1000 \u0111\u1ebfn 1999 c\u00f3 1000 s\u1ed1 kh\u00f4ng c\u00f3 s\u1ed1 8 \u1edf h\u00e0ng ngh\u00ecn. Ta x\u00e9t t\u1eeb 000 \u0111\u1ebfn 999 c\u00f3 1000 s\u1ed1 (999-000+1 = 1000) c\u00f3 s\u1ed1 ch\u1eef s\u1ed1 l\u00e0: 3 x 1000 = 3000 (ch\u1eef s\u1ed1) \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9. S\u1ed1 ch\u1eef s\u1ed1 8 l\u00e0: 3000 : 10 = 300 (ch\u1eef s\u1ed1)","\u1ede s\u1ed1 2008 c\u00f3 1 ch\u1eef s\u1ed1 8. N\u00ean s\u1ed1 ch\u1eef s\u1ed1 8 c\u00f3 trong c\u00e1c s\u1ed1 t\u1eeb 1000 \u0111\u1ebfn 2013 l\u00e0: 300 + 1 = 301 (ch\u1eef s\u1ed1) \u0110\u00e1p s\u1ed1: 301 ch\u1eef s\u1ed1 8 B\u00e0i 56: \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 t\u1eeb 100 \u0111\u1ebfn 999 c\u1ea7n bao nhi\u00eau ch\u1eef s\u1ed1 9 ? Gi\u1ea3i Ta x\u00e9t t\u1eeb 00 (vi\u1ebft cho \u0111\u1ee7 2 ch\u1eef s\u1ed1)\u0111\u1ebfn 99 c\u00f3 100 s\u1ed1. C\u00f3 100 x 2 = 200 (ch\u1eef s\u1ed1) \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 (t\u1eeb 0 \u0111\u1ebfn 9) n\u00ean c\u00f3 200:10=20(ch\u1eef s\u1ed1 9) T\u1eeb 000 \u0111\u1ebfn 999 c\u00f3 1000 s\u1ed1. C\u00f3 3 x 1000= 3 000 (ch\u1eef s\u1ed1) \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 (t\u1eeb 0 \u0111\u1ebfn 9). V\u1eady trong 1000 s\u1ed1 n\u00e0y c\u00f3: 3000 : 10 = 300 (ch\u1eef s\u1ed1 9) T\u1eeb 100 \u0111\u1ebfn 999 c\u00f3: 300 \u2013 20 = 280 (ch\u1eef s\u1ed1 9) B\u00e0i 57: T\u00ecm m\u1ed9t s\u1ed1 c\u00f3 5 ch\u1eef s\u1ed1. Bi\u1ebft n\u1ebfu vi\u1ebft th\u00eam s\u1ed1 2 v\u00e0o \u0111\u1eb1ng sau s\u1ed1 c\u1ea7n t\u00ecm \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 b\u1eb1ng s\u1ed1 c\u1ea7n t\u00ecm vi\u1ebft th\u00eam 2 v\u00e0o tr\u01b0\u1edbc nh\u00e2n 3 S\u1ed1 c\u00f3 d\u1ea1ng: abcde2=2abcde x 3 t\u00ecm abcde=? Gi\u1ea3i abcde x 10 + 2 = (200000+abcde) x 3 abcde x 7 = 599998 abcde = 599998 : 7 abcde = 85714 B\u00e0i 58: Chia c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 100 th\u00e0nh hai l\u1edbp: l\u1edbp s\u1ed1 ch\u1eb5n v\u00e0 l\u1edbp s\u1ed1 l\u1ebb. Nh\u01b0 v\u1eady t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a hai l\u1edbp h\u01a1n k\u00e9m nhau \u2026 \u0111\u01a1n v\u1ecb. Gi\u1ea3i T\u1eeb 0 \u0111\u1ebfn 99 c\u00f3 100 s\u1ed1: 50 s\u1ed1 ch\u1eb5n (k\u1ec3 c\u1ea3 s\u1ed1 0) v\u00e0 50 s\u1ed1 l\u1ebb. -S\u1ed1 ch\u1eb5n: .H\u00e0ng \u0111\u01a1n v\u1ecb \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 5 ch\u1eef s\u1ed1:0;2;4;6;8. T\u1ed5ng ch\u00fang l\u00e0: (2+4+6+8) x 50:5 = 200 .H\u00e0ng ch\u1ee5c: (1+2+3+4+5+6+7+8+9)x10:2 = 225 .S\u1ed1 1 \u1edf s\u1ed1 100 T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a c\u00e1c s\u1ed1 ch\u1eb5n l\u00e0: 200 + 225 + 1 = 426. -S\u1ed1 l\u1ebb: H\u00e0ng \u0111\u01a1n v\u1ecb \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 5 ch\u1eef s\u1ed1:1;3;5;7;9. .T\u1ed5ng ch\u00fang l\u00e0: (1+3+5+7+9) x 50:5 = 250 .H\u00e0ng ch\u1ee5c: (1+2+3+4+5+6+7+8+9)x10:2 = 225 T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a c\u00e1c s\u1ed1 l\u1ebb l\u00e0: 250 + 225 = 475 Ch\u00fang h\u01a1n k\u00e9m nhau: 475 \u2013 426 = 49 B\u00e0i 59: \u0110\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb s\u1ed1 0 \u0111\u1ebfn s\u1ed1 99999 th\u00ec ch\u1eef s\u1ed1 0 xu\u1ea5t hi\u1ec7n bao nhi\u00eau l\u1ea7n? Gi\u1ea3i T\u1eeb 00000 \u0111\u1ebfn 99999 c\u00f3 100000 s\u1ed1 c\u00f3 100000x5= 500000 (ch\u1eef s\u1ed1) chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9. S\u1ed1 ch\u1eef s\u1ed1 0 l\u00e0: 500 000 : 10 = 50 000 (ch\u1eef s\u1ed1 0) Nh\u01b0ng ta ph\u1ea3i b\u1ecf \u0111i nh\u1eefng ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng cao nh\u1ea5t kh\u00f4ng cho ph\u00e9p: -T\u1eeb 00000 \u0111\u1ebfn 09999 c\u00f3: 10 000 ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng ch\u1ee5c ngh\u00ecn.. -T\u1eeb 0000 \u0111\u1ebfn 0999 c\u00f3: 1000 ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng ngh\u00ecn. -T\u1eeb 000 \u0111\u1ebfn 099 c\u00f3 100 ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng tr\u0103m","-T\u1eeb 00 \u0111\u1ebfn 09 c\u00f3 10 ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng ch\u1ee5c. N\u00ean vi\u1ebft t\u1eeb 0 \u0111\u1ebfn 99999 c\u00f3: 50000 \u2013 (10000+1000+100+10) = 38 890 (ch\u1eef s\u1ed1 0) B\u00e0i 60: T\u00ecm s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 abcd bi\u1ebft : abcd + abc +ab +a = 4321 (c\u00f3 g\u1ea1ch ngang tr\u00ean \u0111\u1ea7u nghe th\u1ea7y) Gi\u1ea3i B\u00e0i 62: T\u1eeb 1 \u0111\u1ebfn 2013 c\u00f3 bao nhi\u00eau ch\u1eef s\u1ed1 1? Gi\u1ea3i Ta chia t\u1eebng m\u00f3c nh\u01b0 sau: -T\u1eeb 000 (cho \u0111\u1ee7 3 ch\u1eef s\u1ed1) \u0111\u1ebfn 999 c\u00f3 1000 s\u1ed1. Trong \u0111\u00f3 c\u00f3 1000 x 3 = 3000 (ch\u1eef s\u1ed1) chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9. S\u1ed1 ch\u1eef s\u1ed1 1 c\u00f3: 3000 : 10 = 300 (ch\u1eef s\u1ed1 1) -T\u01b0\u01a1ng t\u1ef1 t\u1eeb 1000 \u0111\u1ebfn 1999 c\u00f3 th\u00eam 1000 ch\u1eef s\u1ed1 \u1edf h\u00e0ng ngh\u00ecn n\u00ean c\u00f3: 1000+300=1300 (ch\u1eef s\u1ed1 1) -T\u1eeb 2000 \u0111\u1ebfn 2013 c\u00f3: 2001 ; 2010 ; 2011 ; 2012 v\u00e0 2013 c\u00f3 t\u1ea5t c\u1ea3 6 ch\u1eef s\u1ed1 1. Ch\u1eef s\u1ed1 1 c\u00f3 t\u1ea5t c\u1ea3 l\u00e0: 300+1300+6=1606 (ch\u1eef s\u1ed1 1) B\u00e0i 63: H\u00e3y cho bi\u1ebft c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau s\u1ed1 ch\u1eb5n c\u00f3 4 ch\u1eef s\u1ed1 nh\u1ecf h\u01a1n 2012? Gi\u1ea3i - S\u1ed1 ch\u1eb5n l\u1edbn nh\u1ea5t c\u00f3 4 ch\u1eef s\u1ed1 nh\u1ecf h\u01a1n 2012 l\u00e0 2010 - S\u1ed1 ch\u1eb5n b\u00e9 nh\u1ea5t c\u00f3 4 ch\u1eef s\u1ed1 l\u00e0 1000 - S\u1ed1 s\u1ed1 h\u1ea1ng: (2010 \u2013 1000) : 2 + 1 = 506 B\u00e0i 64: T\u1eeb c\u00e1c s\u1ed1 3,4,0,1,2 l\u1eadp \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau ? Gi\u1ea3i C\u00e1c ch\u1eef s\u1ed1 3;4;0;1;2 c\u00f3 5 ch\u1eef s\u1ed1. Ch\u1eef s\u1ed1 0 kh\u00f4ng th\u1ec3 \u1edf h\u00e0ng tr\u0103m n\u00ean: C\u00f3 4 c\u00e1ch l\u1ef1a ch\u1ecdn h\u00e0ng tr\u0103m, 4 c\u00e1ch l\u1ef1a ch\u1ecdn h\u00e0ng ch\u1ee5c, 3 c\u00e1nh l\u1ef1a ch\u1ecdn h\u00e0ng \u0111\u01a1n v\u1ecb. S\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau \u0111\u01b0\u1ee3c l\u1eadp b\u1edf 5 ch\u1eef s\u1ed1 tr\u00ean l\u00e0: 4x4x3= 48 (s\u1ed1) \u0110\u00e1p s\u1ed1: 48 s\u1ed1 B\u00e0i 65:","Cho s\u1ed1 1895. S\u1ed1 n\u00e0y s\u1ebd thay \u0111\u1ecfi nh\u01b0 th\u1ebf n\u00e0o n\u1ebfu: a) X\u00f3a \u0111i ch\u1eef s\u1ed1 5 b) X\u00f3a \u0111i hai ch\u1eef s\u1ed1 cu\u1ed1i. c) Vi\u1ebft th\u00eam ch\u1eef s\u1ed1 0 v\u00e0o ch\u00ednh gi\u1eefa s\u1ed1 \u0111\u00f3. Gi\u1ea3i a) X\u00f3a \u0111i ch\u1eef s\u1ed1 5 Th\u00ec sau khi b\u1edbt \u0111i 5 \u0111\u01a1n v\u1ecb r\u1ed3i m\u1edbi gi\u1ea3m \u0111i 10 l\u1ea7n. (1895-5):10=189 b) X\u00f3a \u0111i hai ch\u1eef s\u1ed1 cu\u1ed1i. Th\u00ec sau khi b\u1edbt \u0111i 95 \u0111\u01a1n v\u1ecb r\u1ed3i m\u1edbi gi\u1ea3m \u0111i 100 l\u1ea7n. (1895-95):100=18 c) Vi\u1ebft th\u00eam ch\u1eef s\u1ed1 0 v\u00e0o ch\u00ednh gi\u1eefa s\u1ed1 \u0111\u00f3. 1895 th\u00eam ch\u1eef s\u1ed1 0 v\u00e0o ch\u00ednh gi\u1eefa ta \u0111\u01b0\u1ee3c 18095. Gi\u00e1 tr\u1ecb n\u00f3 ch\u00ednh l\u00e0 (1895-95)x10+95 = 18095 B\u00e0i 66: S\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau m\u00e0 t\u00edch c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 b\u1eb1ng 840 l\u00e0 s\u1ed1.... Gi\u1ea3i Ta th\u1ea5y: 840=1x2x2x2x3x5x7 \u0110\u1ec3 c\u00e1c ch\u1eef s\u1ed1 kh\u00e1c nhau, ta c\u00f3: 840=1x2x3x4x5x7 S\u1ed1 l\u1edbn nh\u1ea5t \u0111\u00f3 l\u00e0: 754321 B\u00e0i 67: C\u00f3 th\u1ec3 vi\u1ebft \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau a) c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 s\u1ed1 l\u1ebb? b) c\u00e1c ch\u0169 s\u1ed1 c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 s\u1ed1 ch\u1eb5n? Gi\u1ea3i a).\u0110\u1ec1u l\u00e0 l\u1ebb: 1; 3; 5; 7; 9 C\u00f3 3 ch\u1eef s\u1ed1 s\u1ebd l\u00e0: C\u00f3 5 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m; 4 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c v\u00e0 3 c\u00e1ch ch\u1ecdn h\u00e0ng \u0111\u01a1n v\u1ecb 5x4x3= 60 (s\u1ed1 l\u1ebb) b)\u0110\u1ec1u ch\u1eb5n: 0; 2; 4; 6; 8 C\u00f3 3 ch\u1eef s\u1ed1 s\u1ebd l\u00e0: C\u00f3 4 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m; 4 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c v\u00e0 3 c\u00e1ch ch\u1ecdn h\u00e0ng \u0111\u01a1n v\u1ecb 4x4x3= 48 (s\u1ed1 ch\u1eb5n) B\u00e0i 68: a) T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau b) T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u00f3 kh\u00e1c nhau Gi\u1ea3i a) S\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0 10002 b) S\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0 99987 B\u00e0i 69: - M\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 2 ch\u1eef s\u1ed1 nh\u1ecf nh\u1ea5t m\u00e0 hi\u1ec7u b\u0103ng 6 ? - M\u1ed9t s\u00f3 t\u1ef1 nhi\u00ean c\u00f3 2 ch\u1eef s\u1ed1 l\u1edbn nh\u1ea5t m\u00e0 t\u1ed5ng b\u1eb1ng 9 ? - M\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 hai ch\u1eef s\u1ed1 l\u1edbn nh\u1ea5t m\u00e0 t\u1ed5ng b\u1eb1ng 13 ?","- C\u00f3 bao nhi\u00eau s\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn h\u01a1n 42 v\u00e0 nh\u1ecf h\u01a1n 78 ? Gi\u1ea3i -S\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 nh\u1ecf nh\u1ea5t th\u00ec h\u00e0ng ch\u1ee5c l\u00e0 1. S\u1ed1 \u0111\u00f3 l\u00e0 17 -S\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 l\u1edbn nh\u1ea5t khi h\u00e0ng ch\u1ee5c l\u1edbn nh\u1ea5t c\u00f3 th\u1ec3. S\u1ed1 \u0111\u00f3 l\u00e0 90 -S\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 l\u1edbn nh\u1ea5t khi h\u00e0ng ch\u1ee5c l\u1edbn nh\u1ea5t c\u00f3 th\u1ec3. S\u1ed1 \u0111\u00f3 l\u00e0 94 -S\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn h\u01a1n 42 v\u00e0 nh\u1ecf h\u01a1n 78 l\u00e0 c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 43 \u0111\u1ebfn 77, c\u00f3: 77-43+1= 35 (s\u1ed1) B\u00e0i 70: C\u00f3 th\u1ec3 vi\u1ebft \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau a) c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 s\u1ed1 l\u1ebb? b) c\u00e1c ch\u0169 s\u1ed1 c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 s\u1ed1 ch\u1eb5n? Gi\u1ea3i a).\u0110\u1ec1u l\u00e0 l\u1ebb: 1; 3; 5; 7; 9 C\u00f3 3 ch\u1eef s\u1ed1 s\u1ebd l\u00e0: C\u00f3 5 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m; 4 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c v\u00e0 3 c\u00e1ch ch\u1ecdn h\u00e0ng \u0111\u01a1n v\u1ecb 5x4x3= 60 (s\u1ed1 l\u1ebb) b)\u0110\u1ec1u ch\u1eb5n: 0; 2; 4; 6; 8 C\u00f3 3 ch\u1eef s\u1ed1 s\u1ebd l\u00e0: C\u00f3 4 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m; 4 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c v\u00e0 3 c\u00e1ch ch\u1ecdn h\u00e0ng \u0111\u01a1n v\u1ecb 4x4x3= 48 (s\u1ed1 ch\u1eb5n) B\u00e0i 71: a) T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau b) T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u00f3 kh\u00e1c nhau Gi\u1ea3i a) S\u1ed1 t\u1ef1 nhi\u00ean b\u00e9 nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0 10002 b) S\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0 99987 B\u00e0i 72: C\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1, trong \u0111\u00f3 m\u1ed7i s\u1ed1 kh\u00f4ng c\u00f3 hai ch\u1eef s\u1ed1 n\u00e0o gi\u1ed1ng nhau? Gi\u1ea3i C\u00f3 9 c\u00e1ch ch\u1ecdn h\u00e0ng ngh\u00ecn; 9 c\u00e1ch ch\u1ecdn h\u00e0ng tr\u0103m; 8 c\u00e1ch ch\u1ecdn h\u00e0ng ch\u1ee5c v\u00e0 7 c\u00e1ch ch\u1ecbn h\u00e0ng \u0111\u01a1n v\u1ecb. V\u1eady c\u00f3: 9x9x8x7= 4536 (s\u1ed1) B\u00e0i 73: T\u1ed5ng c\u1ee7a m\u1ed9t d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp b\u1eb1ng 2016. T\u00ecm s\u1ed1 b\u00e9 nh\u1ea5t trong d\u00e3y s\u1ed1 \u0111\u00f3? S\u1ed1 l\u1edbn nh\u1ea5t trong d\u00e3y s\u1ed1 \u0111\u00f3? Gi\u1ea3i T\u1ed5ng 2016 l\u00e0 ch\u1eb5n n\u00ean d\u00e3y s\u1ed1 c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0 m\u1ed9t s\u1ed1 l\u1ebb (s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 s\u1ed1 h\u1ea1ng cu\u1ed1i l\u00e0 c\u00f9ng ch\u1eb5n ho\u1eb7c c\u00f9ng l\u1ebb). N\u00ean trung b\u00ecnh c\u1ed9ng c\u1ee7a d\u00e3y s\u1ed1 l\u00e0 s\u1ed1 \u1edf gi\u1eefa. Ta th\u1ea5y: 2016 = 3x3x4x7x8 = 672x3 = 288x7 = 224x9 = 96x21 = 32x63 T\u1eebng c\u1eb7p t\u00edch b\u1eb1ng 2016 cho ta th\u1ea5y c\u00f3 th\u1ec3 c\u00f3 3;7;9;21;63 s\u1ed1 h\u1ea1ng v\u00e0 c\u00f3 TBC t\u01b0\u01a1ng \u1ee9ng l\u00e0 672 ;288 ;224 ;96 ;32. *.Tr\u01b0\u1eddng h\u1ee3p : 672x3=2016 cho ta bi\u1ebft d\u00e3y s\u1ed1 c\u00f3 3 s\u1ed1 h\u1ea1ng s\u1ed1 \u1edf gi\u1eefa l\u00e0 672. T\u1ed5ng s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i l\u00e0 : 672 x 2 = 1344 S\u1ed1 \u0111\u1ea7u h\u01a1n s\u1ed1 cu\u1ed1i : 3 \u2013 1 = 2 S\u1ed1 \u0111\u1ea7u (s\u1ed1 b\u00e9) : (1344-2) :2 = 671","S\u1ed1 cu\u1ed1i (s\u1ed1 l\u1edbn) : 1344 \u2013 671 = 673 (671+672+673=2016). D\u00e3y s\u1ed1 \u0111\u00f3 l\u00e0 : 671 ; 672 ; 673 *.Tr\u01b0\u1eddng h\u1ee3p : 288x7=2016 cho ta bi\u1ebft d\u00e3y s\u1ed1 c\u00f3 7 s\u1ed1 h\u1ea1ng s\u1ed1 \u1edf gi\u1eefa l\u00e0 288. *.Tr\u01b0\u1eddng h\u1ee3p : 244x9=2016 cho ta bi\u1ebft d\u00e3y s\u1ed1 c\u00f3 3 s\u1ed1 h\u1ea1ng s\u1ed1 \u1edf gi\u1eefa l\u00e0 244. *.Tr\u01b0\u1eddng h\u1ee3p : 96x21=2016 cho ta bi\u1ebft d\u00e3y s\u1ed1 c\u00f3 3 s\u1ed1 h\u1ea1ng s\u1ed1 \u1edf gi\u1eefa l\u00e0 96. *.Tr\u01b0\u1eddng h\u1ee3p : 32x63=2016 cho ta bi\u1ebft d\u00e3y s\u1ed1 c\u00f3 3 s\u1ed1 h\u1ea1ng s\u1ed1 \u1edf gi\u1eefa l\u00e0 32. B\u1ed1n