{"id":17815,"date":"2024-04-14T19:59:07","date_gmt":"2024-04-14T19:59:07","guid":{"rendered":"https:\/\/soicau5023.minhngocxoso.com\/?p=17815"},"modified":"2024-04-14T19:59:07","modified_gmt":"2024-04-14T19:59:07","slug":"cach-choi-co-ban-ma-lai-de-trung-cua-lo-de","status":"publish","type":"post","link":"https:\/\/soicau168.com\/cach-choi-co-ban-ma-lai-de-trung-cua-lo-de\/","title":{"rendered":"c\u00e1ch ch\u01a1i c\u01a1 b\u1ea3n m\u00e0 l\u1ea1i d\u1ec5 tr\u00fang c\u1ee7a l\u00f4 \u0111\u1ec1"},"content":{"rendered":"\n
Xin ch\u00e0o c\u00e1c b\u1ea1n, h\u00f4m nay ch\u00fang t\u00f4i s\u1ebd n\u00f3i v\u1ec1 m\u1ed9t s\u1ed1 kh\u00e1i ni\u1ec7m c\u01a1 b\u1ea3n v\u1ec1 b\u00f3ng v\u00e0 th\u00eam m\u1ed9t s\u1ed1 tr\u01b0\u1eddng h\u1ee3p kh\u00e1c hay c\u00e1ch t\u00ednh c\u1ea7u \u2013 l\u00f4. C\u00e1c b\u1ea1n ph\u1ea3i t\u1eadp trung hi\u1ec3u ph\u1ea7n b\u00f3ng, s\u1ebd h\u01a1i kh\u00f3 hi\u1ec3u nh\u00e9. Mong r\u1eb1ng v\u1edbi b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y c\u1ee7a ch\u00fang t\u00f4i c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng \u0111\u1ec3 b\u1eaft \u0111\u01b0\u1ee3c nh\u1eefng con b\u1ea1ch th\u1ee7 l\u00f4, b\u1ea1ch th\u1ee7 \u0111\u1ec1 hay c\u00e1c c\u1ea7u l\u00f4 \u0111\u1ec1 kh\u00e1c ch\u00ednh x\u00e1c nh\u1ea5t nh\u00e9.<\/p>\n\n\n\n
1. B\u00f3ng l\u00e0 g\u00ec ?<\/p>\n\n\n\n
1 c\u00f3 b\u00f3ng l\u00e0 6<\/p>\n\n\n\n
2 c\u00f3 b\u00f3ng l\u00e0 7<\/p>\n\n\n\n
3 c\u00f3 b\u00f3ng l\u00e0 8<\/p>\n\n\n\n
4 c\u00f3 b\u00f3ng l\u00e0 9<\/p>\n\n\n\n
5 c\u00f3 b\u00f3ng l\u00e0 0<\/p>\n\n\n\n
T\u01b0\u01a1ng sinh:Kim = 2 M\u1ed9c = 5 H\u1ecfa = 3 Th\u1ee7y =1 Th\u1ed5 = 4<\/p>\n\n\n\n
Nh\u01b0 v\u1eady, chi\u1ebfu theo b\u00f3ng ta c\u00f3 :<\/p>\n\n\n\n
Kim=7 M\u1ed9c=0 H\u1ecfa=8 Th\u1ee7y=6 Th\u1ed5=9<\/p>\n\n\n\n
Kim -> M\u1ed9c -> H\u1ecfa -> Th\u1ee7y -> Th\u1ed5 -> Kim<\/p>\n\n\n\n
T\u1ee9c l\u00e0 : 2 -> 5 -> 3 -> 1 -> 4 -> 2 v\u00e0 7 -> 0 -> 8 -> 6 -> 9 -> 7<\/p>\n\n\n\n
3. B\u1ea1n \u0111\u1ec3 \u00fd 2 s\u1ed1 cu\u1ed1i c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t<\/p>\n\n\n\n
VD: gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 09846 -> 2 s\u1ed1 cu\u1ed1i l\u00e0 46<\/p>\n\n\n\n
4. T\u1ed5ng c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t :<\/p>\n\n\n\n
VD: gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 12536 -> t\u1ed5ng = 1+2+5+3+6 = 17<\/p>\n\n\n\n * Ta x\u00e9t k\u1ebft qu\u1ea3 trong 2 h\u00f4m \u0111\u1ec3 \u0111\u00e1nh h\u00f4m th\u1ee9 3.<\/p>\n\n\n\n Gi\u1ea3 s\u1eed ng\u00e0y \u0110\u1ea7u (ng\u00e0y th\u1ee9 nh\u1ea5t) c\u00f3 2 s\u1ed1 cu\u1ed1i l\u00e0 58 -> ta suy ra ng\u00e0y th\u1ee9 3 s\u1ebd c\u00f3 36 (v\u00ec 5->3 ; 8 -> 6 )<\/p>\n\n\n\n VD c\u1ee5 th\u1ec3 h\u01a1n: h\u00f4m nay \u0110\u1ec1 v\u1ec1 32 -> ng\u00e0y kia s\u1ebd c\u00f3 l\u00f4 15 (v\u00ec 3->1 ; 2->5)<\/p>\n\n\n\n Nh\u01b0ng c\u00e1ch t\u00ednh n\u00e0y ch\u1ec9 t\u01b0\u01a1ng \u0111\u1ed1i ch\u00ednh x\u00e1c v\u1edbi c\u00e1c gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ng\u00e0y th\u1ee9 Nh\u1ea5t c\u00f3 t\u1ed5ng t\u1eeb 20 -> 30.