{"id":18697,"date":"2024-04-08T02:01:02","date_gmt":"2024-04-08T02:01:02","guid":{"rendered":"https:\/\/soicau1087.minhngocxoso.com\/?p=18697"},"modified":"2024-04-08T02:01:02","modified_gmt":"2024-04-08T02:01:02","slug":"cac-cong-thuc-danh-de-co-ban-de-trung-nhat","status":"publish","type":"post","link":"https:\/\/soicau888247.top\/cac-cong-thuc-danh-de-co-ban-de-trung-nhat\/","title":{"rendered":"C\u00e1c c\u00f4ng th\u1ee9c \u0111\u00e1nh \u0111\u1ec1 c\u01a1 b\u1ea3n d\u1ec5 tr\u00fang nh\u1ea5t?"},"content":{"rendered":"
1. B\u00f3ng l\u00e0 g\u00ec (kh\u00e1i ni\u1ec7m n\u00e0y ch\u1eafc th\u1eb1ng n\u00e0o c\u0169ng bi\u1ebft)<\/span><\/p>\n Phi\u1ebfu \u0111\u00e1nh l\u00f4 \u0111\u1ec1 c\u1ee7a m\u1ed9t sinh vi\u00ean<\/span><\/p>\n 1 c\u00f3 b\u00f3ng l\u00e0 6<\/span><\/p>\n 2 c\u00f3 b\u00f3ng l\u00e0 7<\/span><\/p>\n 3 c\u00f3 b\u00f3ng l\u00e0 8<\/span><\/p>\n 4<\/span> c\u00f3 b\u00f3ng l\u00e0 9<\/span><\/p>\n 5 c\u00f3 b\u00f3ng l\u00e0 0<\/span><\/p>\n 2. V\u00f2ng l\u1eb7p \u00e2m d\u01b0\u01a1ng ng\u0169 h\u00e0nh (c\u00e1i n\u00e0y c\u00f3 th\u1eadt, kh\u00f4ng ph\u1ea3i made in T\u1ef1 tui):<\/span>\u1edf \u0111\u00e2y ch\u1ec9 c\u1ea7n quan t\u00e2m \u0111\u1ebfn s\u1ef1 T\u01b0\u01a1ng sinh:<\/span><\/p>\n Kim = 2 M\u1ed9c = 5 H\u1ecfa = 3 Th\u1ee7y =1 Th\u1ed5 = 4<\/span><\/p>\n Nh\u01b0 v\u1eady, chi\u1ebfu theo b\u00f3ng ta c\u00f3 : Kim=7 M\u1ed9c=0 H\u1ecfa=8 Th\u1ee7y=6 Th\u1ed5=9<\/span><\/p>\n Kim -> M\u1ed9c -> H\u1ecfa -> Th\u1ee7y -> Th\u1ed5 -> Kim<\/span> 3. 2 s\u1ed1 cu\u1ed1i c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t:<\/span><\/p>\n VD: gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 09846 -> 2 s\u1ed1 cu\u1ed1i l\u00e0 46<\/span><\/p>\n 4. T\u1ed5ng c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t :<\/span><\/p>\n Vn: gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 12536 -> t\u1ed5ng = 1+2+5+3+6 = 17<\/span><\/p>\n Ta x\u00e9t k\u1ebft qu\u1ea3 trong 2 h\u00f4m \u0111\u1ec3 \u0111\u00e1nh h\u00f4m th\u1ee9 3.<\/span><\/p>\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 ; alt6 )<\/span><\/p>\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)<\/span><\/p>\n Tuy nhi\u00ean, 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.