tr\u01b0\u1eddng h\u1ee3p sau ta c\u00f3 th\u1ec3 t\u00ecm s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i b\u1eb1ng c\u00e1ch t\u00ecm T\u1ed4NG s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i (TBCx2), HI\u1ec6U s\u1ed1 \u0111\u1ea7u s\u1ed1 cu\u1ed1i b\u1eb1ng s\u1ed1 s\u1ed1 h\u1ea1ng tr\u1eeb \u0111i 1. B\u00e0i 74: Cho d\u00e3y s\u1ed1 30 ,32, 34,.....,x.T\u00ecm x \u0111\u1ec3 ch\u1eef s\u1ed1 c\u1ee7a d\u00e3y g\u1ea5p 7\/2 l\u1ea7n x . Gi\u1ea3i Ch\u1eef s\u1ed1 c\u1ee7a d\u00e3y g\u1ea5p 7\/2 = 3,5 l\u1ea7n x. C\u00e1c s\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1 t\u1eeb: 30 \u0111\u1ebfn 98 c\u00f3 (98 \u2013 30) : 2 + 1= 35 s\u1ed1. M\u1ed7i s\u1ed1 c\u00f2n thi\u1ebfu 3,5 \u2013 2 = 1,5 (ch\u1eef s\u1ed1). V\u1eady c\u00f2n thi\u1ebfu 35 x 1,5 = 52,5 (ch\u1eef s\u1ed1) S\u1ed1 ch\u1eb5n c\u00f3 3 ch\u1eef s\u1ed1 t\u1eeb 100 \u0111\u1ebfn 998 c\u00f3 (998-100):2+1 = 450 s\u1ed1, m\u1ed7i s\u1ed1 c\u00f2n thi\u1ebfu 3,5-3= 0,5 (ch\u1eef s\u1ed1) V\u1eady c\u00f2n thi\u1ebfu 450 x 0,5 = 225 (ch\u1eef s\u1ed1) Nh\u01b0 v\u1eady c\u00f2n thi\u1ebfu t\u1ea5t c\u1ea3: 225 + 52,5 = 277,5 (ch\u1eef s\u1ed1) S\u1ed1 ch\u1eb5n c\u00f3 4 ch\u1eef s\u1ed1 t\u1eeb 1000 \u0111\u1ebfn 9998 c\u00f3 4500 s\u1ed1. M\u1ed7i s\u1ed1 th\u1eeba 4 \u2013 3,5 = 0,5 (ch\u1eef s\u1ed1) S\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 c\u1ea7n c\u00f3 l\u00e0: 277,5 : 0,5 = 555 (s\u1ed1) x = (555-1) x 2 + 1000 = 2108 B\u00e0i 75: M\u1ed9t cu\u1ed1n s\u00e1ch c\u00f3 200 trang \u0111\u00e1nh s\u1ed1 th\u1ee9 t\u1ef1 l\u00e0 1,2,3,4,...,199,200 .H\u1ecfi ch\u1eef s\u1ed1 1 \u0111\u01b0\u1ee3c xu\u1ea5t hi\u1ec7n bao nhi\u00eau l\u1ea7n tr\u00ean cu\u1ed1n s\u00e1ch ? Gi\u1ea3i C\u00e1ch 1: Ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a c\u00e1c s\u1ed1: 1;11;21;31;\u2026\u2026.;181;191 c\u00f3 (191 \u2013 1) : 10 + 1 = 20 (s\u1ed1) Ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng ch\u1ee5c c\u1ee7a c\u00e1c ch\u1ee5c s\u1ed1: 10; 110 (m\u1ed7i ch\u1ee5c c\u00f3 10 ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng ch\u1ee5c). C\u00f3 20 (s\u1ed1) Ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng tr\u0103m c\u1ee7a c\u00e1c s\u1ed1: 100;101;102;\u2026\u2026;198;199. C\u00f3 100 s\u1ed1. S\u1ed1 ch\u1eef s\u1ed1 1 c\u00f3 t\u1ea5t c\u1ea3: 20 + 20 + 100 = 140 (ch\u1eef s\u1ed1 1) C\u00e1ch 2: Ta vi\u1ebft t\u1eeb 00 \u0111\u1ebfn 99 c\u00f3 100 s\u1ed1, c\u00f3 2 x 100 = 200 ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1.S\u1ed1 ch\u1eef s\u1ed1 1 l\u00e0 : 200 : 10 = 20 (ch\u1eef s\u1ed1 10 T\u01b0\u01a1ng t\u1ef1 t\u1eeb 100 \u0111\u1ebfn 199 c\u00f3 20 ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng ch\u1ee5c v\u00e0 h\u00e0ng \u0111\u01a1n v\u1ecb v\u00e0 100 ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng tr\u0103m. S\u1ed1 ch\u1eef s\u1ed1 1 c\u00f3 t\u1ea5t c\u1ea3: 20 + 20 + 100 = 140 (ch\u1eef s\u1ed1 1) B\u00e0i 76: T\u00ecm hai s\u1ed1 c\u00f3 t\u1ed5ng b\u1eb1ng 71. Bi\u1ebft r\u1eb1ng n\u1ebfu ta l\u1ea5y s\u1ed1 l\u1edbn gh\u00e9p v\u00e0o b\u00ean ph\u1ea3i, b\u00ean tr\u00e1i s\u1ed1 b\u00e9 ta \u0111\u01b0\u1ee3c hai s\u1ed1 c\u00f3 \u0111\u1ec1u c\u00f3 4 ch\u1eef s\u1ed1 v\u00e0 hi\u1ec7u l\u00e0 2079. Gi\u1ea3i Cho 2 s\u1ed1 c\u1ea7n t\u00ecm c\u00f3 d\u1ea1ng ab v\u00e0 cd (ab>cd). Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: abcd \u2013 cdab = 2079 ab x 100 + cd \u2013 cd x 100 \u2013 ab = 2079","ab x 99 \u2013 cd x 99 = 2079 (ab \u2013 cd) x 99 = 2079 Hi\u1ec7u c\u1ee7a 2 s\u1ed1 l\u00e0: 2079 : 99 = 21 S\u1ed1 l\u1edbn l\u00e0 : (71 + 21) : 2 = 46 S\u1ed1 b\u00e9 l\u00e0 : 71 \u2013 46 = 25 \u0110\u00e1p s\u1ed1 : 46 v\u00e0 25 B\u00e0i 77: Cho s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 hai ch\u1eef s\u1ed1. C\u00f9ng m\u1ed9t l\u00fac ta th\u00eam ch\u1eef s\u1ed1 1 v\u00e0o b\u00ean ph\u1ea3i v\u00e0 b\u00ean tr\u00e1i s\u1ed1 \u0111\u00f3 s\u1ebd \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 c\u00f3 b\u1ed1n ch\u1eef s\u1ed1 g\u1ea5p 23 l\u1ea7n s\u1ed1 \u0111\u00e3 cho. H\u00e3y t\u00ecm s\u1ed1 ban \u0111\u1ea7u. Gi\u1ea3i Xem s\u1ed1 \u0111\u00e3 cho l\u00e0 ab, th\u00eam v\u00e0o ta \u0111\u01b0\u1ee3c 1ab1. 1ab1 = ab x 23 1001 + ab.10 = ab.23 1001 = ab.13 1001 : 13 = ab ab = 1001 : 13 ab = 77 S\u1ed1 ban \u0111\u1ea7u l\u00e0 77 Th\u1eed l\u1ea1i: 1771 : 77 = 23 B\u00e0i 78: B\u1ed1n s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp l\u00e0 c\u00e1c ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn ,tr\u0103m, ch\u1ee5c, \u0111\u01a1n v\u1ecb c\u1ee7a 1 s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1.Vi\u1ebft c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00f3 theo th\u1ee9 t\u1ef1 ng\u01b0\u1ee3c l\u1ea1i ta s\u1ebd \u0111\u01b0\u1ee3c 1 s\u1ed1 m\u1edbi c\u00f3 4 ch\u1eef s\u1ed1 l\u1edbn h\u01a1n s\u1ed1 ban \u0111\u1ea7u bao nhi\u00eau \u0111\u01a1n v\u1ecb? Gi\u1ea3i Hai s\u1ed1 c\u00f3 d\u1ea1ng: abcd v\u00e0 dcba, v\u1edbi d = a+3 ; c = a+2 ; b = a+1 Ph\u00e2n t\u00edch ta \u0111\u01b0\u1ee3c: abcd = 1000a + 100b + 10c + d (1) dcba = 1000d + 100c + 10b + a (2) Thay gi\u00e1 tr\u1ecb \u1edf tr\u00ean v\u00e0o (2) ta \u0111\u01b0\u1ee3c: 1000 x (a+3) + 100 x (a+2) + 10 x (a+1) + a = 1111a + 3210 Thay gi\u00e1 tr\u1ecb \u1edf tr\u00ean v\u00e0o (1) ta \u0111\u01b0\u1ee3c: 1000a + 100 x (a+1) + 10 x (a+2) + a+3 = 1111a + 123 L\u1ea5y (2) tr\u1eeb \u0111i (1) ta \u0111\u01b0\u1ee3c: 3210 \u2013 123 = 3087 \u0110\u00e1p s\u1ed1: 3087 B\u00e0i 79: T\u00ecm m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean. Bi\u1ebft khi c\u1ed9ng s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00f3 v\u1edbi t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 n\u00f3 ta \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0 1159. Gi\u1ea3i S\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 4 ch\u1eef s\u1ed1. V\u00ec n\u1ebfu c\u00f3 3 ch\u1eef s\u1ed1 d\u00f9 l\u1edbn nh\u1ea5t 999+27 = 1026 < 1159. C\u00f3 d\u1ea1ng: abcd+a+b+c+d = 1159 1001.a + 101.b + 11.c + 2.d = 1159 =>a=1 v\u00e0 b=1 (a kh\u00f4ng th\u1ec3 b\u1eb1ng 0 v\u00e0 ph\u1ea3i b\u00e9 h\u01a1n 2; n\u1ebfu b=0 th\u00ec 11.c+2.d < 158 v\u00e0 ph\u1ea3i b\u00e9 h\u01a1n 2). 1001 + 101 + 11.c + 2.d = 1159 11.c + 2.d = 1159 \u2013 (1001+101)","11.c + 2.d = 57 Ta th\u1ea5y 57 l\u00e0 s\u1ed1 l\u1ebb, 2.d l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean 11c ph\u1ea3i l\u1ebb hay c ph\u1ea3i l\u1ebb v\u00e0 \u2264 5. N\u1ebfu c=3 th\u00ec 11x3 + 2.d < 57 (33+18 = 51<57). V\u1eady c=5 => 55 + 2.d = 57 => d=1 Ta \u0111\u01b0\u1ee3c: a=1 ; b=1 ; c=5 v\u00e0 d=1 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 : 1151 (d\u1ea5u ch\u1ea5m (.) thay cho d\u1ea5u nh\u00e2n \u0111\u1ec3 d\u1ec5 nh\u00ecn) B\u00e0i 80: Cho 4 ch\u1eef s\u1ed1 a, b, 1, 2 kh\u00e1c nhau v\u00e0 kh\u00e1c 0. Bi\u1ebft t\u1ed5ng c\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau \u0111\u01b0\u1ee3c l\u1eadp t\u1eeb 4 ch\u1eef s\u1ed1 \u0111\u00e3 cho l\u00e0 7326. T\u00ecm ch\u1eef s\u1ed1 a, b. Gi\u1ea3i V\u1edbi 4 ch\u1eef s\u1ed1 a,b,1,2 kh\u00e1c 0 ta l\u1eadp \u0111\u01b0\u1ee3c 4x3x2 = 24 (s\u1ed1) M\u1ed7i ch\u1eef s\u1ed1 \u1edf m\u1ed7i h\u00e0ng \u0111\u1ec1u \u0111\u01b0\u1ee3c xu\u1ea5t hi\u1ec7n s\u1ed1 l\u1ea7n l\u00e0: 24 : 4 = 6 (l\u1ea7n) T\u1ed5ng ch\u00fang s\u1ebd l\u00e0: (a+b+1+2) x 111 x 6 = 7326 666.a + 666.b + 666 + 1332 = 7326 666 x (a+b) = 5328 => a + b = 8 V\u1edbi a+b=8 m\u00e0 a,b\u22600;1;2. Ta \u0111\u01b0\u1ee3c: a=3 ; b=5 ho\u1eb7c a=5 ; b=3 B\u00e0i 81: T\u01b0\u01a1ng t\u1ef1 b\u00e0i 76 B\u00e0i 2810.Cho hai s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 t\u1ed5ng b\u1eb1ng 54. n\u1ebfu gh\u00e9p s\u1ed1 l\u1edbn v\u00e0o b\u00ean tr\u00e1i s\u1ed1 b\u00e9 ho\u1eb7c gh\u00e9p s\u00f3 l\u1edbn v\u00e0o ben ph\u1ea3i s\u00f3 b\u00e9 ta \u0111\u1ec1u \u0111\u01b0\u1ee3c s\u1ed1 c\u00f3 b\u1ed1n ch\u1eef s\u1ed1. Hi\u1ec7u c\u1ee7a hai s\u1ed1 c\u00f3 b\u1ed1n ch\u1eef s\u1ed1 n\u00e0y b\u1eb1ng 1584. T\u00ecm hai s\u1ed1 \u0111\u00e3 cho. Gi\u1ea3i Hai s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 d\u1ea1ng: ab v\u00e0 cd v\u1edbi ab + cd = 54 (cd>ab) v\u00e0 (1) cdab \u2013 abcd = 1584 Ph\u00e2n t\u00edch ta \u0111\u01b0\u1ee3c; 1000.c + 100.d + 10.a + b \u2013 1000.a \u2013 100.b \u2013 10.c \u2013 d = 1584 (990.c + 99.d) \u2013 (990.a + 99.b) = 1584 99 x (cd) \u2013 99 x (ab) = 99 x (cd \u2013 ab) =1584 cd \u2013 ab = 1584 : 99 = 16 (2) T\u1eeb (1) v\u00e0 (2) cho ta b\u00e0i to\u00e1n T\u1ed4NG v\u00e0 HI\u1ec6U. S\u1ed1 ab l\u00e0: (54 \u2013 16) : 2 = 19 S\u1ed1 cd l\u00e0: 54 \u2013 19 = 35 \u0110\u00e1p s\u1ed1 : 19 v\u00e0 35 B\u00e0i 82: T\u00ecm s\u1ed1 l\u01b0\u1ee3ng c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 b\u1ed1n ch\u1eef s\u1ed1 m\u00e0: a)S\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 \u0111\u1ea7u(theo th\u1ee9 t\u1ef1 \u1ea5y) c\u1ed9ng v\u1edbi s\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 cu\u1ed1i( theo th\u1ee9 t\u1ef1 \u1ea5y) nh\u1ecf h\u01a1n 100. b)S\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 \u0111\u1ea7u(theo th\u1ee9 t\u1ef1 \u1ea5y) l\u1edbn h\u01a1n s\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 cu\u1ed1i( theo th\u1ee9 t\u1ef1 \u1ea5y). Gi\u1ea3i a). S\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 \u0111\u1ea7u(theo th\u1ee9 t\u1ef1 \u1ea5y) c\u1ed9ng v\u1edbi s\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 cu\u1ed1i( theo","th\u1ee9 t\u1ef1 \u1ea5y) nh\u1ecf h\u01a1n 100. G\u1ed3m c\u00e1c s\u1ed1: 99 00 c\u00f3 1 s\u1ed1 98 00 ; 9801 c\u00f3 2 s\u1ed1 97 00 ; 9701 ;9702 c\u00f3 3 s\u1ed1 \u2026\u2026. 11 00 ; 1101; 1102; \u2026..; 88 c\u00f3 89 s\u1ed1 10 00; 1001; 1002; 1003; \u2026; 89 c\u00f3 90 s\u1ed1 S\u1ed1 c\u00e1c s\u1ed1 l\u00e0: 1+2+3+4+ \u2026.. + 90 = (1+90) x 90 : 2 = 4095 (s\u1ed1) b). S\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 \u0111\u1ea7u(theo th\u1ee9 t\u1ef1 \u1ea5y) l\u1edbn h\u01a1n s\u1ed1 t\u1ea1o b\u1edfi hai ch\u1eef s\u1ed1 cu\u1ed1i( theo th\u1ee9 t\u1ef1 \u1ea5y). G\u1ed3m c\u00e1c s\u1ed1: 99 98; 9997; 9996; \u2026; 9901; 9900 c\u00f3 99 s\u1ed1 98 97; 9896; \u2026.; 9800 c\u00f3 98 s\u1ed1 97 96; 9795; \u2026.; 9700 c\u00f3 97 s\u1ed1. \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. 11 10; 1109; 1108; \u2026; 1100 c\u00f3 11 s\u1ed1 10 09; 1008; \u2026\u2026..; 1000 c\u00f3 10 s\u1ed1. S\u1ed1 c\u00e1c s\u1ed1 l\u00e0: 10+11+12+13+ \u2026. + 99 = (10+99) x 90 : 2 = 4905 (s\u1ed1) B\u00e0i 83: Bi\u1ebft ab b\u1eb1ng 1,75 l\u1ea7n ba . H\u1ecfi ba b\u1eb1ng bao nhi\u00eau l\u1ea7n t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3. Gi\u1ea3i ab = ba x 1,75 100.ab = 175.ba 1000a + 100b = 1750b + 175a 825a = 1650b a\/b = 1650\/825 = 2 a g\u1ea5p 2 l\u1ea7n b ba = 10.b + a = 10.b + 2.b = 12.b T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1: a + b = 2.b + b = 3.b ba g\u1ea5p t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 n\u00f3 s\u1ed1 l\u1ea7n l\u00e0: 12b : 3.b = 4 (l\u1ea7n) Th\u1eed l\u1ea1i: Ta c\u00f3 c\u00e1c s\u1ed1: ba = 12 ; 24 ; 36 ; 48 12 : (1+2) = 24 : (2+4) = 36 : (3+60 = 48 : (4+8) = 4 B\u00e0i 84: Cho s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1, khi chia s\u1ed1 \u0111\u00f3 cho t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 th\u00ec \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 4 v\u00e0 d\u01b0 3 . N\u1ebfu \u0111\u1ed5i ch\u1ed7 c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00e3 cho th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi l\u1edbn h\u01a1n 6 l\u1ea7n t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00f3 l\u00e0 5 \u0111\u01a1n v\u1ecb Gi\u1ea3i Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: ab \u2013 3 = (a+b) x 4 ab \u2013 3 = 4.a + 4.b 10.a + b \u2013 3 = 4.a + 4.b 6.a \u2013 3 = 3.b Nh\u00e2n v\u1edbi 4 ta \u0111\u01b0\u1ee3c: 24.a \u2013 12 = 12.b (1)","ba = (b+a)x6 + 5 (2) 10.b + a = 6.b + 6.a + 5 4.b = 5.a + 5 Nh\u00e2n v\u1edbi 3 ta \u0111\u01b0\u1ee3c: 12.b = 15.a + 15 Thay gi\u00e1 tr\u1ecb 12.b t\u1eeb (1) v\u00e0o (2). 