<\/p>\n\n\n\n * M\u1ed9t s\u1ed1 quy \u0111\u1ecbnh nh\u01b0 sau:<\/strong><\/p>\n\n\n\n \u2013 G\u1ecdi \u0110\u1ec1 ng\u00e0y 1 = MN<\/p>\n\n\n\n \u2013 \u0110\u1ec1 ng\u00e0y 2 = CD<\/p>\n\n\n\n \u2013 L\u00f4 ng\u00e0y 3 (d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c) = AB<\/p>\n\n\n\n Trong c\u00f9ng m\u1ed9t s\u1ed1 th\u00ec A, B l\u00e0 kh\u00e1c nhau. Nh\u01b0ng \u1edf 2 ng\u00e0y kh\u00e1c nhau th\u00ec c\u00f3 th\u1ec3 gi\u1ed1ng nhau !<\/p>\n\n\n\n VD : MN = 35 (3#5) CD = 39 (3#9, C=M) , xin nh\u1edb l\u00e0 \u1edf \u0111\u00e2y l\u00e0 CD ch\u01b0a c\u00f3 quan h\u1ec7 g\u00ec v\u1edbi nhau.<\/p>\n\n\n\n Sau \u0111\u00e2y ta s\u1ebd \u0111i v\u00e0o t\u1eebng ng\u00e0y c\u1ee5 th\u1ec3 :<\/strong><\/p>\n\n\n\n 1. T\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t = 20 ; 25<\/p>\n\n\n\n Ng\u00e0y 1 c\u00f3 \u0110\u1ec1 = MN -> L\u00f4 ng\u00e0y 3 = AB => \u0110\u00e1nh BA<\/p>\n\n\n\n 2. T\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t = S\u1ed1 ch\u1eb5n (22, 24, 26, 28)<\/p>\n\n\n\n Ng\u00e0y 1 \u0110\u1ec1 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3 = AB<\/p>\n\n\n\n L\u1ea5y 10 \u2013 B = I , gi\u1eef nguy\u00ean A => \u0110\u00e1nh AI<\/p>\n\n\n\n 3. T\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t = 30 v\u00e0 s\u1ed1 l\u1ebb (21, 23, 29, 30)<\/p>\n\n\n\n Ng\u00e0y 1 \u0110\u1ec1 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3 = AB => \u0110\u00e1nh AB<\/p>\n\n\n\n III. M\u1ed9t s\u1ed1 tr\u01b0\u1eddng h\u1ee3p kh\u00f4ng th\u1ecfa m\u00e3n (t\u1ee9c l\u00e0 kh\u00f4ng n\u00ean \u0111\u00e1nh khi c\u00f3 c\u00e1c \u0111i\u1ec1u ki\u1ec7n sau x\u1ea3y ra)<\/strong><\/p>\n\n\n\n 1. N\u1ebfu d\u1ef1 \u0111o\u00e1n \u0111c L\u00f4 ng\u00e0y 3 l\u00e0 AB. \u0110\u1ec1 ng\u00e0y 2 v\u1ec1 CD -> AB, BA ( ho\u1eb7c b\u00f3ng c\u1ee7a AB, BA)<\/p>\n\n\n\n <=> \u0110\u1ec1 ng\u00e0y 1 v\u00e0 2 l\u00e0 b\u00f3ng c\u1ee7a nhau. Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n VD: \u0110\u1ec1 ng\u00e0y 1 l\u00e0 37 -> L\u00f4 ng\u00e0y 3 l\u00e0 10. Nh\u01b0ng \u0110\u1ec1 ng\u00e0y 2 v\u1ec1 28 -> 56 (b\u00f3ng l\u00e0 01) => Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n 2. N\u1ebfu d\u1ef1 \u0111o\u00e1n \u0111c L\u00f4 ng\u00e0y 3 l\u00e0 AB. T\u1ed5ng \u0110\u1eb7c bi\u1ec7t ng\u00e0y 2 = AB hay BA ( ho\u1eb7c b\u00f3ng AB, BA) => Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n 3. N\u1ebfu d\u1ef1 \u0111o\u00e1n \u0111c L\u00f4 ng\u00e0y 3 l\u00e0 AB. Nh\u01b0ng \u0110\u1ec1 ng\u00e0y 2 = AB hay BA (ho\u1eb7c b\u00f3ng AB, BA) => Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n 4. T\u1ed5ng \u0110\u1eb7c bi\u1ec7t = 27 => Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n 5. T\u1eeb k\u1ebft qu\u1ea3 ng\u00e0y 1 -> L\u00f4 ng\u00e0y 3 l\u00e0 AB. T\u1eeb k\u1ebft qu\u1ea3 ng\u00e0y 2 -> L\u00f4 ng\u00e0y 4 l\u00e0 AB, BA (ho\u1eb7c b\u00f3ng AB, BA) => Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n V\u1edbi nh\u1eefng b\u00ed k\u00edp m\u00e0 ch\u00fang t\u00f4i \u0111\u00e3 n\u00eau ra \u1edf tr\u00ean, hi v\u1ecdng s\u1ebd gi\u00fap c\u00e1c b\u1ea1n c\u00f3 th\u00eam cho m\u00ecnh nh\u1eefng kinh nghi\u1ec7m ch\u01a1i l\u00f4 \u0111\u1ec1 hi\u1ec7u qu\u1ea3 nh\u1ea5t nh\u00e9. Ch\u00fac c\u00e1c b\u1ea1n may m\u1eafn!<\/p>\n\r\nII. C\u00e1ch t\u00ednh:<\/strong><\/h3>\n\n\n\n
SOI C\u1ea6U CAO C\u1ea4P<\/h3>\r\n