24.a \u2013 12 = 15.a + 15 9.a = 15 + 12 a = 27 : 9 = 3 Thay a=3 v\u00e0o (2) ta \u0111\u01b0\u1ee3c: 12.b = 15 x 3 + 15 = 60 b = 60 : 12 = 5 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 35 (thay d\u1ea5u (x) b\u1eb1ng d\u1ea5u ch\u1ea5m (.) cho d\u1ec5 nh\u00ecn) B\u00e0i 85: Cho hai s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 t\u1ed5ng b\u1eb1ng 1 111 110. S\u1ed1 th\u1ee9 nh\u1ea5t c\u00f3 c\u00e1c ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng tr\u0103m \u0111\u1ec1u l\u00e0 8, s\u1ed1 th\u1ee9 hai c\u00f3 c\u00e1c ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng tr\u0103m \u0111\u1ec1u l\u00e0 2. N\u1ebfu d\u00f9ng ch\u1eef s\u1ed1 0 thay cho c\u00e1c ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng tr\u0103m c\u1ee7a hai s\u1ed1 \u0111\u00f3 th\u00ec ta \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi m\u00e0 s\u1ed1 n\u00e0y g\u1ea5p 9 s\u1ed1 kia. T\u00ecm hai s\u1ed1 \u0111\u00e3 cho Gi\u1ea3i Khi thay hai ch\u1eef s\u1ed1 \u1edf h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng tr\u0103m t\u1eeb 8 th\u00e0nh 0 th\u00ec s\u1ed1 \u0111\u00f3 gi\u1ea3m \u0111i 8800 \u0111\u01a1n v\u1ecb, t\u1eeb 2 th\u00e0nh 0 th\u00ec s\u1ed1 \u0111\u00f3 gi\u1ea3m \u0111i 2200 \u0111\u01a1n v\u1ecb. T\u1ed5ng hai s\u1ed1 l\u00fac n\u00e0y c\u00f2n: 1 111 110 \u2013 (8800 + 2200) = 1 100 110 T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 1 + 9 = 10 (ph\u1ea7n) S\u1ed1 b\u00e9 sau khi \u0111\u00e3 \u0111\u1ed5i h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng tr\u0103m: 1 100 110 : 10 = 110 011 S\u1ed1 l\u1edbn sau khi \u0111\u00e3 \u0111\u1ed5i h\u00e0ng ngh\u00ecn v\u00e0 h\u00e0ng tr\u0103m: 1 100 110 \u2013 110 011 = 990099 Ta c\u00f3 2 k\u1ebft qu\u1ea3 : Hai s\u1ed1 \u0111\u00f3 l\u00e0: 112211 v\u00e0 998899 ho\u1eb7c : 118811 v\u00e0 992299 B\u00e0i 86: T\u00ecm m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean , khi vi\u1ebft th\u00eam m\u1ed9t s\u1ed1 b\u00e9 h\u01a1n 100 v\u00e0o b\u00ean ph\u1ea3i s\u1ed1 \u0111\u00e3 cho th\u00ec s\u1ed1 \u0111\u00f3 t\u0103ng th\u00eam 1234 \u0111\u01a1n v\u1ecb . T\u00ecm s\u1ed1 \u0111\u00e3 cho v\u00e0 s\u1ed1 vi\u1ebft th\u00eam Gi\u1ea3i S\u1ed1 b\u00e9 h\u01a1n 100 c\u00f3 2 ch\u1eef s\u1ed1. Khi vi\u1ebft th\u00eam v\u00e0o b\u00ean ph\u1ea3i s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 th\u00ec s\u1ed1 \u0111\u00f3 t\u0103ng th\u00eam 1234 \u0111\u01a1n v\u1ecb cho ta bi\u1ebft s\u1ed1 \u0111\u00f3 b\u00e9 h\u01a1n 20. G\u1ecdi s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 1a, s\u1ed1 th\u00eam v\u00e0o l\u00e0 s\u1ed1 nk, ta \u0111\u01b0\u1ee3c: 1ank = 1a + 1234 1000 + 100.a + nk = 10 + a + 1234 = a + 1244 99.a + nk = 244","Suy ra: a<3 (99x 3 = 297>244). Ta th\u1ea5y: 244 : 99 = 2 (d\u01b0 46) V\u1eady a=2 v\u00e0 nk = 46 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 12 S\u1ed1 vi\u1ebft th\u00eam l\u00e0 46. (thay d\u1ea5u nh\u00e2n (x) b\u1eb1ng d\u1ea5u ch\u1ea5m(.) cho d\u1ec5 nh\u00ecn). B\u00e0i 87: T\u00ecm a,b,c,d. Bi\u1ebft: abcd+abc+ab+a = 2238 Gi\u1ea3i T\u1eeb abcd+abc+ab+a = 2238 cho ta bi\u1ebft a<=2 1111a + 111b + 11c + d = 2238 *.N\u1ebfu a=1 th\u00ec: 1111 + 111b + 11c + d = 2238 111b + 11c + d = 2238 \u2013 1111 =1127 Do b,c,d l\u00e0 s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1 n\u00ean: 111b + 11c + d < 1127 (lo\u1ea1i) *.N\u1ebfu a=2 th\u00ec: 2222 + 111b + 11c + d = 2238 111b + 11c + d = 2238 \u2013 2222 =16 => b=0 Ta \u0111\u01b0\u1ee3c: 11c + d = 16 => c=1 v\u00e0 d=5 \u0110\u00e1p s\u1ed1: a=2; b=0; c=1; d=5 (Th\u1eed l\u1ea1i: 2015+201+20+2 = 2238) B\u00e0i 88: T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean bi\u1ebft r\u1eb1ng c\u1ed9ng s\u1ed1 \u0111\u00f3v\u1edbi c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 th\u00ec b\u1eb1ng 2006 Gi\u1ea3i G\u1ecdi s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 abcd. Ta \u0111\u01b0\u1ee3c: abcd + a + b + c + d = 2006 1001a + 101b + 11c + 2d = 2006 (0 < a \u2264 2). *.Tr\u01b0\u1eddng h\u1ee3p a=2 => b=c=0 v\u00e0 2d = 2006 \u2013 2002 = 4 Hay d=2 S\u1ed1 c\u1ea7n t\u00ecm: 2002 *.Tr\u01b0\u1eddng h\u1ee3p a=1 1001 + 101b + 11c + 2d =2006 101b + 11c + 2d = 1005 V\u1edbi gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a c v\u00e0 d (99+18=117) th\u00ec 101b = 888 (b>8). V\u1eady b = 9. Ta \u0111\u01b0\u1ee3c: 11c + 2d = 1005 \u2013 909 = 96 (c>6) 96 l\u00e0 s\u1ed1 ch\u1eb5n, 2d ch\u1eb5n n\u00ean 11c ph\u1ea3i l\u00e0 s\u1ed1 ch\u1eb5n m\u00e0 l\u1edbn h\u01a1n 6 v\u1eady: c=8 2d = 96 \u2013 88 = 8 d=8:2","d=4 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 1984 B\u1ed4 SUNG B\u00e0i 89 : Cho m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean g\u1ed3m c\u00e1c ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c vi\u1ebft li\u00ean ti\u1ebfp t\u1eeb 12345678\u2026\u2026\u202619981999.H\u00e3y t\u00ednh t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00f3. Gi\u1ea3i X\u00e9t d\u00e3y s\u1ed1 t\u1eeb 000 \u0111\u1ebfn 999 c\u00f3 1000 s\u1ed1. M\u1ed7i s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 n\u00ean c\u00f3 t\u1ea5t c\u1ea3 3 x 1000 = 3000 (s\u1ed1) \u0111\u01b0\u1ee3c chia \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1. S\u1ed1 l\u1ea7n xu\u1ea5t hi\u1ec7n c\u1ee7a m\u1ed7i ch\u1eef s\u1ed1: 3 000 : 10 = 300 (l\u1ea7n) T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 t\u1eeb 000 \u0111\u1ebfn 999 (hay t\u1eeb 1 \u0111\u1ebfn 999). (1+2+3+4+5+6+7+8+9) x 300 = 13 500 T\u1eeb 1000 \u0111\u1ebfn 1999 th\u00ec c\u00f3 1999-1000+1 = 1000 (s\u1ed1) C\u00f3 th\u00eam 1000 ch\u1eef s\u1ed1 1. Nh\u01b0 v\u1eady t\u1ed5ng s\u1ebd b\u1eb1ng. 13 500 + 1 000 = 14 500 T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 12345678\u2026\u2026\u202619981999 l\u00e0: 13 500 + 14 500 = 28 000 \u0110\u00e1p s\u1ed1: 28 000 B\u00e0i 90 : H\u00e3y cho bi\u1ebft c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 c\u00f3 m\u1eb7t ch\u1eef s\u1ed1 0? Gi\u1ea3i Ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng ch\u1ee5c th\u00ec c\u00f3: 9 x 10 = 90 (s\u1ed1) Ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb c\u0169ng c\u00f3 90 s\u1ed1. C\u00f3 t\u1ea5t c\u1ea3: 90 + 90 = 180 (s\u1ed1) B\u00e0i 91: H\u00e3y cho bi\u1ebft trong d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp : 1;2;3;...;2013;2014 c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau ch\u1eef s\u1ed1 1? Gi\u1ea3i T\u1eeb 000 \u0111\u1ebfn 999 c\u00f3 1000 s\u1ed1 n\u00ean c\u00f3 3 x 1000 = 3000 (ch\u1eef s\u1ed1) chi \u0111\u1ec1u cho 10 ch\u1eef s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9. V\u1eady c\u00f3: 3000 : 10 = 300 (ch\u1eef s\u1ed1 1) T\u1eeb 1000 \u0111\u1ebfn 1999 c\u00f3 th\u00eam 1000 ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng ngh\u00ecn. N\u00ean c\u00f3: 300 + 1000 = 1300 (ch\u1eef s\u1ed1 1) T\u1eeb 2001 \u0111\u1ebfn 2014 c\u00f3: 2001; 2010; 2011, 2012, 2013, 2014 c\u00f3 7 ch\u1eef s\u1ed1 1. C\u00f3 t\u1ea5t c\u1ea3 l\u00e0: 300 + 1300 + 7 = 1607 (ch\u1eef s\u1ed1 1) B\u00e0i 92: Cho 4 ch\u1eef s\u1ed1: 0, 2, 3, 5. H\u00e3y l\u1eadp t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 m\u00e0 m\u1ed7i s\u1ed1 c\u00f3 \u0111\u1ee7 4 ch\u1eef s\u1ed1 \u0111\u00e3 cho. T\u00ednh t\u1ed5ng. Gi\u1ea3i S\u1ed1 c\u00e1c s\u1ed1 t\u1ea1o \u0111\u01b0\u1ee3c: 3 x 3 x 2 = 18 (s\u1ed1) T\u1ed5ng c\u00e1c s\u1ed1: (2+3+5) x 6 x 1000 + (2+3+5) x (6:3) x (3-1) x 111 = 64440 B\u00e0i 93:","Cho 4 ch\u1eef s\u1ed1: 1, 3, 3, 4. H\u00e3y l\u1eadp t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 m\u00e0 m\u1ed7i s\u1ed1 c\u00f3 \u0111\u1ee7 4 ch\u1eef s\u1ed1 \u0111\u00e3 cho. T\u00ednh t\u1ed5ng. Gi\u1ea3i Ch\u1eef s\u1ed1 1 \u1edf h\u00e0ng ngh\u00ecn: l\u1eadp \u0111\u01b0\u1ee3c 3 s\u1ed1. T\u1ed5ng l\u00e0: 1 x 3 x 1000 + 3 x 2 x 111 + 4 x 111 = 4110 Ch\u1eef s\u1ed1 4 \u1edf h\u00e0ng ngh\u00ecn: l\u1eadp \u0111\u01b0\u1ee3c 3 s\u1ed1. T\u1ed5ng l\u00e0: 4 x 3 x 1000 + 3 x 2 x 111 + 1 x 111 = 12777 Ch\u1eef s\u1ed1 3 \u1edf h\u00e0ng ngh\u00ecn: l\u1eadp \u0111\u01b0\u1ee3c 6 s\u1ed1. T\u1ed5ng l\u00e0: 3 x 6 x 1000 + (1+3+4) x 2 x 111 = 19776 S\u1ed1 c\u00e1c s\u1ed1 t\u1ea1o \u0111\u01b0\u1ee3c : 3 + 3 + 6 = 12 (s\u1ed1) T\u1ed5ng 12 s\u1ed1 \u0111\u00f3 l\u00e0 : 4110 + 12777 + 19776 = 36663 B\u00e0i 94: Cho 5 ch\u1eef s\u1ed1: 0, 1, 3, 2, 4. H\u00e3y l\u1eadp t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 c\u00f3 5 ch\u1eef s\u1ed1 m\u00e0 m\u1ed7i s\u1ed1 c\u00f3 \u0111\u1ee7 5ch\u1eef s\u1ed1 \u0111\u00e3 cho. T\u00ednh t\u1ed5ng. Gi\u1ea3i S\u1ed1 c\u00e1c s\u1ed1 l\u1eadp \u0111\u01b0\u1ee3c: 4 x 4 x 3 x 2 = 96 (s\u1ed1) T\u1ed5ng c\u00e1c s\u1ed1: (1+2+3+4) x 24 x 10000 + (1+2+3+4) x (24 : 4) x (4-1) x 1111 = 2599980 B\u00e0i 95: Cho 5 ch\u1eef s\u1ed1 0, 1, 2, 3, 4. a, C\u00f3 th\u1ec3 vi\u1ebft \u0111\u01b0\u1ee3cbao nhi\u00eau s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 kh\u00e1c nhau t\u1eeb 5 ch\u1eef s\u1ed1 \u0111\u00e3 cho? Trong c\u00e1c s\u1ed1 vi\u1ebft \u0111\u01b0\u1ee3c c\u00f3 bao nhi\u00eau s\u1ed1 ch\u1eb5n? b, T\u00ecm s\u1ed1 ch\u1eb5n l\u1edbn nh\u1ea5t, s\u1ed1 l\u1ebb nh\u1ecf nh\u1ea5t c\u00f3 4 ch\u1eef s\u1ed1 kh\u00e1c nhau \u0111\u01b0\u1ee3c vi\u1ebft t\u1eeb 5 ch\u1eef s\u1ed1 \u0111\u00e3 cho Gi\u1ea3i S\u1ed1 c\u00e1c s\u1ed1 vi\u1ebft \u0111\u01b0\u1ee3c l\u00e0: 4x4x3x2 = 96 (s\u1ed1) V\u1edbi m\u1ed7i ch\u1eef s\u1ed1 1 v\u00e0 3 \u1edf h\u00e0ng ch\u1ee5c ngh\u00ecn th\u00ec s\u1ed1 s\u1ed1 l\u1ebb s\u1ebd l\u00e0: 24 : 4 = 6 (s\u1ed1 l\u1ebb) V\u1edbi m\u1ed7i ch\u1eef s\u1ed1 2 v\u00e0 4 \u1edf h\u00e0ng ch\u1ee5c ngh\u00ecn th\u00ec s\u1ed1 s\u1ed1 l\u1ebb s\u1ebd l\u00e0: 24 : 4 x 2 = 12 (s\u1ed1 l\u1ebb) S\u1ed1 s\u1ed1 l\u1ebb t\u1ea5t c\u1ea3 l\u00e0: (6 + 12) x 2 = 36 (s\u1ed1 l\u1ebb) S\u1ed1 s\u1ed1 ch\u1eb5n l\u00e0 : 96 \u2013 36 = 60 (s\u1ed1 ch\u1eb5n) S\u1ed1 ch\u1eb5n l\u1edbn nh\u1ea5t l\u00e0 4320 S\u1ed1 l\u1ebb nh\u1ecf nh\u1ea5t l\u00e0 1023 B\u00e0i 96: C\u00f3 th\u1ec3 vi\u1ebft \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 kh\u00e1c nhau, bi\u1ebft r\u1eb1ng: a, C\u00e1c ch\u1eef s\u1ed1 c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 nh\u1eefng s\u1ed1 l\u1ebb? b, C\u00e1c ch\u1eef s\u1ed1 c\u1ee7a ch\u00fang \u0111\u1ec1u l\u00e0 nh\u1eefng s\u1ed1 ch\u1eb5n? Gi\u1ea3i C\u00f3 5 ch\u1eef s\u1ed1 l\u1ebb l\u00e0 1;3;5;7;9. S\u1ed1 c\u00e1c s\u1ed1 vi\u1ebft \u0111\u01b0\u1ee3c l\u00e0 : 5x4x3 = 60 (s\u1ed1) C\u00f3 5 ch\u1eef s\u1ed1 ch\u1eb5n l\u00e0 0 ;2 ;4 ;6 ;8 S\u1ed1 c\u00e1c s\u1ed1 vi\u1ebft \u0111\u01b0\u1ee3c l\u00e0 : 4x4x3 = 48 (s\u1ed1) B\u00e0i 97:","a, T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c vi\u1ebft t\u1eef 3 ch\u1eef s\u1ed1 kh\u00e1c nhau. b, T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau. Gi\u1ea3i S\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0 10002 S\u1ed1 t\u1ef1 nhi\u00ean l\u1edbn nh\u1ea5t c\u00f3 5 ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c vi\u1ebft t\u1eeb 3 ch\u1eef s\u1ed1 kh\u00e1c nhau l\u00e0 99987 B\u00e0i 98: Vi\u1ebft li\u00ean ti\u1ebfp c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn 15 \u0111\u1ec3 \u0111\u01b0\u1ee3c 1 s\u1ed1 t\u1ef1 nhi\u00ean. H\u00e3y xo\u00e1 \u0111i 10 ch\u1eef s\u1ed1 v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c m\u00e0 v\u1eabn gi\u1eef nguy\u00ean th\u1ee9 t\u1ef1 c\u1ee7a c\u00e1c ch\u1eef s\u1ed1 c\u00f2n l\u1ea1i \u0111\u1ec3 \u0111\u01b0\u1ee3c: a, S\u1ed1 l\u1edbn nh\u1ea5t; b, S\u1ed1 nh\u1ecf nh\u1ea5t; Vi\u1ebft c\u00e1c s\u1ed1 \u0111\u00f3. Gi\u1ea3i a\/.S\u1ed1 l\u1edbn nh\u1ea5t l\u00e0 91112131415 (123456789101112131415) b\/.S\u1ed1 nh\u1ecf nh\u1ea5t l\u00e0 01112131415 (123456789101112131415) B\u00e0i 99: Vi\u1ebft li\u00ean ti\u1ebfp 10 s\u1ed1 ch\u1eb5n kh\u00e1c 0 \u0111\u1ea7u ti\u00ean \u0111\u1ec3 \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean. H\u00e3y xo\u00e1 \u0111i 10 ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c m\u00e0 v\u1eabn gi\u1eef nguy\u00ean th\u1ee9 t\u1ef1 c\u1ee7a c\u00e1c ch\u1eef s\u1ed1 c\u00f2n l\u1ea1i \u0111\u1ec3 \u0111\u01b0\u1ee3c: a, S\u1ed1 ch\u1eb5n l\u1edbn nh\u1ea5t; b, S\u1ed1 l\u1ebb nh\u1ecf nh\u1ea5t. Gi\u1ea3i S\u1ed1 \u0111\u00f3 l\u00e0 2468101214161820 a\/.S\u1ed1 ch\u1eb5n l\u1edbn nh\u1ea5t l\u00e0 861820 (2468101214161820) b\/.S\u1ed1 l\u1ebb nh\u1ecf nh\u1ea5t l\u00e0 012111 (2468101214161820) B\u00e0i 100: Thay c\u00e1c ch\u1eef b\u1eb1ng c\u00e1c s\u1ed1 th\u00edch h\u1ee3p \u0111\u1ec3: abc - ca = ca - ac Gi\u1ea3i T\u1eeb: abc \u2013 ca = ca \u2013 ac Ph\u00e2n t\u00edch ta \u0111\u01b0\u1ee3c: 100a + 10b + c \u2013 10c \u2013 a = 10c + a \u2013 10a \u2013 c 99a + 10b \u2013 9c = 9c \u2013 9a 108a + 10b = 18c V\u1edbi c c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c=9 th\u00ec 18c = 18 x 9 = 162 => a = 1 Ta \u0111\u01b0\u1ee3c: 108 + 10b = 18c 18c >= 108 => c >= 6 M\u00e0 10b l\u00e0 s\u1ed1 tr\u00f2n ch\u1ee5c n\u00ean 8 x c c\u00f3 ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 8 V\u1eady c=6 Ta \u0111\u01b0\u1ee3c: 108 + 10b = 18 x 6 = 108 V\u1eady b=0 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 : 106 Th\u1eed l\u1ea1i : 106 \u2013 61 = 61 \u2013 16 = 45 B\u00e0i 101: T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 3 ch\u1eef s\u1ed1 bi\u1ebft r\u1eb1ng khi nh\u00e2n s\u1ed1 \u0111\u00f3 v\u1edbi 2 th\u00ec ta \u0111\u01b0\u1ee3c 1 s\u1ed1 b\u1eb1ng s\u1ed1 c\u00e1c ch\u1eef s\u1ed1 \u0111\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1 \u0111\u1ebfn s\u1ed1 ph\u1ea3i t\u00ecm .","Gi\u1ea3i S\u1ed1 ch\u1eef s\u1ed1 \u0111\u01b0\u1ee3c vi\u1ebft t\u1eeb 1 \u0111\u1ebfn 100 l\u00e0: 9 + 90 x 2 + 3 = 192. 192 b\u00e9 h\u01a1n 100x2 l\u00e0: 200 \u2013 192 = 8 Khi vi\u1ebft th\u00eam 1 s\u1ed1 th\u00ec \u0111\u01b0\u1ee3c th\u00eam 3 ch\u1eef s\u1ed1. L\u00fac n\u00e0y s\u1ed1 c\u1ea7n t\u00ecm th\u00eam 1 \u0111\u01a1n v\u1ecb nh\u00e2n v\u1edbi 2. N\u00ean m\u1ed7i khi vi\u1ebft th\u00eam m\u1ed9t s\u1ed1 th\u00ec \u0111\u01b0\u1ee3c th\u00eam: 3 \u2013 1x2 = 1 (ch\u1eef s\u1ed1) S\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 c\u1ea7n ph\u1ea3i v\u1ebft th\u00eam l\u00e0: 8 : 1 = 8 (s\u1ed1) S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 100 + 8 = 108 Th\u1eed l\u1ea1i : 108 x 2 = 216 9 + 90 x 2 + (108-100+1) x 3 = 216 10.HI\u1ec6U - T\u1ec8 B\u00e0i 1: N\u1ebfu b\u1edbt 1 c\u1ea1nh c\u1ee7a h\u00ecnh vu\u00f4ng \u0111i 7m v\u00e0 b\u1edbt 1 c\u1ea1nh kh\u00e1c \u0111i 25m th\u00ec \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p 3 l\u1ea7n chi\u1ec1u r\u1ed9ng .T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng. Sau khi b\u1edbt s\u1ebd th\u00e0nh h\u00ecnh ch\u1eef nh\u1eadt v\u00e0 hi\u1ec7u 2 c\u1ee7a d\u00e0i v\u00e0 r\u1ed9ng l\u00e0: 25 \u2013 7 = 18(m) S\u01a1 \u0111\u1ed3: D\u00e0i: |___|___|___| R\u1ed9ng: |___| \u202618m\u2026 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 \u2013 1 = 2 (ph\u1ea7n) C\u1ea1nh h\u00ecnh vu\u00f4ng ban \u0111\u1ea7u: 18 : 2 + 25 = 34 (m) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng: 34 x 34 = 1156 (m2) B\u00e0i 2: M\u1ed9t nh\u00e0 m\u00e1y s\u1ea3n xu\u1ea5t trong qu\u00fd I \u0111\u01b0\u1ee3c s\u1ed1 s\u1ea3n ph\u1ea9m k\u00e9m qu\u00fd II 3 l\u1ea7n. N\u1ebfu qu\u00fd I s\u1ea3n xu\u1ea5t t\u0103ng l\u00ean 152 s\u1ea3n ph\u1ea9m th\u00ec ch\u1ec9 k\u00e9m qu\u00fd II 500 s\u1ea3n ph\u1ea9m. T\u00ednh s\u1ed1 s\u1ea3n ph\u1ea9m nh\u00e0 m\u00e1y s\u1ea3n xu\u1ea5t trong m\u1ed7i qu\u00fd? S\u1ed1 s\u1ea3n ph\u1ea9m qu\u00fd II h\u01a1n qu\u00fd I l\u00e0: 152 + 500 = 652 (sp) Ta c\u00f3 s\u01a1 \u0111\u1ed3: Qu\u00fd I: |___| \u2026652\u2026 Qu\u00fd II: |___|___|___| Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 \u2013 1 = 2 (ph\u1ea7n) S\u1ea3n ph\u1ea9m qu\u00fd I: 652 : 2 =326 (sp) S\u1ea3n ph\u1ea9m qu\u00fd II: 326 x 3 = 978 (sp) B\u00e0i 3: Hi\u1ec7u hai s\u1ed1 th\u1eadp ph\u00e2n l\u00e0 22,2.Khi c\u1ed9ng hai s\u1ed1 th\u1eadp ph\u00e2n \u0111\u00f3 v\u1edbi nhau, b\u1ea1n L\u00fd d\u00e3 d\u1ecbch nh\u1ea7m d\u1ea5u ph\u1ea9y c\u1ee7a s\u1ed1 b\u00e9 sang tr\u00e1i 1 ch\u1eef s\u1ed1 n\u00ean t\u1ed5ng t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 92,93.T\u00ecm t\u1ed5ng \u0111\u00fang c\u1ee7a hai s\u1ed1 th\u1eadp ph\u00e2n \u0111\u00f3. Do d\u1ecbch chuy\u1ec3n d\u1ea5u ph\u1ea9y sang tr\u00e1i 1 ch\u1eef s\u1ed1 th\u00ec s\u1ed1 \u0111\u00f3 gi\u1ea3m \u0111i 10 l\u1ea7n. S\u1ed1 l\u1edbn: |________________________|..22,2..| S\u1ed1 b\u00e9: |__|__|__|__|__|__|__|__|__|__|","\u0110\u00e3 gi\u1ea3m s\u1ed1 b\u00e9 10 l\u1ea7n T\u1ed5ng 92,93 S\u1ed1 l\u1edbn: |________________________|..22,2..| S\u1ed1 b\u00e9: |__| T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 10 + 1 = 11 (ph\u1ea7n) S\u1ed1 b\u00e9 l\u00e0: (92,93-22,2) : 11 x10 = 64,3 S\u1ed1 l\u1edbn: 64,3 + 22,2 = 86,5 T\u1ed5ng \u0111\u00fang: 86,5 + 64,3 = 150,8 B\u00e0i 4: T\u00ecm 2 s\u1ed1 Cho hai s\u1ed1 c\u00f3 hi\u1ec7u b\u1eb1ng 30. Bi\u1ebft n\u1ebfu b\u1edbt m\u1ed7i s\u1ed1 \u0111i 4 \u0111\u01a1n v\u1ecb th\u00ec \u0111\u01b0\u1ee3c hai s\u1ed1 m\u1edbi m\u00e0 s\u1ed1 n\u1ea7y b\u1eb1ng 2\/5 s\u1ed1 kia. Tm2 hai s\u1ed1 \u0111\u00f3. B\u1edbt \u0111i \u1edf m\u1ed7i s\u1ed1 4 \u0111\u01a1n v\u1ecb th\u00ec hi\u1ec7u kh\u00f4ng \u0111\u1ed5i v\u00e0 v\u1eabn b\u1eb1ng 30 S\u1ed1 b\u00e9: 30 : (5 - 2) x 2 + 4 = 24 S\u1ed1 l\u1edbn: 24 + 30 = 54 B\u00e0i 5: C\u00f3 3 nh\u00f3m h\u1ecdc sinh \u0111i lao \u0111\u1ed9ng. N\u1ebfu l\u1ea5y 2\/5 s\u1ed1 h\u1ecdc sinh c\u1ee7a nh\u00f3m th\u1ee9 nh\u1ea5t chia \u0111\u1ec1u cho 2 nh\u00f3m kia th\u00ec s\u1ed1 h\u1ecdc sinh c\u1ee7a 3 nh\u00f3m s\u1ebd b\u1eb1ng nhau. N\u1ebfu b\u1edbt \u1edf nh\u00f3m th\u1ee9 nh\u1ea5t 3 em th\u00ec s\u1ed1 h\u1ecdc sinh c\u00f2n l\u1ea1i c\u1ee7a nh\u00f3m n\u00e0y s\u1ebd b\u1eb1ng t\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a 2 nh\u00f3m kia. T\u00ednh s\u1ed1 h\u1ecdc sinh \u0111i lao \u0111\u1ed9ng? C\u00e1ch 1: Ph\u00e2n s\u1ed1 ch\u1ec9 s\u1ed1 hs c\u00f2n l\u1ea1i c\u1ee7a nh\u00f3m 1 sau khi chia \u0111\u1ec1u cho 2 nh\u00f3m kia: 1 \u2013 2\/5 = 3\/5 (nh\u00f3m 1) M\u1ed7i nh\u00f3m 2 v\u00e0 3 nh\u1eadn \u0111\u01b0\u1ee3c: 2\/5 : 2 = 1\/5 (nh\u00f3m 1) Ph\u00e2n s\u1ed1 ch\u1ec9 s\u1ed1 hs c\u1ee7a nh\u00f3m 2 v\u00e0 3: 3\/5 \u2013 1\/5 = 2\/5 (nh\u00f3m 1) Ph\u00e2n s\u1ed1 ch\u1ec9 t\u1ed5ng s\u1ed1 hs c\u1ee7a 2 nh\u00f3m 2 v\u00e0 3: 2\/5 x 2 = 4\/5 (nh\u00f3m 1) Ph\u00e2n s\u1ed1 ch\u1ec9 3 hs: 1 \u2013 4\/5 = 1\/5 (nh\u00f3m 1) S\u1ed1 hs nh\u00f3m 1: 3 x 5 = 15 (em) T\u1ed5ng s\u1ed1 hs 2 nh\u00f3m 2 v\u00e0 3: 15 x 4\/5 = 12 (em) S\u1ed1 h\u1ecdc sinh \u0111i lao \u0111\u1ed9ng: 15 + 12 = 27 (h\u1ecdc sinh) \u0110\u00e1p s\u1ed1: 27 h\u1ecdc sinh C\u00e1ch 2: Nh\u00f3m 1 c\u00f3 5 ph\u1ea7n, b\u1edbt \u0111i 2 ph\u1ea7n chia cho m\u1ed7i nh\u00f3m 1 ph\u1ea7n th\u00ec 3 nh\u00f3m b\u1eb1ng nhau. N\u00ean nh\u00f3m 2 v\u00e0 3 c\u00f3 2 ph\u1ea7n. T\u1ed5ng nh\u00f3m 2 v\u00e0 3 c\u00f3 4 ph\u1ea7n. Ta c\u00f3 s\u01a1 \u0111\u1ed3: 3 hs Nh\u00f3m 1: |___|___|___|___|___| Nh\u00f3m 2: |___|___|\u2026..| Nh\u00f3m 3: |___|___|\u2026..| T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 + 2 + 2 = 9 (ph\u1ea7n) S\u1ed1 h\u1ecdc sinh \u0111i lao \u0111\u1ed9ng: 3 x 9 = 27 (h\u1ecdc sinh) \u0110\u00e1p s\u1ed1: 27 h\u1ecdc sinh B\u00e0i 6: gi\u1ea3i to\u00e1n l\u1edbp 3 Hi\u1ec7u hai s\u1ed1 b\u1eb1ng 4104, bi\u1ebft r\u1eb1ng n\u1ebfu x\u00f3a ch\u1eef s\u1ed1 0 \u1edf h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a s\u1ed1 b\u1ecb tr\u1eeb th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 tr\u1eeb. t\u00ecm hai s\u1ed1 \u0111\u00f3? X\u00f3a ch\u1eef s\u1ed1 0 \u1edf s\u1ed1 b\u1ecb tr\u1eeb \u0111\u01b0\u1ee3c s\u1ed1 tr\u1eeb, cho ta bi\u1ebft s\u1ed1 b\u1ecb tr\u1eeb g\u1ea5p 10 l\u1ea7n s\u1ed1 tr\u1eeb.","Ta c\u00f3 s\u01a1 \u0111\u1ed3: S\u1ed1 b\u1ecb tr\u1eeb: |___|___|___|___|___|___|___|___|___|___| S\u1ed1 tr\u1eeb: |___| ....................4104..........................| Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 10 - 1 = 9 (ph\u1ea7n) S\u1ed1 tr\u1eeb l\u00e0: 4104 : 9 = 456 S\u1ed1 b\u1ecb tr\u1eeb l\u00e0: 456 + 4104 = 4560 \u0110\u00e1p s\u1ed1: 4560 v\u00e0 456 B\u00e0i 7: M\u1ed9t h\u1ecdc sinh \u0111ang gi\u1ea3i to\u00e1n, \u0111\u00e1ng l\u1ebd ph\u1ea3i chia m\u1ed9t s\u1ed1 chia h\u1ebft cho 3 r\u1ed3i c\u1ed9ng th\u01b0\u01a1ng v\u1edbi 8, th\u00ec l\u1ea1i l\u1ea5y s\u1ed1 \u0111\u00f3 nh\u00e2n v\u1edbi 3 r\u1ed3i l\u1ea5y t\u00edch tr\u1eeb \u0111i 8.M\u1eb7c d\u00f9 v\u1eady, \u0111\u00e1p s\u1ed1 v\u1eabn \u0111\u00fang. H\u00e3y t\u00ecm s\u1ed1 \u0111\u00e3 cho. Chia cho 3 th\u00ec s\u1ed1 \u0111\u00f3 gi\u1ea3m \u0111i 3 l\u1ea7n. Nh\u00e2n v\u1edbi 3 th\u00ec s\u1ed1 \u0111\u00f3 t\u0103ng l\u00ean 3 l\u1ea7n. Nh\u01b0 v\u1eady nh\u00e2n v\u1edbi 3 s\u1ebd g\u1ea5p chia cho 3 l\u00e0: 3 x 3 = 9 (l\u1ea7n) Ta c\u00f3 s\u01a1 \u0111\u1ed3: S\u1ed1 c\u1ea7n t\u00ecm: |__|__|__| Chia 3: |__| <\u2026.8\u2026..> <\u2026..8\u2026..> Nh\u00e2n 3: |__|__|__|__|__|__|__|__|__| Khi nh\u00e2n v\u1edbi 3 th\u00ec h\u01a1n chia cho 3: 8 + 8 = 16 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau. 9 \u2013 1 = 8 (ph\u1ea7n) S\u1ed1 c\u1ea7n t\u00ecm: (16 : 8) x 3 = 6 \u0110\u00e1p s\u1ed1: 6 D\u00f9ng PP \u0110\u1ea1i s\u1ed1. G\u1ecdi s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 a. Ta c\u00f3: a:3+8=ax3-8 a x 1\/3 + 8 = a x 3 - 8 a x (3 - 1\/3) = 16 a x 8\/3 = 16 a = 16 : 8\/3 a=6 V\u1eady s\u1ed1 \u0111\u00e3 cho l\u00e0 6 B\u00e0i 8: Hi\u1ec7u c\u1ee7a 2 s\u1ed1 b\u1eb1ng 41. Bi\u1ebft n\u1ebfu gi\u1ea3m s\u1ed1 b\u00e9 \u0111i 19 \u0111\u01a1n v\u1ecb v\u00e0 th\u00eam v\u00e0o s\u1ed1 l\u1edbn 12 \u0111\u01a1n v\u1ecb th\u00ec s\u1ed1 b\u00e9 b\u1eb1ng 3\/5 s\u1ed1 l\u1edbn. T\u00ecm s\u1ed1 l\u1edbn. N\u1ebfu gi\u1ea3m s\u1ed1 b\u00e9 \u0111i 19 \u0111\u01a1n v\u1ecb v\u00e0 th\u00eam v\u00e0o s\u1ed1 l\u1edbn 12 \u0111\u01a1n v\u1ecb th\u00ec hi\u1ec7u l\u00fac n\u00e0y s\u1ebd l\u00e0: 41 + 19 + 12 = 72 Ta c\u00f3 s\u01a1 \u0111\u1ed3: S\u1ed1 l\u1edbn: |-----|-----|-----|-----|-----| S\u1ed1 b\u00e9: |-----|-----|-----| ....72\u2026. Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 \u2013 3 = 2 (ph\u1ea7n)","Gi\u00e1 tr\u1ecb m\u1ed7i ph\u1ea7n : 72 : 2 = 36 36 x 5 = 180 S\u1ed1 l\u1edbn sau khi th\u00eam v\u00e0o 12 \u0111\u01a1n v\u1ecb: S\u1ed1 l\u1edbn c\u1ea7n t\u00ecm: 180 \u2013 12 = 168 \u0110\u00e1p s\u1ed1: 168 B\u00e0i 9: Hi\u1ec7u c\u1ee7a 2 s\u1ed1 b\u1eb1ng 161. Bi\u1ebft r\u1eb1ng n\u1ebfu t\u0103ng s\u1ed1 l\u1edbn th\u00eam 8 \u0111\u01a1n v\u1ecb v\u00e0 gi\u1ea3m s\u1ed1 b\u00e9 \u0111i 3 \u0111\u01a1n v\u1ecb th\u00ec s\u1ed1 l\u1edbn g\u1ea5p 3 l\u1ea7n s\u1ed1 b\u00e9. T\u00ecm s\u1ed1 b\u00e9. Hi\u1ec7u 2 s\u1ed1 sau th\u00eam v\u00e0 b\u1edbt: 161 + 8 + 3 = 172 Ta c\u00f3 s\u01a1 \u0111\u1ed3 sau khi th\u00eam v\u00e0 b\u1edbt: S\u1ed1 l\u1edbn: |---------|---------|---------| S\u1ed1 b\u00e9: |---------|\u2026\u2026.172\u2026\u2026 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 \u2013 1 = 2 (ph\u1ea7n) S\u1ed1 b\u00e9 sau khi gi\u1ea3m: 172 : 2 = 86 S\u1ed1 b\u00e9 l\u00e0: 86 + 3 = 89 \u0110\u00e1p s\u1ed1: 89 B\u00e0i 10: Hi\u1ec7u c\u1ee7a 2 s\u1ed1 b\u1eb1ng 41. Bi\u1ebft n\u1ebfu gi\u1ea3m s\u1ed1 b\u00e9 \u0111i 19 \u0111\u01a1n v\u1ecb v\u00e0 th\u00eam v\u00e0o s\u1ed1 l\u1edbn 12 \u0111\u01a1n v\u1ecb th\u00ec s\u1ed1 b\u00e9 b\u1eb1ng 3\/5 s\u1ed1 l\u1edbn. T\u00ecm s\u1ed1 l\u1edbn. Khi gi\u1ea3m s\u1ed1 b\u00e9 \u0111i 19 \u0111\u01a1n v\u1ecb v\u00e0 th\u00eam v\u00e0o s\u1ed1 l\u1edbn 12 \u0111\u01a1n v\u1ecb th\u00ec hi\u1ec7u m\u1edbi s\u1ebd l\u00e0: 41+19+12= 72 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 -3 = 2 (ph\u1ea7n) S\u1ed1 l\u1edbn sau khi th\u00eam b\u1edbt: 72 : 2 x 5 = 180 S\u1ed1 l\u1edbn l\u00e0: 180 \u2013 12 = 168 \u0110\u00e1p s\u1ed1: 168 B\u00e0i 11: Hi\u1ec7u c\u1ee7a hai s\u1ed1 l\u00e0 12 bi\u1ebft n\u1ebfu nh\u00e2n s\u1ed1 b\u00e9 cho 3 r\u1ed3i tr\u1eeb \u0111i s\u1ed1 l\u1edbn ta \u0111\u01b0\u1ee3c 18. h\u1ecfi t\u1ed5ng c\u1ee7a hai s\u1ed1? S\u01a1 \u0111\u1ed3: S\u1ed1 l\u1edbn: |------------|--12--| \u202618 \u2026\u2026\u2026. S\u1ed1 b\u00e9: |------------|\u2026\u2026\u2026\u2026...|\u2026\u2026\u2026\u2026.| Hai l\u1ea7n s\u1ed1 b\u00e9: 12 + 18 = 30 S\u1ed1 b\u00e9: 30 : 2 = 15 S\u1ed1 l\u1edbn: 15 + 12 = 27 T\u1ed5ng 2 s\u1ed1: 15 + 27 = 42 B\u00e0i 12: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi l\u00e0 320cm. bi\u1ebft n\u1ebfu vi\u1ebft th\u00eam ch\u1eef s\u1ed1 1 v\u00e0o b\u00ean tr\u00e1i s\u1ed1 \u0111\u00f3 \u0111o chi\u1ec1u r\u1ed9ng ta \u0111\u01b0\u1ee3c s\u1ed1 \u0111o chi\u1ec1u d\u00e0i. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3? N\u1eeda chu vi: 320 : 2 = 160 (cm) Chi\u1ec1u d\u00e0i l\u00e0 s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1, chi\u1ec1u r\u1ed9ng c\u00f3 2 ch\u1eef s\u1ed1.(n\u1ebfu chi\u1ec1u r\u1ed9ng c\u00f3 1 ch\u1eef s\u1ed1 th\u00ec khi th\u00eam ch\u1eef s\u1ed1 1 v\u00e0o b\u00ean tr\u00e1i s\u1ebd c\u00f3 d\u1ea1ng 1*. N\u1eeda chu vi s\u1ebd l\u00e0 s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1). Khi th\u00eam v\u00e0o b\u00ean tr\u00e1i s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 m\u1ed9t ch\u1eef s\u1ed1 1 th\u00ec s\u1ed1 \u0111\u00f3 th\u00eam 100 \u0111\u01a1n v\u1ecb. Ta c\u00f3 s\u01a1 \u0111\u1ed3:","D\u00e0i: |----------|------100------| R\u1ed9ng: |----------| T\u1ed5ng 160 Chi\u1ec1u r\u1ed9ng l\u00e0: (160-100):2= 30 (cm) Chi\u1ec1u d\u00e0i: 160 \u2013 30 = 130 (cm) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt; 130 x 30 = 3900 (cm2) \u0110\u00e1p s\u1ed1: 3900 cm2. B\u00e0i 13: C\u00f3 hai kho th\u00f3c, kho A ch\u1ee9a nhi\u1ec1u h\u01a1n kho B 60 t\u1ea5n. Bi\u1ebft sau khi chuy\u1ec3n t\u1eeb kho A sang kho B 50 t\u1ea5n th\u00ec s\u1ed1 th\u00f3c c\u00f2n l\u1ea1i trong kho A b\u1eb1ng 4\/5 s\u1ed1 th\u00f3c trong kho B. V\u1eady l\u00fac \u0111\u1ea7u m\u1ed7i kho ch\u01b0a bao nhi\u00eau t\u1ea5n th\u00f3c? Sau khi chuy\u1ec3n 50 t\u1ea7n th\u00f3c t\u1eeb kho A sang kho B th\u00ec kho B s\u1ebd nhi\u1ec1u h\u01a1n kho A l\u00e0: 50 x 2 \u2013 60 = 40 (t\u1ea5n) Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 \u2013 4 = 1 (ph\u1ea7n) 1 ph\u1ea7n \u1ee9ng v\u1edbi 40 t\u1ea5n v\u1eady s\u1ed1 th\u00f3c kho A hi\u1ec7n nay l\u00e0; 40 x 4 = 160 (t\u1ea5n) S\u1ed1 th\u00f3c kho A ban \u0111\u1ea7u l\u00e0: 160 + 50 = 210 (t\u1ea5n) S\u1ed1 th\u00f3c kho B l\u00e0: 210 \u2013 60 = 150 (t\u1ea5n) \u0110\u00e1p s\u1ed1: Kho A 210 t\u1ea5n Kho B 150 t\u1ea5n. B\u00e0i 14: C\u00f3 hai kho th\u00f3c.Kho A ch\u1ee9a nhi\u1ec1u h\u01a1n kho B 50 t\u1ea5n.Sau khi chuy\u1ec3n t\u1eeb kho A sang kho B 10 t\u1ea5n th\u00ec s\u1ed1 th\u00f3c c\u00f2n l\u1ea1i trong kho A b\u1eb1ng 4\/3 s\u1ed1 th\u00f3c kho B.V\u1eady l\u00fac \u0111\u1ea7u kho A ch\u1ee9a...t\u1ea5n th\u00f3c. Sau khi chuy\u1ec1n 10 t\u1ea5n th\u00f3c t\u1eeb kho A sang kho B th\u00ec l\u00fac n\u00e0y kho A c\u00f2n nhi\u1ec1u h\u01a1n kho B l\u00e0: 50 \u2013 10x2 = 30 (t\u1ea5n) (tr\u1edf v\u1ec1 b\u00e0i to\u00e1n HI\u1ec6U & T\u1ec8) Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau l\u00e0: 4 \u2013 3 = 1 (ph\u1ea7n) 1 ph\u1ea7n t\u01b0\u01a1ng \u1ee9ng v\u1edbi 30 t\u1ea5n. V\u1eady s\u1ed1 th\u00f3c \u1edf kho A l\u00fac n\u00e0y l\u00e0: 30 x 4 = 120 (t\u1ea5n) S\u1ed1 th\u00f3c kho A l\u00fac \u0111\u1ea7u l\u00e0: 120 + 10 = 130 (t\u1ea5n) \u0110\u00e1p s\u1ed1: 130 t\u1ea5n. B\u00e0i 15: T\u00ecm m\u1ed9t s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 bi\u1ebft r\u1eb1ng n\u1ebfu vi\u1ebft th\u00eam ch\u1eef s\u1ed1 0 v\u00e0o b\u00ean ph\u1ea3i s\u1ed1 \u0111\u00f3 ta \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 h\u01a1n s\u1ed1 \u0111\u00e3 cho 504 \u0111\u01a1n v\u1ecb. Th\u00eam ch\u1eef s\u1ed1 0 v\u00e0o b\u00ean ph\u1ea3i m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi h\u01a1n s\u1ed1 c\u0169 g\u1ea5p 10 l\u1ea7n. Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 10 \u2013 1 = 9 (ph\u1ea7n) S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 504 : 9 = 56 \u0110\u00e1p s\u1ed1: 56 B\u00e0i 16: T\u00ecm 1 s\u1ed1 t\u1ef1 nhi\u00ean bi\u1ebft r\u1eb1ng n\u1ebfu vi\u1ebft th\u00eam m\u1ed9t s\u1ed1 v\u00e0o b\u00ean ph\u1ea3i th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi l\u1edbn h\u01a1n s\u1ed1 ph\u1ea3i t\u00ecm 1678 \u0111\u01a1n v\u1ecb","Khi vi\u1ebft th\u00eam b\u00ean ph\u1ea3i 1 ch\u1eef s\u1ed1 th\u00ec s\u1ed1 \u0111\u00f3 l\u1edbn h\u01a1n s\u1ed1 c\u0169 10 l\u1ea7n v\u00e0 s\u1ed1 \u0111\u01a1n v\u1ecb b\u1eb1ng s\u1ed1 th\u00eam v\u00e0o. G\u1ecdi s\u1ed1 th\u00eam v\u00e0o l\u00e0 T, ta c\u00f3 s\u01a1 \u0111\u1ed3: S\u1ed1 m\u1edbi: |---|---|---|---|---|---|---|---|---|---|..T.. S\u1ed1 c\u0169: |---| \u2026\u2026\u2026.1678 \u2026\u2026\u2026\u2026\u2026\u2026.. Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 10 \u2013 1 = 9 (ph\u1ea7n) Hi\u1ec7u b\u1edbt \u0111i T s\u1ebd chia h\u1ebft cho 9. T\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a hi\u1ec7u: 1+6+7+8= 22 V\u1eady T=4 v\u00ec 22 \u2013 4 = 18 chia h\u1ebft cho 9. S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: (1678-4):9 = 186 \u0110\u00e1p s\u1ed1: 186 B\u00e0i 17 Hi\u1ec7u hai s\u1ed1 b\u1eb1ng 111.T\u00ecm 2 s\u1ed1 \u0111\u00f3, n\u1ebfu b\u1edbt \u0111i 9 \u0111\u01a1n v\u1ecb \u1edf s\u1ed1 b\u00e9 th\u00ec s\u1ed1 m\u1edbi b\u1eb1ng 3\/5 s\u1ed1 l\u1edbn. Tr\u01b0\u1eddng h\u1ee3p b\u1edbt \u0111i s\u1ed1 b\u00e9. L\u00fac n\u00e0y hi\u1ec7u m\u1edbi s\u1ebd l\u00e0\u201d 111 + 9 = 120 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 \u2013 3 = 2 (ph\u1ea7n) S\u1ed1 b\u00e9 l\u00e0: 120 : 2 x 3 + 9 = 189 S\u1ed1 l\u1edbn l\u00e0: 189 + 111 = 300 \u0110\u00e1p s\u1ed1: 189 v\u00e0 300 B\u00e0i 18 Hi\u1ec7u c\u1ee7a hai s\u1ed1 \u0111\u00f3 b\u1eb1ng 93 bi\u1ebft s\u1ed1 th\u1ee9 nh\u1ea5t nh\u00e2n v\u1edbi 4 b\u1eb1ng s\u00f3 th\u1ee9 hai nh\u00e2n v\u1edbi 5.T\u00ecm hai s\u1ed1 \u0111\u00f3. S\u1ed1 th\u1ee9 nh\u1ea5t c\u00f3 5 ph\u1ea7n s\u1ed1 th\u1ee9 hai c\u00f3 4 ph\u1ea7n. Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 \u2013 4 = 1 (ph\u1ea7n) S\u1ed1 th\u1ee9 nh\u1ea5t: 93 x 5 = 465 S\u1ed1 th\u1ee9 hai: 465 - 93 = 372 \u0110\u00e1p s\u1ed1: 465 v\u00e0 372 B\u00e0i 19 Hi\u1ec7u 2 s\u1ed1 l\u00e0 41, bi\u1ebft r\u1eb1ng n\u1ebfu th\u00eam 12 \u0111\u01a1n v\u1ecb v\u00e0o s\u1ed1 l\u1edbn v\u00e0 b\u1edbt 19 \u0111\u01a1n v\u1ecb s\u1ed1 b\u00e9 th\u00ec s\u1ed1 b\u00e9 b\u1eb1ng 3\/5 s\u1ed1 l\u1edbn. T\u00ecm s\u1ed1 l\u1edbn. Trong ph\u00e9p tr\u1eeb n\u1ebfu th\u00eam v\u00e0o s\u1ed1 b\u1ecb tr\u1eeb v\u00e0 b\u1edbt \u0111i s\u1ed1 tr\u1eeb th\u00ec hi\u1ec7u s\u1ebd t\u0103ng th\u00eam v\u00e0o b\u1eb1ng t\u1ed5ng s\u1ed1 them v\u00e0o v\u00e0 b\u1edbt ra. Hi\u1ec7u m\u1edbi l\u00e0: 41+12+19= 72 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 5 \u2013 3 = 2 (ph\u1ea7n) Gi\u00e1 tr\u1ecb 1 ph\u1ea7n l\u00e0: 72 : 2 = 36 S\u1ed1 l\u1edbn sau khi th\u00eam v\u00e0o l\u00e0: 36 x 5 = 180 \u0110\u00e1p s\u1ed1: 180 B\u00e0i 20","Hi\u1ec7u hai s\u1ed1 b\u1eb1ng 97.T\u00ecm s\u1ed1 b\u00e9 bi\u1ebft n\u1ebfu c\u00f9ng th\u00eam v\u00e0o m\u1ed7i s\u1ed1 21 \u0111\u01a1n v\u1ecb th\u00ec s\u1ed1 b\u00e9 b\u1eb1ng 3\/4 s\u1ed1 l\u1edbn. Khi c\u00f9ng th\u00eam 2 s\u1ed1 m\u1ed9t s\u1ed1 \u0111\u01a1n v\u1ecb nh\u01b0 nhau th\u00ec hi\u1ec7u v\u1eabn kh\u00f4ng \u0111\u1ed5i. Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 4 \u2013 3 = 1 (ph\u1ea7n) S\u1ed1 b\u00e9 sau khi th\u00eam: 97 x 3 = 291 S\u1ed1 b\u00e9 l\u00e0: 291 \u2013 21 = 270 \u0110\u00e1p s\u1ed1: 270 B\u00e0i 21 M\u1ed9t c\u1eeda h\u00e0ng c\u00f3 s\u1ed1 b\u00fat ch\u00ec xanh g\u1ea5p 3 l\u1ea7n s\u1ed1 b\u00fat ch\u00ec \u0111\u1ecf. Sau khi c\u1eeda h\u00e0ng b\u00e1n \u0111i 12 b\u00fat ch\u00ec xanh v\u00e0 7 b\u00fat ch\u00ec \u0111\u1ecf th\u00ec ph\u1ea7n c\u00f2n l\u1ea1i s\u1ed1 b\u00fat ch\u00ec xanh h\u01a1n s\u1ed1 b\u00fat ch\u00ec \u0111\u1ecf l\u00e0 51 c\u00e2y. H\u1ecfi tr\u01b0\u1edbc khi b\u00e1n m\u1ed7i lo\u1ea1i b\u00fat ch\u00ec c\u00f3 bao nhi\u00eau chi\u1ebfc? S\u1ed1 b\u00fat ch\u00ec xanh h\u01a1n b\u00fat ch\u1ec9 \u0111\u1ecf l\u00e0: 51 + (12 \u2013 7) = 56 (c\u00e2y) Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 \u2013 1 = 2 (ph\u1ea7n) S\u1ed1 b\u00fat ch\u00ec \u0111\u1ecf: 56 : 2 = 28 (ch\u00ec \u0111\u1ecf) S\u1ed1 b\u00fat ch\u00ec xanh: 28 x 3 = 84 (ch\u00ec xanh) \u0110\u00e1p s\u1ed1: 28 \u0111\u1ecf; 84 xanh B\u00e0i 22 Cho hai s\u1ed1 c\u00f3 hi\u1ec7u b\u1eb1ng 12,8 v\u00e0 bi\u1ebft n\u1ebfu \u0111em s\u1ed1 th\u1ee9 nh\u1ea5t chia cho 0,5 ; s\u1ed1 th\u1ee9 hai nh\u00e2n v\u1edbi 3 th\u00ec \u0111\u01b0\u1ee3c hai k\u1ebft qu\u1ea3 b\u1eb1ng nhau. T\u00ecm 2 s\u1ed1 \u0111\u00f3. Chia v\u1edbi 0,5 t\u1ee9c l\u00e0 nh\u00e2n v\u1edbi 2. Cho ta bi\u1ebft s\u1ed1 th\u1ee9 nh\u1ea5t c\u00f3 3 ph\u1ea7n th\u00ec s\u1ed1 th\u1ee9 hai c\u00f3 2 ph\u1ea7n T\u1ec9 s\u1ed1 c\u1ee7a ch\u00fang l\u00e0 3\/2 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau : 3 \u2013 2 = 1 (ph\u1ea7n) S\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 : 12,8 x 3 = 38,4 S\u1ed1 th\u1ee9 hai l\u00e0 : 38,4 \u2013 12,8 = 25,6 \u0110\u00e1p s\u1ed1 : 38,4 v\u00e0 25,6 PH\u1ea6N B\u1ed4 SUNG B\u00e0i 23: T\u00ecm m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean. Bi\u1ebft n\u1ebfu x\u00f3a \u0111i ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a n\u00f3 th\u00ec ta \u0111\u01b0\u1ee3c s\u1ed1 m\u1edbi k\u00e9m h\u01a1n s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 1814 \u0111\u01a1n v\u1ecb. Khi x\u00f3a ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb c\u1ee7a m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean th\u00ec s\u1ed1 \u0111\u00f3 m\u1ea5t \u0111i s\u1ed1 \u0111\u01a1n v\u1ecb \u0111\u00f3 r\u1ed3i gi\u1ea3m \u0111i 10 l\u1ea7n.","S\u1ed1 c\u1ea7n t\u00ecm: |---|---|---|---|---|---|---|---|---|---|\u0111v| S\u1ed1 \u0111\u00e3 x\u00f3a: |---| \u2026\u2026\u2026\u20261814\u2026\u2026\u2026\u2026.|\u2026| Hi\u1ec7u s\u1ed1 ph\u1ea7n: 10 - 1 = 9 (ph\u1ea7n) S\u1ed1 sau khi \u0111\u00e3 x\u00f3a ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb: 1814 : 9 = 201 (d\u01b0 5) S\u1ed1 ban \u0111\u1ea7u: 201 x 10 + 5 = 2015 \u0110\u00e1p s\u1ed1: 2015 B\u00e0i 24: T\u00ecm m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 hai ch\u1eef s\u1ed1. Bi\u1ebft n\u1ebfu vi\u1ebft th\u00eam ch\u1eef s\u1ed1 6 v\u00e0o b\u00ean ph\u1ea3i s\u1ed1 \u0111\u00f3 th\u00ec ta \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 m\u1edbi h\u01a1n s\u1ed1 phai t\u00ecm l\u00e0 771 \u0111\u01a1n v\u1ecb Khi vi\u1ebft th\u00eam 1 ch\u1eef s\u1ed1 6 v\u00e0o b\u00ean ph\u1ea3i m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean th\u00ec s\u1ed1 \u0111\u00f3 t\u0103ng l\u00ean g\u1ea5p 10 l\u1ea7n v\u00e0 th\u00eam 6 \u0111\u01a1n v\u1ecb. S\u1ed1 c\u1ea7n t\u00ecm: |---| \u2026\u2026\u2026\u2026.771 \u0111v \u2026\u2026\u2026...| Sau khi th\u00eam: |---|---|---|---|---|---|---|---|---|---|6 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 10 \u2013 1 = 9 (ph\u1ea7n) 9 l\u1ea7n s\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 771 \u2013 6 = 765 S\u1ed1 c\u1ea7n t\u00ecm l\u00e0: 765 : 9 = 85 B\u00e0i 25: Hi\u1ec7u hai s\u1ed1 l\u00e0 2013. N\u1ebfu vi\u1ebft th\u00eam s\u1ed1 33 v\u00e0o t\u1eadn c\u00f9ng b\u00ean ph\u1ea3i c\u1ee7a s\u1ed1 b\u00e9 th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 l\u1edbn. T\u00ecm hai s\u1ed1 \u0111\u00f3 Gi\u1ea3i Khi th\u00eam 33 v\u00e0o b\u00ean ph\u1ea3i th\u00ec s\u1ed1 b\u00e9 t\u0103ng l\u00ean g\u1ea5p 100 l\u1ea7n v\u00e0 33 \u0111\u01a1n v\u1ecb. Gi\u1ea3m s\u1ed1 l\u1edbn \u0111i 33 \u0111\u01a1n v\u1ecb th\u00ec s\u1ed1 l\u1edbn g\u1ea5p 100 l\u1ea7n s\u1ed1 b\u00e9 v\u00e0 hi\u1ec7u s\u1ebd l\u00e0: 2013 \u2013 33 = 1980 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 100 \u2013 1 = 99 (ph\u1ea7n) S\u1ed1 b\u00e9 l\u00e0: 1980 : 99 = 20 S\u1ed1 l\u1edbn l\u00e0 : 20 + 2013 = 2033 \u0110\u00e1p s\u1ed1 : 20 v\u00e0 2033 B\u00e0i 26: Cho ba s\u1ed1 t\u1ef1 nhi\u00ean, trong \u0111\u00f3 2 l\u1ea7n s\u1ed1 th\u1ee9 nh\u1ea5t b\u1eb1ng 3 l\u1ea7n s\u1ed1 th\u1ee9 hai v\u00e0 b\u1eb1ng 5 l\u1ea7n s\u1ed1 th\u1ee9 ba. T\u00ecm s\u1ed1 th\u1ee9 hai, bi\u1ebft r\u1eb1ng hi\u1ec7u c\u1ee7a s\u1ed1 l\u1edbn nh\u1ea5t v\u00e0 s\u1ed1 b\u00e9 nh\u1ea5t b\u1eb1ng 72 Gi\u1ea3i S\u1ed1 th\u1ee9 nh\u1ea5t b\u1eb1ng 3\/2 s\u1ed1 th\u1ee9 hai v\u00e0 b\u1eb1ng 5\/2 s\u1ed1 th\u1ee9 ba. => 15\/15 s\u1ed1 th\u1ee9 nh\u1ea5t b\u1eb1ng 15\/10 s\u1ed1 th\u1ee9 hai v\u00e0 b\u1eb1ng 15\/6 s\u1ed1 th\u1ee9 ba. Hay s\u1ed1 th\u1ee9 nh\u1ea5t c\u00f3 15 ph\u1ea7n, s\u1ed1 th\u1ee9 hai c\u00f3 10 ph\u1ea7n v\u00e0 s\u1ed1 th\u1ee9 ba c\u00f3 6 ph\u1ea7n b\u1eb1ng nhau. Hi\u1ec7u s\u1ed1 ph\u1ea7n c\u1ee7a s\u1ed1 l\u1edbn nh\u1ea5t v\u00e0 b\u00e9 nh\u1ea5t l\u00e0: 15 \u2013 6 = 9 (ph\u1ea7n) Gi\u00e1 tr\u1ecb 1 ph\u1ea7n l\u00e0: 72 : 9 = 8 S\u1ed1 th\u1ee9 hai l\u00e0:","8 x 10 = 80 8.H\u00ccNH H\u1eccC B\u00e0i 1: Cho tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m I, sao cho IB=IC. N\u1ed1i AI, tr\u00ean \u0111o\u1ea1n AI l\u1ea5y \u0111i\u1ec3m M \u0111\u1ec3 c\u00f3 MI=1\/2AM. N\u1ed1i v\u00e0 k\u00e9o d\u00e0i \u0111o\u1ea1n CM c\u1eaft c\u1ea1nh AB t\u1ea1i N. So s\u00e1nh di\u1ec7n t\u00edch 2 h\u00ecnh tam gi\u00e1c AMN v\u00e0 BMN. Gi\u1ea3i Ta c\u00f3 SMIC= 1\/2 SMCA (2 tam gi\u00e1c c\u00f3 IM= 1\/2 AM; c\u00f9ng \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C). SMIC=SMIB (2 tam gi\u00e1c c\u00f3 IB=IC; c\u00f9ng \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M). Cho ta: SAMC=SBMC (SBMC=SMIC+SMIB). Hai tam gi\u00e1c AMC v\u00e0 BMC c\u00f3 chung \u0111\u00e1y MC. N\u00ean 2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A v\u00e0 t\u1eeb B xu\u1ed1ng c\u1ea1nh \u0111\u00e1y MC b\u1eb1ng nhau. Hai \u0111\u01b0\u1eddng c\u00e0o n\u00e0y c\u0169ng ch\u00ednh l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c AMN v\u00e0 BMN. Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 c\u1ea1nh \u0111\u00e1y chung l\u00e0 MN. V\u1eady: SAMN=SBMN B\u00e0i 2: Cho tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m N sao cho NA < NC. T\u00ecm \u0111i\u1ec3m M tr\u00ean BC \u0111\u1ec3 \u0111o\u1ea1n th\u1eb3ng NM chia h\u00ecnh tam gi\u00e1c ABC l\u00e0m 2 ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau?","H\u01b0\u1edbng d\u1eabn t\u00ecm c\u00e1ch gi\u1ea3i N\u1ebfu N l\u00e0 \u0111i\u1ec3m K trung \u0111i\u1ec3m c\u1ee7a AC th\u00ec NB (KB) s\u1ebd chia h\u00ecnh tam gi\u00e1c ABC l\u00e0m 2 h\u00ecnh tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Do NA < NC n\u00ean \u0111i\u1ec3m M ph\u1ea3i n\u1eb1m tr\u00ean BC. Qua h\u00ecnh v\u1ebd cho ta th\u1ea5y \u0111i\u1ec3m M tr\u00ean BC th\u1ebf n\u00e0o \u0111\u1ec3 NM v\u00e0 KB k\u1ebft h\u1ee3p v\u1edbi 2 c\u1ea1nh c\u1ee7a ABC \u0111\u1ec3 c\u00f3 2 h\u00ecnh tam gi\u00e1c nh\u1ecf c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau th\u00ec M ch\u00ednh l\u00e0 \u0111i\u1ec3m c\u1ea7n t\u00ecm. Gi\u1ea3i L\u1ea5y K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AC. N\u1ed1i BK. Ta c\u00f3 SABK = SCBK (K trung \u0111i\u1ec3m AC) ==> SABK = 1\/2 SABC T\u1eeb K k\u1ebb \u0111o\u1ea1n th\u1eb3ng song song v\u1edbi NB c\u1eaft BC t\u1ea1i M. Trong h\u00ecnh thang NBMK c\u1eb7p tam gi\u00e1c NOK v\u00e0 BOM c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau.","(SNBK=SNBM ; SNOK=SNBK \u2013 SNBO ; SBOM= SNBM \u2013 SNBO ==> SNOK=SBOM ) T\u1ee9 gi\u00e1c ABMN c\u00f3: SABMN = SABK + SBOM \u2013 SNOK = SABK = SABC V\u1eady M ch\u00ednh l\u00e0 \u0111i\u1ec3m c\u1ea7n t\u00ecm. B\u00e0i 3: M\u1ed9t mi\u1ebfng v\u01b0\u1eddn tr\u1ed3ng c\u00e2y \u0103n tr\u00e1i c\u00f3 chi\u1ec1u d\u00e0i 25m, chi\u1ec1u r\u1ed9ng b\u1eb1ng 3\/5 chi\u1ec1u d\u00e0i. Trong v\u01b0\u1eddn ng\u01b0\u1eddi ta x\u1ebb 2 l\u1ed1i \u0111i c\u00f3 chi\u1ec1u r\u1ed9ng l\u00e0 1m (nh\u01b0 h\u00ecnh v\u1ebd). T\u00ednh ph\u1ea7n di\u1ec7n t\u00edch c\u00f2n l\u1ea1i \u0111\u1ec3 tr\u1ed3ng c\u00e2y? C\u00e1ch 1: Chi\u1ec1u r\u1ed9ng mi\u1ebfng v\u01b0\u1eddn: 25 : 5 x 3 = 15 (m) Chi\u1ec1u d\u00e0i m\u1ed7i h\u00ecnh ch\u1eef nh\u1eadt nh\u1ecf: ( 25 - 1 ) : 2 = 12 (m) Chi\u00eau r\u1ed9ng m\u1ed7i h\u00ecnh ch\u1eef nh\u1eadt nh\u1ecf: ( 15 - 1 ) : 2 = 7 (m) Di\u1ec7n t\u00edch ph\u1ea7n c\u00f2n l\u1ea1i \u0111\u1ec3 tr\u1ed3ng c\u00e2y: 12 x 7 x 4 = 336 (m\u00e9t vu\u00f4ng) \u0110\u00e1p s\u1ed1 : 336 m\u00e9t vu\u00f4ng C\u00e1ch 2: Chi\u1ec1u r\u1ed9ng mi\u1ebfng v\u01b0\u1eddn : 25 : 5 x 3 = 15 (m) Di\u1ec7n t\u00edch mi\u1ebfng v\u01b0\u1eddn : 25 x 15 = 375 (m\u00e9t vu\u00f4ng) Di\u1ec7n t\u00edch l\u1ed1i \u0111i theo chi\u1ec1u d\u00e0i : 25 x 1 = 25 (m\u00e9t vu\u00f4ng) Di\u1ec7n t\u00edch l\u1ed1i \u0111i theo chi\u00eau r\u1ed9ng : 15 x 1 - 1 = 14 (m\u00e9t vu\u00f4ng) Di\u1ec7n t\u00edch ph\u1ea7n \u0111\u1ea5t c\u00f2n l\u1ea1i \u0111\u1ec3 tr\u1ed3ng c\u00e2y: 375 - ( 25 + 14 ) = 336 (m\u00e9t vu\u00f4ng) \u0110\u00e1p s\u1ed1 : 336 m\u00e9t vu\u00f4ng C\u00e1ch 3: Gi\u1ea3 s\u1eed ta d\u1eddi 2 l\u1ed1i \u0111i ra s\u00e1t b\u00eca ranh mi\u1ebfng v\u01b0\u1eddn, l\u00fac n\u00e0y l\u1ed1i \u0111i s\u1ebd c\u00f3 h\u00ecnh ch\u1eef L (nh\u01b0 h\u00ecnh v\u1ebd) v\u00e0 ph\u1ea7n \u0111\u1ea5t c\u00f2n l\u1ea1i l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt tr\u1ecdn v\u1eb9n. Chi\u1ec1u r\u1ed9ng mi\u1ebfng v\u01b0\u1eddn : 25 : 5 x 3 = 15 (m) Chi\u1ec1u r\u1ed9ng ph\u1ea7n \u0111\u1ea5t c\u00f2n l\u1ea1i : 15 - 1 = 14 (m) Chi\u1ec1u d\u00e0i ph\u1ea7n \u0111\u1ea5t c\u00f2n l\u1ea1i : 25 - 1 = 24 (m) Di\u1ec7n t\u00edch ph\u1ea7n \u0111\u1ea5t c\u00f2n l\u1ea1i \u0111\u1ec3 tr\u1ed3ng c\u00e2y : 24 x 14 = 336 (m\u00e9t vu\u00f4ng) \u0110\u00e1p s\u1ed1 : 336 m\u00e9t vu\u00f4ng B\u00e0i 4 Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD. Tr\u00ean c\u1ea1nh AB l\u1ea5y hai \u0111i\u1ec3m M, N sao cho AM = MN = NB. P l\u00e0 \u0111i\u1ec3m chia c\u1ea1nh DC th\u00e0nh 2 ph\u1ea7n b\u1eb1ng nhau. ND c\u1eaft MP t\u1ea1i O, n\u1ed1i PN (h\u00ecnh","v\u1ebd). Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c DOP l\u1edbn h\u01a1n di\u1ec7n t\u00edch tam gi\u00e1c MON l\u00e0 3,5 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD. (\u0110\u1ec1 thi h\u1ecdc sinh gi\u1ecfi Qu\u1ea3ng Ninh - TTT s\u1ed1 35) Gi\u1ea3i 2 tam gi\u00e1c MPN v\u00e0 NPD c\u00f3 ph\u1ea7n chung l\u00e0 tam gi\u00e1c NOP. M\u00e0 SDOP - SMON = 3,5cm2. N\u00ean SNPD - SMPN = 3,5cm2 . M\u1eb7t kh\u00e1c SNPD = \u00bc SABCD (NDP c\u00f3 \u0111\u00e1y b\u1eb1ng \u00bd chi\u1ec1u d\u00e0i v\u00e0 \u0111\u01b0\u1eddng cao b\u1eb1ng chi\u1ec1u r\u1ed9ng h\u00ecnh ABCD) v\u00e0 SMPN = 1\/6 SABCD (MPN c\u00f3 \u0111\u00e1y b\u1eb1ng 1\/3 chi\u1ec1u d\u00e0i v\u00e0 \u0111\u01b0\u1eddng cao b\u1eb1ng chi\u1ec1u r\u1ed9ng h\u00ecnh ABCD). Hay: \u00bc SABCD - 1\/6 SABCD = 1\/12 SABCD = 3,5cm2 Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt: 3,5 x 12 = 42 (cm2) \u0110\u00e1p s\u1ed1: 42 cm2 B\u00e0i 5 Trong h\u00ecnh v\u1ebd, ABCD v\u00e0 CEFG l\u00e0 hai h\u00ecnh vu\u00f4ng. Bi\u1ebft EF = 12 cm. H\u00e3y t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AEG. (\u0110\u1ec1 thi to\u00e1n qu\u1ed1c t\u1ebf Ti\u1ec3u h\u1ecdc \u1edf H\u1ed3ng K\u00f4ng) Gi\u1ea3i","N\u1ed1i AC. Ta c\u00f3 SACE = SACG (\u0111\u00e1y CE=CG c\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf, \u0111\u01b0\u1eddng cao AB=AD c\u1ea1nh h\u00ecnh vu\u00f4ng l\u1edbn). Hai tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung l\u00e0 ACI. Suy ra SCIE = SAIG M\u00e0 SAEG = SAIG + SGIE = SCIE + SGIE = SGEC Di\u1ec7n t\u00edch tg GEC b\u1eb1ng v\u1edbi di\u1ec7n t\u00edch tg. AEG 12 x 12 : 2 = 72 (cm2) \u0110\u00e1p s\u1ed1: 72 cm2 B\u00e0i 6: Nu\u00f4i c\u00e1 s\u1ea5u M\u1ed9t tr\u1ea1i nu\u00f4i c\u00e1 s\u1ea5u c\u00f3 m\u1ed9t h\u1ed3 n\u01b0\u1edbc h\u00ecnh vu\u00f4ng, \u1edf gi\u1eefa h\u1ed3 ng\u01b0\u1eddi ta ch\u1eeba m\u1ed9t \u0111\u1ea3o nh\u1ecf h\u00ecnh vu\u00f4ng cho c\u00e1 s\u1ea5u b\u00f2 l\u00ean ph\u01a1i n\u1eafng. Ph\u1ea7n m\u1eb7t n\u01b0\u1edbc c\u00f2n l\u1ea1i r\u1ed9ng 2000m2. T\u1ed5ng chu vi h\u1ed3 n\u01b0\u1edbc v\u00e0 chu vi \u0111\u1ea3o l\u00e0 200m. T\u00ednh c\u1ea1nh h\u1ed3 n\u01b0\u1edbc v\u00e0 c\u1ea1nh c\u1ee7a \u0111\u1ea3o? Gi\u1ea3i Gi\u1ea3 s\u1eed ta d\u1eddi h\u00f2n \u0111\u1ea3o s\u00e1t v\u1edbi g\u00f3c c\u1ee7a h\u1ed3 n\u01b0\u1edbc. N\u1ed1i g\u00f3c \u0111\u1ea3o v\u00e0 g\u00f3c h\u1ed3 (nh\u01b0 h\u00ecnh v\u1ebd). M\u1eb7t n\u01b0\u1edbc c\u00f2n l\u1ea1i l\u00e0 2 h\u00ecnh thang vu\u00f4ng c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau (2 \u0111\u00e1y b\u1eb1ng nhau v\u00e0 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau _ B\u1eb1ng hi\u1ec7u c\u1ee7a c\u1ea1nh h\u1ed3 v\u00e0 c\u1ea1nh \u0111\u1ea3o). Di\u1ec7n t\u00edch m\u1ed7i h\u00ecnh thang l\u00e0: 2000 : 2 = 1000 (m2) T\u1ed5ng 2 \u0111\u00e1y l\u00e0: 200 : 4 = 50 (m) Chi\u1ec1u cao h\u00ecnh thang c\u0169ng l\u00e0 hi\u1ec7u c\u1ea3u c\u1ea1nh h\u1ed3 v\u00e0 c\u1ea1nh \u0111\u1ea3o: 1000 x 2 : 50 = 40 (m) C\u1ea1nh c\u1ee7a \u0111\u1ea3o l\u00e0: (50 \u2013 40) : 2 = 5 (m)","C\u1ea1nh c\u1ee7a h\u1ed3 l\u00e0: 50 \u2013 5 = 45 (m) \u0110\u00e1p s\u1ed1: C\u1ea1nh \u0111\u1ea3o 5 m\u00e9t ; C\u1ea1nh h\u1ed3 45 m\u00e9t. B\u00e0i 7: T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng Cho h\u00ecnh v\u1ebd: Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u00e0 251,2cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng. Gi\u1ea3i H\u01b0\u1edbng gi\u1ea3i: r x r = 251,2 : 3,14 = 80 r x r ch\u00ednh l\u00e0 di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng nh\u1ecf (h\u00ecnh vu\u00f4ng 1\/4) 80 x 4 = 320 Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u1edbn: (cm2) B\u00e0i 8: Di\u1ec7n t\u00edch h\u00ecnh t\u1ee9 gi\u00e1c Cho h\u00ecnh tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh AB ta l\u1ea5y \u0111i\u1ec3m E sao cho BE g\u1ea5p \u0111\u00f4i AE; tr\u00ean c\u1ea1nh AC ta l\u1ea5y \u0111i\u1ec3m D sao cho CD g\u1ea5p \u0111\u00f4i AD. N\u1ed1i E v\u1edbi D ta \u0111\u01b0\u1ee3c h\u00ecnh tam gi\u00e1c AED c\u00f3 di\u1ec7n t\u00edch 5 cm2. H\u00e3y t\u00ednh di\u1ec7n t\u00edch h\u00ecnh t\u1ee9 gi\u00e1c BCDE. Gi\u1ea3i H\u01b0\u1edbng gi\u1ea3i: SBDE = 5 x 2 = 10 (cm2) SABD = 10 + 5 = 15 (cm2) SBDC = 15 x 2 = 30 (cm2) SBCDE = SBDE + SBDC = 10 + 30 = 40 cm2 B\u00e0i 9: So s\u00e1nh di\u1ec7n t\u00edch 2 tam gi\u00e1c. Cho h\u00ecnh vu\u00f4ng ABCD, g\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh AD. \u0110o\u1ea1n th\u1eb3ng AC c\u1eaft BM t\u1ea1i N. a, Di\u1ec7n t\u00edch tam gi\u00e1c BMC g\u1ea5p m\u1ea5y l\u1ea7n Di\u1ec7n t\u00edch tam gi\u00e1c AMB? b, Di\u1ec7n t\u00edch tam gi\u00e1c BNC g\u1ea5p m\u1ea5y l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c ANB ? T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ANB b\u1eb1ng 1,5 dm2 Gi\u1ea3i","a) Theo \u0111\u1ec1 b\u00e0i : AM = 1\/2 AD n\u00ean AM = 1\/2 BC Ta c\u00f3 : sAMB = 1\/2 sBMC ( v\u00ec c\u1ea1nh \u0111\u00e1y AM = 1\/2BC, chi\u1ec1u cao t\u1eeb M xu\u1ed1ng BC b\u1eb1ng chi\u1ec1u cao BA) hay sBMC = 2 x sAMB b) T\u1eeb c\u00e2u a: sBMC = 2 x sAMB m\u00e0 hai tam gi\u00e1c n\u00e0y chung \u0111\u00e1y MB n\u00ean chi\u1ec1u cao CI g\u1ea5p \u0111\u00f4i chi\u1ec1u cao AH M\u1eb7t kh\u00e1c tam gi\u00e1c BNC v\u00e0 ANC c\u00f3 chung \u0111\u00e1y NB, chi\u1ec1u cao CI = 2 x AH Suy ra sBNC = 2 x sANB sABC = 1\/2 sABCD ( .....) sABC = 1.5 x (1+2) = 4,5 (dm2) sABCD = 4,5 x 2 = 9 (dm2) B\u00e0i 10: T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng Cho tam gi\u00e1c ABC c\u00f3 BC = 8 cm. Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m ch\u00ednh gi\u1eefa D. N\u1ed1i B v\u1edbi D. Tr\u00ean BD l\u1ea5y \u0111i\u1ec3m E sao cho BE g\u1ea5p \u0111\u00f4i ED. N\u1ed1i AE, k\u00e9o d\u00e0i c\u1eaft BC \u1edf M. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n BM. Gi\u1ea3i SAED = SEDC (AD=DC ; chung d\u01b0\u1eddng cao k\u1ebb t\u1eeb E) SAED = \u00bd SAEB (ED = \u00bd BE ; chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A) Suy ra SABE = SAEC M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y AE n\u00ean d\u01b0\u1eddng cao k\u1ebb t\u1eeb B v\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng AM b\u1eb1ng nhau. 2 \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c BEM v\u00e0 CEM v\u00e0 c\u00f3 chung \u0111\u00e1y EM. Suy ra SBEM = SCEM V\u1eady BM = MC = 8 : 2 = 4 (cm) B\u00e0i 11: T\u00ednh S ch\u1eef nh\u1eadt ban \u0111\u1ea7u. 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. 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 : Chi\u1ec1u r\u1ed9ng c\u0169: !---! Chi\u1ec1u d\u00e0i c\u0169: !---!---!---!---! Chi\u1ec1u r\u1ed9ng m\u1edbi !---!---!---!---! Chi\u1ec1u d\u00e0i m\u1edbi: !---!---!---!---!---!---!---!---!---!---!---!---!---!---!---!---! ( - - - - - - -- - - - - - 45m - - - -- - - - - - -) 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 : 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 12: Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c Cho h\u00ecnh thang ABCD nh\u01b0 h\u00ecnh b\u00ean. Bi\u1ebft di\u1ec7n t\u00edch 2 tam gi\u00e1c AED v\u00e0 BCF l\u1ea7n l\u01b0\u1ee3c b\u1eb1ng 5,2cm2 v\u00e0 4,8cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh t\u1ee9 gi\u00e1c MFNE. Gi\u1ea3i N\u1ed1i M v\u1edbi N, ta c\u00f3: S(ADN) = S(MDN) ( v\u00ec hai tam gi\u00e1c c\u00f3 chung \u0111\u00e1y DN, \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb A v\u00e0 M xu\u1ed1ng \u0111\u00e1y DN b\u1eb1ng nhau). V\u00ec hai tam gi\u00e1c tr\u00ean c\u00f3 chung ph\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c EDN, n\u00ean : S(ADE) = S(MEN) = 5,2 ( cm2). T\u01b0\u01a1ng t\u1ef1 nh\u01b0 v\u1eady ta c\u0169ng c\u00f3 S(BFC) = S(MNF) = 4,8 (cm2). V\u1eady di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c MENF l\u00e0: 5,2 + 4,8 = 10 ( cm2). \u0110\u00e1p s\u1ed1: 10 cm2 B\u00e0i 13: Hi\u1ec7u 2 di\u1ec7n t\u00edch Cho h\u00ecnh vu\u00f4ng c\u1ea1nh 20cm v\u00e0 h\u00ecnh tr\u00f2n c\u00f3 b\u00e1n k\u00ednh 10cm (h\u00ecnh v\u1ebd). T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n kh\u00f4ng t\u00f4 \u0111\u1eadm c\u1ee7a h\u00ecnh vu\u00f4ng v\u00e0 ph\u1ea7n kh\u00f4ng t\u00f4 \u0111\u1eadm c\u1ee7a h\u00ecnh tr\u00f2n.","Gi\u1ea3i Hai h\u00ecnh \u0111\u00e3 cho c\u00f3 chung ph\u1ea7n di\u1ec7n t\u00edch t\u00f4 \u0111\u1eadm, n\u00ean hi\u1ec7u di\u1ec7n t\u00edch ph\u1ea7n kh\u00f4ng t\u00f4 \u0111\u1eadm c\u1ee7a h\u00ecnh vu\u00f4ng v\u00e0 di\u1ec7n t\u00edch ph\u1ea7n kh\u00f4ng t\u00f4 \u0111\u1eadm c\u1ee7a h\u00ecnh tr\u00f2n ch\u00ednh b\u1eb1ng hi\u1ec7u di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh vu\u00f4ng v\u00e0 h\u00ecnh tr\u00f2n. Hi\u1ec7u di\u1ec7n t\u00edch c\u1ea7n t\u00ecm l\u00e0: (20 x 20) \u2013 (10 x 10 x 3,14) = 86 ( cm2). B\u00e0i 14: Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c Cho t\u1ee9 gi\u00e1c ABCD, M l\u00e0 \u0111i\u1ec3m \u1edf tr\u00ean c\u1ea1nh AB sao cho AM = 1\/3 BM. T\u00ednh di\u1ec7n t\u00edch tam g\u00e1ic MCD bi\u1ebft r\u1eb1ng di\u1ec7n t\u00edch tam gi\u00e1c ACD v\u00e0 tam gi\u00e1c BCD t\u01b0\u01a1ng \u1ee9ng l\u00e0 24cm2 v\u00e0 16cm2. Gi\u1ea3i Chi\u1ec1u cao AI v\u00e0 BK l\u1ea7n l\u01b0\u1ee3t c\u1ee7a 2 tam gi\u00e1c ACD v\u00e0 BCD c\u00f3 t\u1ec9 l\u1ec7 24\/16 = 3\/2 Xem AI = 3 \u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i th\u00ec BK = 2 (\u0111v d\u00e0i) X\u00e9t 2 tam gi\u00e1c BMN v\u00e0 MAN c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb N v\u00e0 BM=3MA N\u00ean S_BMN = 3S_MNA v\u00e0 c\u00f3 chung \u0111\u00e1y MN. Suy ra: \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B g\u1ea5p 3 l\u1ea7n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A xu\u1ed1ng MN. Hay KN=3NI Xem KN = 3 (\u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i) v\u00e0 NI= 1 (\u0111\u01a1n v\u1ecb \u0111\u1ed9 d\u00e0i) th\u00ec KI=4 (\u0111v d\u00e0i) Di\u1ec7n t\u00edch h\u00ecnh thang BAIK = (2+3):2x4 = 10 (\u0111\u01a1nv\u1ecb2) KBM c\u00f3 \u0111\u00e1y KB, cao t\u1eeb M SKBM = 2x3:2=3 (\u0111v 2) T\u01b0\u01a1ng t\u1ef1: SMAI = 1x3:2 = 1,5 (\u0111v2) SKMI = SKBAI \u2013 (SKBM+SMAI) = 10 \u2013 (3+1,5) = 5,5 (\u0111v2)","Chi\u1ec1u cao MN = 5,5 x 2 : 4 = 2,75 (\u0111v d\u00e0i) Tam gi\u00e1c MCD v\u00e0 ACD c\u00f3 chung \u0111\u00e1y. T\u1ec9 l\u1ec7 \u0111\u01b0\u1eddng cao ch\u00ednh l\u00e0 t\u1ec9 l\u1ec7 di\u1ec7n t\u00edch. SMCD\/SACD = 2,75\/3 SMCD\/24 = 2,75\/3 => SMCD = 24 x 2,75 :3 = 22 (cm2) B\u00e0i 15: Di\u1ec7n t\u00edch h\u00ecnh thang Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y nh\u1ecf AB = 2\/3 CD. AC v\u00e0 BD c\u1eaft nhau t\u1ea1i O. Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c BOC l\u00e0 15 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD ? Gi\u1ea3i X\u00e9t tam gi\u00e1c ABC v\u00e0 ACD c\u00f3 chi\u1ec1u cao b\u1eb1ng nhau v\u00e0 c\u00f9ng b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang m\u00e0 \u0111\u00e1y AB = 2\/3 \u0111\u00e1y CD => S_ABC = 2\/3 S_ACD. M\u1eb7t kh\u00e1c 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y AC => s\u1ed1 \u0111o chi\u1ec1u cao t\u1eeb \u0111\u1ec9nh B = 2\/3 s\u1ed1 \u0111o chi\u1ec1u cao t\u1eeb \u0111\u1ec9nh D. X\u00e9t tam gi\u00e1c BOC v\u00e0 DOC c\u00f3 chung \u0111\u00e1y OC chi\u1ec1u cao t\u1eeb \u0111\u1ec9nh B = 2\/3 chi\u1ec1u cao t\u1eeb \u0111\u1ec9nh D => S_BOC = 2\/3 S_DOC. => S_DOC = 15 : 2 x 3 = 22,5 (cm2) V\u1eady S_BCD = 15 + 22,5 = 37,5 (cm2) S_ABD = 37,5 x 2\/3 = 25 (cm2) V\u1eady S_ABCD l\u00e0 : 37,5 + 25 = 62,5 (cm2). B\u00e0i 16: T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n BM Cho tam gi\u00e1c ABC c\u00f3 BC = 8 cm. Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m ch\u00ednh gi\u1eefa D. N\u1ed1i B v\u1edbi D. Tr\u00ean BD l\u1ea5y \u0111i\u1ec3m E sao cho BE g\u1ea5p \u0111\u00f4i ED. N\u1ed1i AE, k\u00e9o d\u00e0i c\u1eaft BC \u1edf M. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n BM. Gi\u1ea3i","SAED = SEDC (AD=DC ; chung d\u01b0\u1eddng cao k\u1ebb t\u1eeb E) SAED = \u00bd SAEB (ED = \u00bd BE ; chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A) Suy ra SABE = SAEC M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y AE n\u00ean d\u01b0\u1eddng cao k\u1ebb t\u1eeb B v\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng AM b\u1eb1ng nhau. 2 \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c BEM v\u00e0 CEM v\u00e0 c\u00f3 chung \u0111\u00e1y EM. Suy ra SBEM = SCEM V\u1eady BM = MC = 8 : 2 = 4 (cm) B\u00e0i 17: Cho h\u00ecnh thang vu\u00f4ng ABCD , AD= 6cm ; DC = 12cm ; AB = 2\/3 DC. a) T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD. b) K\u00e9o d\u00e0i c\u1ea1nh b\u00ean AD v\u00e0 CB, ch\u00fang g\u1eb7p nhau t\u1ea1i M . T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh AM. Gi\u1ea3i a) C\u1ea1nh AB l\u00e0 : 12 x 2\/3 = 8 (cm) Di\u1ec7n t\u00edch ABCD l\u00e0 : (8 + 12) : 2 x 6 = 60 (cm2) b) -X\u00e9t tam gi\u00e1c ABC \u0111\u00e1y AB v\u00e0 DBC \u0111\u00e1y CD c\u00f3 chi\u1ec1u cao b\u1eb1ng nhau = 6cm m\u00e0 \u0111\u00e1y AB = 2\/3 CD => S_ABC = 2\/3 S_DBC. V\u1eabn x\u00e9t 2 tam gi\u00e1c ABC v\u00e0 DBC chung \u0111\u00e1y BC v\u00ec S_ABC = 2\/3 S_DBC => chi\u1ec1u cao AK = 2\/3 DH. -X\u00e9t tam gi\u00e1c AMC v\u00e0 DMC chung \u0111\u00e1y MC m\u00e0 chi\u1ec1u cao AK = 2\/3 DH => S_AMC = 2\/3 S_DMC. M\u00e0 S_DMC l\u1edbn h\u01a1n S_AMC l\u00e0 : 12 x 6 : 2 = 36 (cm2) S_AMC l\u00e0 : 36 : (3-2) x 2 = 72 (cm2) (To\u00e1n Hi\u1ec7u - T\u1ec9) X\u00e9t tam gi\u00e1c AMC \u0111\u00e1y AM, chi\u1ec1u cao CD => AM = 72 x 2 : 12 = 12 (cm)","B\u00e0i 18: Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3 di\u1ec7n t\u00edch 360cm2. Tr\u00ean c\u1ea1nh AB l\u1ea5y 2 \u0111i\u1ec3m M v\u00e0 N sao cho AM=1\/2AB, AN=1\/3AB. G\u1ecdi giao \u0111i\u1ec3m c\u1ee7a DM v\u00e0 CN l\u00e0 O. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c MON. Ta c\u00f3: MN = 1\/2 AB - 1\/3 AB = 1\/6 AB X\u00e9t tam gi\u00e1c NMD v\u00e0 MCD c\u00f3 chi\u1ec1u cao = chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt m\u00e0 \u0111\u00e1y NM = 1\/6 CD => S_NMD = 1\/6 S_MCD. M\u00e0 S_MCD = 360 : 2 = 180 (cm2) => S_NMD = 180 : 6 = 30 (cm2) M\u1eb7t kh\u00e1c 2 tam gi\u00e1c n\u00e0y chugn \u0111\u00e1y MD => Chi\u1ec1u cao tam gi\u00e1c NMD \u0111\u1ec9nh N = 1\/6 chi\u1ec1u cao tam gi\u00e1c MCD \u0111\u1ec9nh C X\u00e9t tam gi\u00e1c NMD v\u00e0 NMC chung \u0111\u00e1y NM chi\u1ec1u cao b\u1eb1ng nhau => S_NMD = S_NMC = 30 (cm2) X\u00e9t tam gi\u00e1c NMO v\u00e0 MCO c\u00f3 chung \u0111\u00e1y MO chi\u1ec1u cao tam gi\u00e1c NMO = 1\/6 chi\u1ec1u cao MCO => S_NMO = 1\/6 S_MCO V\u1eady di\u1ec7n t\u00edch NMO l\u00e0 : 30 : (1 + 6) = 30\/7 (cm2) B\u00e0i 19: Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD, tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m M sao cho BM = MC, tr\u00ean c\u1ea1nh CD l\u1ea5y N sao cho NC = 1\/3xDC. H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c AMN v\u1edbi di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c ADN AB=a ; BC=b Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt: S=a.b S_ADN= 2\/3a x b : 2 = 1\/3 ab = 1\/3S Ta c\u00f3: S_AMN = (S_AMC + S_ANC) \u2013 S_MCN= (MC x AB :2 + NC x AD : 2) \u2013 (NC x MC : 2) = (1\/2b x a : 2 + 1\/3a x b : 2) \u2013 (1\/3a x 1\/2b : 2) = \u00bcS + 1\/6S - 1\/12S = 5\/12 S \u2013 1\/12 S = 4\/12 S = 1\/3 S B\u00e0i 20:","HCN c\u00f3 di\u1ec7n t\u00edch 360 cm2.T\u00ednh di\u1ec7n t\u00edch HCN v\u1edbi s\u1ed1 \u0111o chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng t\u01b0\u01a1ng \u1ee9ng l\u00e0 3\/2s\u1ed1 \u0111o HCN \u0111\u00e3 cho G\u1ecdi S=a x b S_t\u0103ng = 3\/2a x 3\/2b = 9\/4 S Di\u1ec7n t\u00edch m\u1edbi: 360 x 9\/4 = 810 (cm2) B\u00e0i 21: Cho h\u00ecnh tam gi\u00e1c ABC. Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = 1\/3 AB. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m N sao cho AN = 1\/4 AC. N\u1ed1i M v\u1edbi C, n\u1ed1i N v\u1edbi B c\u1eaft nhau t\u1ea1i O. H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch tam gi\u00e1c BOC v\u00e0 di\u1ec7n tich tam gi\u00e1c ABC. N\u1ed1i A v\u1edbi O. Ta c\u00f3: SABN = 1\/3 SBNC n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A v\u00e0 C xu\u1ed1ng NB c\u00f3 t\u1ec9 l\u1ec7 1\/3 Suy ra SABO = 1\/3 SBOC (chung \u0111\u00e1y OB) T\u01b0\u01a1ng t\u1ef1: SAMC = 1\/2SBMC n\u00ean d\u01b0\u1eddng cao k\u1ebb t\u1eeb A v\u00e0 B xu\u1ed1ng MC c\u00f3 t\u1ec9 l\u1ec7 1\/2 Suy ra SAOC = 1\/2 SBOC (chung \u0111\u00e1y OC) T\u1eeb \u0111\u00f3 ta c\u00f3: SAOC + SAOB = (1\/3+1\/2)SBOC = 5\/6 SBOC SAOC + SAOB c\u00f3 5 ph\u1ea7n th\u00ec SBOC c\u00f3 6 ph\u1ea7n v\u00e0 SABC c\u00f3 (5+6) 11 ph\u1ea7n V\u1eady: AOCB = 6\/11 SABC B\u00e0i 22: T\u00ednh \u0111\u1ed9 d\u00e0i Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 900 cm2 v\u00e0 c\u1ea1nh BC = 45 cm. M l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean AB sao cho MB = 1\/3 AB. T\u1eeb M k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi BC c\u1eaft AC t\u1ea1i N. T\u00ednh \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n MN. Ta c\u00f3: SCMB = 1\/2 SAMC (chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C, \u0111\u00e1y MB=1\/2AM) => SCMB = 300 cm2 => \u0110\u01b0\u1eddng cao MI = 300 x 2 : 45 = 13 1\/3 (cm) (h\u1ed7n s\u1ed1)","H\u00ecnh thang NMBC cho ta SCMB = SCNB = 300 cm2 (chung \u0111\u00e1y CB, \u0111\u01b0\u1eddng cao b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang) =>SANB = 900 \u2013 300 = 600 (cm2) M\u1eb7t kh\u00e1c SNMB = 1\/2 SNMA => SNMB = 600 : 3 = 200 (cm2) M\u00e0 tam gi\u00e1c NMB c\u00f3 \u0111\u00e1y NM v\u00e0 \u0111\u01b0\u1eddng cao b\u1eb1ng \u0111\u01b0\u1eddng cao MI. \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n MN = 200 x 2 : 13 1\/3 = 30 (cm) \u0110\u00e1p s\u1ed1: MN = 30cm B\u00e0i 23: T\u00ednh c\u1ea1nh h\u00ecnh vu\u00f4ng C\u00f3 hai t\u1edd gi\u1ea5y h\u00ecnh vu\u00f4ng m\u00e0 s\u1ed1 \u0111o c\u00e1c c\u1ea1nh h\u01a1n k\u00e9m nhau 8 cm . \u0110em \u0111\u1eb7t t\u1edd gi\u1ea5y h\u00ecnh vu\u00f4ng nh\u1ecf n\u1eb1m tr\u1ecdn trong t\u1edd gi\u1ea5y h\u00ecnh vu\u00f4ng l\u1edbn th\u00ec ph\u1ea7n di\u1ec7n t\u00edch c\u00f2n l\u1ea1i kh\u00f4ng b\u1ecb che c\u1ee7a t\u1edd gi\u1ea5y l\u1edbn l\u00e0 96 cm2. T\u00ednh c\u1ea1nh m\u1ed7i t\u1edd gi\u1ea5y ? Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng (3) 8 x 8 = 64 (cm2) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt (1). (96 \u2013 64) : 2 = 16 (cm2) C\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf: C\u1ea1nh h\u00ecnh vu\u00f4ng l\u1edbn: 16 : 8 = 2 (cm) 2 + 8 = 10 (cm) B\u00e0i 24: T\u00ednh S h\u00ecnh thang Cho h\u00ecnh thang ABCD c\u00f3 hai \u0111\u00e1y AB v\u00e0 CD, hai \u0111\u01b0\u1eddng ch\u00e9o c\u1eaft nhau t\u1ea1i O,bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AOB b\u1eb1ng 4 cm2, di\u1ec7n t\u00edch tam gi\u00e1c BOC b\u1eb1ng 9 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD. Trong h\u00ecnh thang ABCD cho ta: SAOD = SBOC = 9 cm2 X\u00e9t 2 tam gi\u00e1c AOB v\u00e0 AOD c\u00f3 ch\u00fang \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A n\u00ean 2 \u0111\u00e1y OB v\u00e0 OD s\u1ebd t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch. Suy ra OB\/OD = 4\/9 M\u1eb7t kh\u00e1c, 2 tam gi\u00e1c BOC v\u00e0 DOC c\u00f3 ch\u00fang \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C n\u00ean 2 di\u1ec7n t\u00edch s\u1ebd t\u1ec9 l\u1ec7 v\u1edbi 2 \u0111\u00e1y. M\u00e0 OB\/OD = 4\/9 n\u00ean SBOC\/SDOC = 4\/9 Di\u1ec7n t\u00edch tam gi\u00e1c DOC: 9 : 4 x 9 = 20,25 (cm2) Di\u1ec7n t\u00edch h\u00ecnh thang ABCD: 4 + 9 + 9 + 20,25 = 42,25 (cm2)","B\u00e0i 25: Ng\u01b0\u1eddi ta \u0111\u01b0a cho Mai v\u00e0 Minh m\u1ed7i b\u1ea1n m\u1ed9t t\u1edd b\u00eca h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi l\u00e0 100cm v\u00e0 c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 nhau r\u1ed3i y\u00eau c\u1ea7u c\u1eaft th\u00e0nh 3 h\u00ecnh ch\u1eef nh\u1eadt b\u1eb1ng nhau. Sau khi c\u1eaft t\u1ed5ng chu vi c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u1ee7a Mai c\u1eaft \u0111\u01b0\u1ee3c h\u01a1n t\u1ed5ng chu vi c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt c\u1ee7a Minh c\u1eaft \u0111\u01b0\u1ee3c l\u00e0 40cm. Em h\u00e3y t\u00ednh di\u1ec7n t\u00edch c\u1ee7a t\u1edd b\u00eca ban \u0111\u1ea7u. Khi c\u1eaft th\u00e0nh 3 h\u00ecnh ch\u1eef nh\u1eadt b\u1eb1ng nhau th\u00ec t\u1ed5ng chu vi 3 h\u00ecnh s\u1ebd d\u00e0i h\u01a1n chu vi c\u0169 4 l\u1ea7n \u0111\u01b0\u1eddng c\u1eaft. Chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng: 40 : 4 = 10 (cm) N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt: 100 : 2 = 50 (cm) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt: (50 \u2013 10) : 2 = 20 (cm) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt: 50 \u2013 20 = 30 (cm) Di\u1ec7n t\u00edch t\u1edd b\u00eca h\u00ecnh ch\u1eef nh\u1eadt: 30 x 20 = 600 (cm2) B\u00e0i 26: \u0110\u01b0\u1eddng k\u00ednh c\u1ee7a m\u1ed9t h\u00ecnh tr\u00f2n t\u0103ng 10% th\u00ec di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n \u0111\u00f3 t\u0103ng bao nhi\u00eau %? \u0110\u01b0\u1eddng k\u00ednh t\u0103ng 10% th\u00ec b\u00e1n k\u00ednh c\u0169ng t\u0103ng 10% C\u00f4ng th\u1ee9c t\u00ednh S= r x r x 3,14. B\u00e1n k\u00ednh t\u0103ng 10% th\u00ec: S(t\u0103ng) = 110%r x 110%r x 3,14 = 121% x r x r x 3,14 = 121%S Di\u1ec7n t\u00edch t\u0103ng: 121% - 100% = 21%. B\u00e0i 27: X\u1ebfp h\u00ecnh l\u1eadp ph\u01b0\u01a1ng Ng\u01b0\u1eddi ta x\u1ebfp nh\u1eefng h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf c\u1ea1nh 1 cm th\u00e0nh 1 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc 1,6 dm ; 1,2dm ; 8 cm. Sau \u0111\u00f3 ng\u01b0\u1eddi ta s\u01a1n 6 m\u1eb7t c\u1ee7a h\u00ecnh v\u1eeba x\u1ebfp \u0111\u01b0\u1ee3c . T\u00ednh s\u1ed1 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf \u0111\u01b0\u1ee3c s\u01a1n 1 m\u1eb7t, 2 m\u1eb7t.","C\u00e1c h\u00ecnh l\u1eadp ph\u01b0\u01a1ng s\u01a1n 1 m\u1eb7t kh\u00f4ng k\u1ec1 b\u00ean g\u00f3c v\u00e0 c\u1ea1nh (m\u1eb7t ch\u00ednh) C\u00e1c h\u00ecnh l\u1eadp ph\u01b0\u01a1ng s\u01a1n 2 m\u1eb7t n\u1eb1m \u1edf c\u1ea1nh kh\u00f4ng l\u00e0 g\u00f3c. Ta c\u00f3: *.2 h\u00ecnh ch\u1eef nh\u1eadt 16 x 12 2 h\u00ecnh ch\u1eef nh\u1eadt 12 x 8 2 h\u00ecnh ch\u1eef nh\u1eadt 16 x 8 S\u1ed1 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng s\u01a1n 1 m\u1eb7t: (16-2)x(12-2)x2 + (12-2)x(8-2)x2 + (16-2)x(8-2)x2 = 568 (h\u00ecnh s\u01a1n 1 m\u1eb7t) *.4 c\u1ea1nh 16cm 4 c\u1ea1nh 12cm 4 c\u1ea1nh 8cm S\u1ed1 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng s\u01a1n 2 m\u1eb7t: (16-2)x4 + (12-2)x4 + (8-2)x4 = 120 (h\u00ecnh s\u01a1n 2 m\u1eb7t) B\u00e0i 28: Di\u1ec7n t\u00edch tam gi\u00e1c Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 \u0111i\u1ec3m N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AC . Tr\u00ean h\u00ecnh \u0111\u00f3 c\u00f3 h\u00ecnh thangBMNE. N\u1ed1i B v\u1edbi N, n\u1ed1i E v\u1edbi M, hai \u0111o\u1ea1n th\u1eb3ng n\u00e0y g\u1eb7p nhau t\u1ea1i \u0111i\u1ec3m O a\/ So s\u00e1nh di\u1ec7n t\u00edch 2 h\u00ecnh tam gi\u00e1c OMB v\u00e0 OEN b\/ So s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EMC v\u1edbi di\u1ec7n t\u00edch h\u00ecnh AEMB ( \u0110\u1ec1 thi HSG to\u00e0n qu\u1ed1c 1984 - 1985 ) (Ch\u01b0a bi\u1ebft 2 \u0111i\u1ec3m M v\u00e0 E c\u1ee7a h\u00ecnh thang BMNE) \u0110i\u1ec3m E n\u1eb1m tr\u00ean \u0111o\u1ea1n AN , \u0111i\u1ec3m M n\u1eb1m tr\u00ean BC, BE l\u00e0 \u0111\u00e1y l\u1edbn MN l\u00e0 \u0111\u00e1y b\u00e9, BN v\u00e0 ME l\u00e0 2 \u0111\u01b0\u1eddng ch\u00e9o h\u00ecnh thang. a). BMNE l\u00e0 h\u00ecnh thang n\u00ean SMBE=SNBE (c\u00f3 ch\u00fang \u0111\u00e1y BE, \u0111\u01b0\u1eddng cao b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang), 2 tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung l\u00e0 OBE n\u00ean SOMB=SOEN b). Do AN=NC n\u00ean SABN=SCBN (cm tr\u00ean) S_EMC=S_CBN \u2013 S_OMB + S_OEN m\u00e0 S_OMB = S_OEN Suy ra: S_EMC=S_CBN T\u01b0\u01a1ng t\u1ef1: S_AEMB=S_ABN \u2013 S_OEN + S_OMB m\u00e0 S_OEN = S_OMB (cm tr\u00ean) Suy ra: S_AEMB=S_ABN Ta \u0111\u00e3 c\u00f3 SABN=SCBN V\u1eady: S_EMC=S_AEMB (\u0111i\u1ec1u ph\u1ea3i ch\u1ee9ng minh) b).Nhanh h\u01a1n Do AN=NC n\u00ean SABN=SCBN= 1\/2 SABC (cm tr\u00ean) S_EMC=S_CBN \u2013 S_OMB + S_OEN m\u00e0 S_OMB = S_OEN Suy ra: S_EMC=S_CBN = 1\/2SABC","V\u1eady: S_EMC=S_AEMB (\u0111i\u1ec1u ph\u1ea3i ch\u1ee9ng minh) B\u00e0i 29: 1).Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 600cm2. D l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh BC. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m E sao cho AE = 1\/3 AC. AD c\u1eaft BE t\u1ea1i M. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AME. Ta c\u00f3: -S_ABD=S_ACD (c\u00f3 CD=BD, \u0111\u01b0\u1eddng cao ch\u00fang t\u1eeb A v\u00e0 c\u00f3 ch\u00fang \u0111\u00e1y AD n\u00ean 2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B v\u00e0 C b\u1eb1ng nhau) -AE=1\/3AC hay AE=1\/2EC -S_ABE=1\/2S_CBE (AE=1\/2EC, \u0111\u01b0\u1eddng cao chung t\u1eeb B v\u00e0 c\u00f3 chung \u0111\u00e1y EB n\u00ean \u0111\u01b0\u1eddng cao t\u1eeb C g\u1ea5p 2 l\u1ea7n \u0111\u01b0\u1eddng cao t\u1eeb A). N\u00ean: S_ABM=S_ACM (chung \u0111\u00e1y AM, 2 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau \u2013cmt-) (1) S_CMD=S_BMD (chung \u0111\u00e1y MD, 2 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau \u2013cmt-) (2) S_MBC=2S_MBA (chung \u0111\u00e1y MB, cao t\u1eeb C g\u1ea5p 2 l\u1ea7n cao t\u1eeb A) (3) T\u1eeb (1), (2) v\u00e0 (3) cho ta: S_ABM=S_ACM = S_CMD=S_BMD = 600 : 4 = 150 (cm2) M\u00e0: S_ABE=1\/3S_ABC= 600:3 = 200 (cm2) S_AME = S_ABE-SABM = 200-150= 50 (cm2) B\u00e0i 30: Cho tam gi\u00e1c ABC. \u0110i\u1ec3m M tr\u00ean AC sao cho AM = 1\/4 AC. \u0110i\u1ec3m N tr\u00ean BC sao cho di\u1ec7n t\u00edch tam gi\u00e1c MCN b\u1eb1ng di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AMNB. T\u00ednh t\u1ec9 s\u1ed1 gi\u1eefa BN v\u00e0 BC? Ch\u1ecdn \u0111i\u1ec3m N tr\u00ean BC v\u00e0 gi\u1ea3 s\u1eed S_MCN=S_AMNB. N\u1ed1i AN. Do AM=1\/4AC hay AM=1\/3MC Ta c\u00f3: S_MNC=3S_AMN (MC=3AM, chung \u0111\u01b0\u1eddng cao t\u1eeb N)","\u0110\u1ec3 S_AMNB=SMNC th\u00ec S_ANB=(3-1)S_AMN=2S_AMN Di\u1ec7n t\u00edch ABC c\u00f3 3+1+2=6 (ph\u1ea7n) th\u00ec S_ANB c\u00f3 2 ph\u1ea7n hay S_ANB=1\/3S_ABC. Suy ra: BN=1\/3BC B\u00e0i 31: T\u00ednh k\u00edch th\u01b0\u1edbc t\u1ea5m k\u00ednh. C\u00f3 hai t\u1ea5m k\u00ednh h\u00ecnh ch\u1eef nh\u1eadt. Chi\u1ec1u r\u1ed9ng c\u1ee7a m\u1ed7i t\u1ea5m k\u00ednh b\u1eb1ng 1\/2 chi\u1ec1u d\u00e0i c\u1ee7a n\u00f3 v\u00e0 chi\u1ec1u d\u00e0i c\u1ee7a t\u1ea5m k\u00ednh nh\u1ecf \u0111\u00fang b\u1eb1ng chi\u1ec1u r\u1ed9ng c\u1ee7a t\u1ea5m k\u00ednh to. Gh\u00e9p hai t\u1ea5m k\u00ednh s\u00e1t v\u00e0o nhau v\u00e0 \u0111\u1eb7t l\u00ean b\u00e0n c\u00f3 di\u1ec7n t\u00edch 90 dm2 th\u00ec v\u1eeba kh\u00edt. H\u00e3y t\u00ednh k\u00edch th\u01b0\u1edbc c\u1ee7a m\u1ed7i t\u1ea5m k\u00ednh \u0111\u00f3. D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p gh\u00e9p h\u00ecnh ta c\u00f3 : N\u1ebfu g\u1ecdi chi\u1ec1u r\u1ed9ng t\u1ea5m k\u00ednh nh\u1ecf l\u00e0 m\u1ed9t ph\u1ea7n th\u00ec chi\u1ec1u d\u00e0i t\u1ea5m k\u00ednh nh\u1ecf ( c\u0169ng l\u00e0 chi\u1ec1u r\u1ed9ng t\u1ea5m k\u00ednh l\u1edbn ) l\u00e0 hai ph\u1ea7n v\u00e0 chi\u1ec1u d\u00e0i t\u1ea5m k\u00ednh l\u1edbn l\u00e0 b\u1ed1n ph\u1ea7n b\u1eb1ng nhau. Gh\u00e9p 2 t\u1ea5m k\u00ednh l\u1ea1i ( nh\u01b0 \u0111\u1ec1 b\u00e0i ) ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i l\u00e0 5 ph\u1ea7n v\u00e0 chi\u1ec1u r\u1ed9ng l\u00e0 2 ph\u1ea7n. Ta chia h\u00ecnh ch\u1eef nh\u1eadt v\u1eeba gh\u00e9p n\u00e0y th\u00e0nh 10 h\u00ecnh vu\u00f4ng nh\u1ecf b\u1eb1ng nhau m\u1ed7i h\u00ecnh vu\u00f4ng nh\u1ecf c\u00f3 c\u1ea1nh l\u00e0 1 ph\u1ea7n . Di\u1ec7n t\u00edch 1 h\u00ecnh vu\u00f4ng nh\u1ecf l\u00e0 : 90 : 10 = 9 dm2 C\u1ea1nh m\u1ed7i h\u00ecnh vu\u00f4ng nh\u1ecf l\u00e0 3 dm2 ( 3 x 3 = 9 ) ; C\u0169ng l\u00e0 chi\u1ec1u r\u1ed9ng t\u1ea5m k\u00ednh nh\u1ecf. Chi\u1ec1u d\u00e0i t\u1ea5m k\u00ednh nh\u1ecf , hay chi\u1ec1u r\u1ed9ng t\u1ea5m k\u00ednh l\u1edbn : 3 x 2 = 6 dm Chi\u1ec1u d\u00e0i t\u1ea5m k\u00ednh l\u1edbn : 6 x 2 = 12 dm \u0110\u00e1p s\u1ed1 : T\u1ea5m k\u00ednh nh\u1ecf : 3dm v\u00e0 6 dm T\u1ea5m k\u00ednh l\u1edbn : 6dm v\u00e0 12 dm B\u00e0i 32: Qu\u00e3ng \u0111\u01b0\u1eddng AB d\u00e0i 96km. C\u00f9ng m\u1ed9t l\u00fac, xe \u00f4 t\u00f4 \u0111i t\u1eeb A v\u00e0 xe g\u1eafn m\u00e1y \u0111i t\u1eeb B, ch\u1ea1y ng\u01b0\u1ee3c chi\u1ec1u, g\u1eb7p nhau c\u00e1ch A l\u00e0 64km. N\u1ebfu xe g\u1eafn m\u00e1y \u0111i tr\u01b0\u1edbc 45 ph\u00fat th\u00ec hai xe g\u1eb7p nhau c\u00e1ch A 52km. T\u00ednh v\u1eadn t\u1ed1c m\u1ed7i xe. 45\u2019 = 3\/4 gi\u1edd N\u1ebfu kh\u1edfi h\u00e0nh c\u00f9ng l\u00fac g\u1eb7p nhau c\u00e1ch B: 96-64= 32 (km) T\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c c\u1ee7a xe \u00f4 t\u00f4 v\u00e0 xe m\u00e1y: 64\/32 = 2 (v\u1eadn t\u1ed1c xe \u00f4 t\u00f4 2 l\u1ea7n v\u1eadn t\u1ed1c xe m\u00e1y). Xe m\u00e1y \u0111i tr\u01b0\u1edbc 45\u2019 th\u00ec \u0111\u1ebfn C, 2 xe g\u1eb7p nhau \u1edf K. \u0110o\u1ea1n KC d\u00e0i: 52 : 2 = 26 (km) 45\u2019 xe m\u00e1y ch\u1ea1y \u0111\u01b0\u1ee3c: 96 \u2013 (52+26) = 18 (km) V\u1eadn t\u1ed1c xe m\u00e1y: 18 : 3 x 4 = 24 (km\/gi\u1edd) V\u1eadn t\u1ed1c \u00f4 t\u00f4: 24 x 2 = 48 (km\/gi\u1edd) \u0110\u00e1p s\u1ed1: 24 km\/gi\u1edd v\u00e0 48 km\/gi\u1edd B\u00e0i 33: M\u1ed9t t\u1edd gi\u1ea5y h\u00ecnh vu\u00f4ng c\u00f3 di\u1ec7n t\u00edch l\u00e0 72 cm2 th\u00ec \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a t\u1edd gi\u1ea5y \u0111\u00f3 d\u00e0i bao nhi\u00eau?","C\u1eaft v\u00e0 gh\u00e9p th\u00e0nh 2 h\u00ecnh vu\u00f4ng nh\u1ecf, m\u1ed7i h\u00ecnh c\u00f3 di\u1ec7n t\u00edch: 72 : 2 = 36 (cm2) V\u00ec 36 = 6 x 6 n\u00ean c\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf b\u1eb1ng 6cm. C\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf b\u1eb1ng \u00bd \u0111\u01b0\u1eddng ch\u00e9o h\u00ecnh vu\u00f4ng l\u1edbn. \u0110\u01b0\u1eddng ch\u00e9o h\u00ecnh vu\u00f4ng l\u1edbn l\u00e0: 6 x 2 = 12 (cm) \u0110\u00e1p s\u1ed1: 12 cm B\u00e0i 34: H\u00ecnh vu\u00f4ng ABCD v\u00e0 h\u00ecnh ch\u1eef nh\u1eadt MNPQ c\u00f3 chu vi b\u1eb1ng nhau. H\u00e3y so s\u00e1nh c\u1ea1nh h\u00ecnh vu\u00f4ng v\u00e0 c\u1ea1nh c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt. H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng v\u00e0 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt. Chu vi: Do chu vi 2 h\u00ecnh b\u1eb1ng nhau n\u00ean n\u1eeda chu vi 2 h\u00ecnh c\u0169ng b\u1eb1ng nhau. G\u1ecdi a l\u00e0 c\u1ea1nh h\u00ecnh vu\u00f4ng; b v\u00e0 c l\u00e0 c\u1ea1nh h\u00ecnh ch\u1eef nh\u1eadt. Ta c\u00f3 a+a = b+c => (a+a)\/2 = (b+c)\/2 Hay a = (b+c)\/2 a l\u00e0 trung b\u00ecnh c\u1ed9ng c\u1ee7a b v\u00e0 c. Di\u1ec7n t\u00edch: Gi\u1ea3 s\u1eed c\u1ea1nh h\u00ecnh vu\u00f4ng l\u00e0 10m th\u00ec c\u1ea1nh h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 th\u1ec3 l\u00e0: 11 v\u00e0 9; 12 v\u00e0 8; \u2026 Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u00e0: 10 x 10 = 100 (m2) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 th\u1ec3 l\u00e0: *. 11 x 9 = 99 (m2) *. 12 x 8 = 96 (m2) \u2026\u2026\u2026\u2026\u2026\u2026\u2026 C\u1ea1nh h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 \u0111\u1ed9 l\u1ec7ch v\u1edbi c\u1ea1nh h\u00ecnh vu\u00f4ng c\u00e0ng l\u1edbn th\u00ec di\u1ec7n t\u00edch c\u00e0ng gi\u1ea3m (gi\u1ea3m v\u1ec1 \u0111\u1ebfn 0 n\u1ebfu c\u1ea1nh h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 20 v\u00e0 0. Kh\u00f4ng c\u00f2n l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt) \u0110\u1ebfn \u0111\u00e2y xin n\u00f3i th\u00eam: N\u1ebfu chu vi c\u00e1c h\u00ecnh b\u1eb1ng nhau th\u00ec di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u1edbn nh\u1ea5t. Chu vi h\u00ecnh tr\u00f2n l\u00e0 40m th\u00ec b\u00e1n k\u00ednh l\u00e0: 40 : 3,14 : 2 = 6,369427 (m) Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n: 6,369 x 6,369 x 3,14 = 127,3714\u2026 (m2) B\u00e0i 35: M\u1ed9t mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt n\u1ebfu b\u1edbt chi\u1ec1u d\u00e0i 8m.Chi\u1ec1u r\u1ed9ng t\u0103ng 5m ta \u0111\u01b0\u1ee3c mi\u1ebfng \u0111\u1ea5t h\u00ecnh vu\u00f4ng Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng \u00edt h\u01a1n di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt 122m .T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u00fac \u0111\u1ea7u ?","Chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng: 8 + 5 = 13 (m) SOBCK = SMNOA + 122 => SNPCK = SMNOA + 122 + 8x5 = SMNOA + 162 M\u00e0 NPCK v\u00e0 MNOA c\u00f3 MN = NK (c\u1ea1nh h\u00ecnh vu\u00f4ng) v\u00e0 NP h\u01a1n NO l\u00e0 : 8 \u2013 5 = 3 (m) C\u1ea1nh h\u00ecnh vu\u00f4ng: 162 : 3 = 54 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt: 54 + 8 = 62 (m) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt: 54 \u2013 3 = 51 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt: 62 x 51 = 3162 (m2) \u0110\u00e1p s\u1ed1: 3162 m2. B\u00e0i 36: M\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i l\u00e0 8m. Ng\u01b0\u1eddi ta chia m\u1ea3nh \u0111\u1ea5t l\u00e0m 2 ph\u1ea7n, m\u1ed9t ph\u1ea7n \u0111\u1ec3 l\u00e0m v\u01b0\u1eddn, m\u1ed9t ph\u1ea7n \u0111\u1ec3 \u0111\u00e0o ao nu\u00f4i c\u00e1. Di\u1ec7n t\u00edch ph\u1ea7n \u0111\u1ea5t l\u00e0m v\u01b0\u1eddn b\u1eb1ng 1\/2 m\u1ea3nh \u0111\u1ea5t. Chu vi ph\u1ea7n \u0111\u1ea5t l\u00e0m v\u01b0\u1eddn b\u1eb1ng 2\/3 chu vi m\u1ea3nh \u0111\u1ea5t. T\u00ednh di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt. Chu vi v\u01b0\u1eddn b\u1eb1ng 2\/3 chu vi m\u1ea3nh \u0111\u1ea5t th\u00ec n\u1eeda chu vi v\u01b0\u1eddn c\u0169ng b\u1eb1ng 2\/3 n\u1eeda chu vi m\u1ea3nh \u0111\u1ea5t. G\u1ecdi chi\u1ec1u d\u00e0i mi\u1ebfng \u0111\u1ea5t l\u00e0 a = 8m, chi\u1ec1u r\u1ed9ng l\u00e0 b (b>0 v\u00e0 b<8). *.Tr\u01b0\u1eddng h\u1ee3p 1: N\u1ebfu m\u1ed7i c\u1ea1nh v\u01b0\u1eddn b\u1eb1ng 2\/3 c\u1ea1nh c\u1ee7a mi\u1ebfng \u0111\u1ea5t th\u00ec chu vi c\u0169ng b\u1eb1ng 2\/3 chu vi mi\u1ebfng \u0111\u1ea5t nh\u01b0ng di\u1ec7n t\u00edch s\u1ebd b\u1eb1ng 2\/3 x 2\/3 = 4\/9 di\u1ec7n t\u00edch mi\u1ebfng \u0111\u1ea5t. (lo\u1ea1i) *.Tr\u01b0\u1eddng h\u1ee3p 2: C\u1eaft 2 mi\u1ebfng \u0111\u1ea5t theo chi\u1ec1u d\u00e0i \u0111\u1ec3 c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng \u00bd di\u1ec7n t\u00edch mi\u1ebfng \u0111\u1ea5t th\u00ec: P\u0111\u1ea5t\/2 = 8 + b (P l\u00e0 chu vi) Pv\u01b0\u1eddn\/2 = 8 + b\/2 M\u00e0: 8 + b\/2 = 2\/3 (8 + b) = 16\/3 + 2\/3b => b\/6 = 8 \u2013 16\/3 = 8\/3 => b = 8\/3 x 6 = 16 (lo\u1ea1i) b=16 > 8 (8m l\u00e0 chi\u1ec1u d\u00e0i)","*.Tr\u01b0\u1eddng h\u1ee3p 3: C\u1eaft 2 mi\u1ebfng \u0111\u1ea5t theo chi\u1ec1u r\u1ed9ng th\u00ec: P\u0111\u1ea5t\/2 = 8 + b Pv\u01b0\u1eddn\/2 = 8\/2 + b M\u00e0: 8\/2 + b = 2\/3 (8 + b) = 16\/3 + 2\/3b => b\/3 = 16\/3 \u2013 4 = 4\/3 => b = 4\/3 x 3 = 4 Di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t: 8 x 4 = 32 (m2) \u0110\u00e1p s\u1ed1 : 32 (m2) B\u00e0i 37 Cho tam gi\u00e1c ABC. D l\u00e0 \u0111i\u1ec3m tr\u00ean c\u1ea1nh BC sao cho BD = 2\/3 DC. M v\u00e0 E l\u00e0 hai \u0111i\u1ec3m tr\u00ean \u0111o\u1ea1n th\u1eb3ng AD sao cho AM = ME = ED. a) Em h\u00e3y t\u00ecm tr\u00ean h\u00ecnh v\u1ebd nh\u1eefng tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau ? Gi\u1ea3i th\u00edch t\u1ea1i sao ? b) K\u00e9o d\u00e0i BE c\u1eaft \u1edf AC \u1edf N. Cho bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c BED = 4 cm2 .H\u00e3y t\u00ednh di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c DEC v\u00e0 ABC; r\u1ed3i so s\u00e1nh \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n th\u1eb3ng AN v\u00e0 CN. a)C\u00e1c tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau: BED, BME, BAM (c\u1ea1nh \u0111\u00e1y ED=ME=AM, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B) BAE, BMD (c\u1ea1nh \u0111\u00e1y AE=MC=2AM, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B). b)Hai tam gi\u00e1c EBD v\u00e0 DEC c\u00f3 BD=2\/3DC chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb E. N\u00ean SEBD = 2\/3 SECD => SDEC = 4 : 2 x 3 = 6 (cm2) *.Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 AD = ED x 3 (AM=ME=ED) 2 tam gi\u00e1c ABD v\u00e0 EBD c\u00f3: AD = ED x 3, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B. N\u00ean SABD = SEBD x 3 = 4 x 3 = 12 (cm2) M\u00e0 BD= 2\/3 DC hay BD = 2\/5 BC V\u1eady SABC = SABD : 2 x 5 = 12 : 2 x 5 = 30 (cm2) *.SAEC = SABC \u2013 SABD \u2013 SEDC = 30 \u2013 12 \u2013 6 = 12 (cm2) X\u00e9t 2 tam gi\u00e1c ABE (Dt=4+4=8 cm2) v\u00e0 CBE (Dt= 4+6=10cm2). C\u00f3: Chung \u0111\u00e1y BE n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B v\u00e0 t\u1eeb C xu\u1ed1ng BE c\u00f3 t\u1ec9 l\u1ec7 8\/10 (4\/5). Di\u1ec7n t\u00edch AEN = 12 : (4+5) x 4 = 16\/3 (cm2) Di\u1ec7n t\u00edch ACN = 12 : (4+5) x 5 = 20\/3 (cm2) 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb E n\u00ean 2 \u0111\u00e1y t\u1ec9 l\u1ec7 v\u1edbi 2 di\u1ec7n t\u00edch T\u1ec9 l\u1ec7 c\u1ee7a AN v\u00e0 NC l\u00e0 16\/3 : 20\/3 = 16\/20 = 4\/5","B\u00e0i 38 Cho tam gi\u00e1c ABC c\u00f3 M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC ;N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AC , K\u1ebb AM v\u00e0 BN c\u1eaft nhau t\u1ea1i O . Ch\u1ee9ng minh r\u1eb1ng OA = 2 x OM SABN = SCBN (c\u00f3 AN=NC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B) N\u1ebfu xemNB l\u00e0 c\u1ea1nh \u0111\u00e1y th\u00ec 2 \u0111\u01b0\u1eddng cao t\u1eeb A v\u00e0 C xu\u1ed1ng NB b\u1eb1ng nhau. Hai \u0111\u01b0\u1eddng cao n\u00e0y ch\u00ednh l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c AOB v\u00e0 COB c\u00f3 chung \u0111\u00e1y OB. Suy ra: SAOB = SCOB. M\u00e0 SOBM = SOMC = \u00bd SOBC = \u00bd SAOB (CM=MB, chung \u0111\u01b0\u1eddng cao t\u1eeb O). Suy ra: SAOB = SOBM x 2. 2 tam gi\u00e1c AOB v\u00e0 MOB c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B. N\u00ean 2 \u0111\u00e1y OA v\u00e0 OM t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch. OA = OM x 2 B\u00e0i 39 T\u0103ng \u0111\u1ed9 d\u00e0i c\u1ea1nh m\u1ed9t h\u00ecnh vu\u00f4ng th\u00eam 4cm th\u00ec di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng t\u0103ng th\u00eam 664cm2. T\u00ecm di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng \u0111\u00f3. Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng nh\u1ecf \u1edf g\u00f3c: 4 x 4 = 16 (cm2) Di\u1ec7n t\u00edch 1 h\u00ecnh ch\u1eef nh\u1eadt. (664 \u2013 16) : 2 = 324 (cm2) C\u1ea1nh h\u00ecnh vu\u00f4ng ban \u0111\u1ea7u: 324 : 4 = 81 (cm) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ban \u0111\u1ea7u: 81 x 81 = 6561 (cm2) \u0110ap s\u1ed1: 6561 cm2. B\u00e0i 40 Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 12 cm2, c\u1ea1nh \u0111\u00e1y BC = 6 cm. N l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh AC. T\u1eeb N k\u1ebb song song v\u1edbi BC c\u1eaft AB t\u1ea1i M. T\u00ednh:","a)\u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng MN. b)Di\u1ec7n t\u00edch h\u00ecnh thang NMBC. (T\u00ednh \u0111\u01b0\u1eddng cao t\u1eeb A c\u1ee7a tam gi\u00e1c ABC. N\u1ed1i NB, d\u1ef1a v\u00e0o AN=NC, t\u00ednh \u0111\u01b0\u1ee3c SABN=SCBN , t\u00ednh \u0111\u01b0\u1ee3c SCBN ; t\u00ednh \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb N xu\u1ed1ng CB. Suy ra \u0111\u01b0\u1eddng cao t\u1eeb A xu\u1ed1ng NM. SNBC=SMBC =1\/2 SABC. => SAMC=1\/2SABC (6cm2). SAMN=SCMN (6:2=3 (cm2)). T\u00ednh \u0111\u01b0\u1ee3c MN l\u00e0 c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c AMN. H\u00ecnh thang NMBC \u0111\u00e3 bi\u1ebft \u0111\u01b0\u1ee3c NM; CB v\u00e0 chi\u1ec1u cao n\u00ean t\u00ednh \u0111\u01b0\u1ee3c di\u1ec7n t\u00edch.) \u0110\u01b0\u1eddng cao k\u1ebb t\u1eeb A xu\u1ed1ng BC: 12 x 2 : 6 = 4 (cm) SABN = SNBC = SABC : 2 = 12 : 2 = 6 (cm2) (AN=NC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B) \u0110\u01b0\u1eddng cao k\u1ebb t\u1eeb N xu\u1ed1ng BC: 6 x 2 : 6 = 2 (cm) \u0110\u01b0\u1eddng cao k\u1ebb t\u1eeb A xu\u1ed1ng NM: 4 \u2013 2 = 2 (cm) Ta l\u1ea1i c\u00f3: SMBC = SNBC = 6 (cm2) (Ch\u00fang \u0111\u00e1y BC, b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang). =>SAMC = SMBC = 6 (cm2) (12 \u2013 6 = 6 (cm2)) =>SAMN = SNMC = 6 : 2 = 3 (cm2) C\u1ea1nh \u0111\u00e1y MN c\u1ee7a tam gi\u00e1c AMN: 3 x 2 : 2 = 3 (cm) (ho\u1eb7c 12 - 3 = 9 (cm2)) Di\u1ec7n t\u00edch h\u00ecnh thang NMBC: (6 + 3) x 4 : 2 = 9 (cm2) \u0110\u00e1p s\u1ed1: 9 cm2 B\u00e0i 41 Gi\u1ea3m chi\u1ec1u d\u00e0i 1 h\u00ecnh ch\u1eef nh\u1eadt 5m t\u0103ng chi\u1ec1u r\u1ed9ng l\u00ean 5m th\u00ec \u0111\u01b0\u1ee3c 1 h\u00ecnh vu\u00f4ng c\u00f3 di\u1ec7n t\u00edch l\u1edbn h\u01a1n di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt 25m2.T\u00ecm di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u? D\u00e0i h\u01a1n r\u1ed9ng: 5 + 5 = 10 (m) G\u1ecdi a d\u00e0i, b r\u1ed9ng => a = b+10 DT ban \u0111\u1ea7u S = a x b = (b+10) x b = b.b + 10b DT \u0111\u00e3 thay \u0111\u1ed5i:"]