<\/span><\/p>\n # \u0110\u1ec3 ti\u1ec7n cho vi\u1ec7c tr\u00ecnh b\u00e0y, t\u00f4i s\u1ebd \u0111\u01b0a ra m\u1ed9t s\u1ed1 quy \u0111\u1ecbnh: <\/span>g\u1ecdi \u0110\u1ec1 ng\u00e0y 1 = MN<\/span><\/p>\n \u0110\u1ec1 ng\u00e0y 2 = CD<\/span><\/p>\n L\u00f4 ng\u00e0y 3 (d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c) = AB<\/span> Sau \u0111\u00e2y ta s\u1ebd \u0111i v\u00e0o t\u1eebng ng\u00e0y c\u1ee5 th\u1ec3 :<\/span><\/p>\n 1. T\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t = 20 ; 25<\/span><\/p>\n Ng\u00e0y 1 c\u00f3 \u0110\u1ec1 = MN -> L\u00f4 ng\u00e0y 3 = AB. \u0110\u00e1nh BA<\/span><\/p>\n 2. T\u1ed5ng \u0111b= s\u1ed1 ch\u1eb5n ( 22, 24, 26,28)<\/span><\/p>\n ng\u00e0y 1 \u0111\u1ec1= MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3= AB<\/span><\/p>\n l\u1ea5y 10-B= I , gi\u1eef nguy\u00ean A. \u0110\u00e1nh AI<\/span><\/p>\n 3. T\u1ed5ng = s\u1ed1 l\u1ebb v\u00e0 30 (21, 23, 29, 30)<\/span><\/p>\n ng\u00e0y 1 \u0111\u1ec3 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3 = AB. \u0110\u00e1nh AB<\/span>M\u1ed9t s\u1ed1 tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t :<\/span><\/p>\n 1. \u0110\u1ec1 ng\u00e0y 1 = MN -> l\u00f4 ng\u00e0y 3 l\u00e0 AB. \u0110\u1ec1 ng\u00e0y 2 l\u00e0 CD. M\u00e0 MN -> CD ( MN suy ra CD t\u01b0\u01a1ng t\u1ef1 nh\u01b0 MN-> AB)<\/span><\/p>\n t\u1ee9c l\u00e0 \u1edf \u0111\u00e2y AB=CD ( k\u1ebft qu\u1ea3 ng\u00e0y 2 = v\u1edbi l\u00f4 d\u1ef1 t\u00ednh ng\u00e0y 3)<\/span><\/p>\n \u0110\u00e1nh BA<\/span><\/p>\n 2. \u0110\u1ec1 ng\u00e0y 1 -> l\u00f4 ng\u00e0y 3 l\u00e0 AB. T\u1ed5ng ng\u00e0y 2 = xy -> AB<\/span><\/p>\n \u0110\u00e1nh BA<\/span><\/p>\n vd: \u0111\u1ec1 ng\u00e0y 1 = 35 -> l\u00f4 ng\u00e0y 3 = 13. Nh\u01b0ng t\u1ed5ng ng\u00e0y 2 c\u0169ng = 35 -> 13 => ng\u00e0y 3 \u0111\u00e1nh l\u00f4 31<\/span><\/p>\n 3. \u0110\u1ec1 ng\u00e0y 1 = MN -> l\u00f4 d\u1ef1 \u0111o\u00e1n ng\u00e0y 3 = AB. \u0110\u1ec1 ng\u00e0y 2= CD. M\u00e0 C= M+x, D=N+x ( x l\u00e0 s\u1ed1 \u00e2m ho\u1eb7c d\u01b0\u01a1ng)<\/span><\/p>\n \u0110\u00e1nh BA<\/span><\/p>\n 4. M\u1ed9t c\u00e1ch t\u00ecm kh\u00e1c c\u00f3 \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y :<\/span> VD: ng\u00e0y 1 c\u00f3 l\u00f4 35 l\u00e0 2 nh\u00e1y, l\u00f4 30 c\u00f3 m\u1ed9t ph\u00e1t -> l\u00f4 ng\u00e0y th\u1ee9 3 = 18 ( v\u00ec 3->1 ; 0->8)<\/span><\/p>\n \u0110\u1eb7c bi\u1ec7t : n\u1ebfu ng\u00e0y 1 c\u00f3 2 con th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n nh\u01b0 v\u1eady -> l\u00f4 ng\u00e0y 3 d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c 2 con. X\u1ebfp con th\u1ea5p trong 2 con \u0111\u00f3 l\u00ean tr\u01b0\u1edbc ( vd HI , TU v\u00ec HI \u0110\u00e1nh HT<\/span><\/p>\n K\u1ebft qu\u1ea3 \u0111\u00ea!<\/span><\/p>\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)<\/span><\/p>\n \u0110\u1ec1 ng\u00e0y 1 v\u00e0 2 l\u00e0 b\u00f3ng c\u1ee7a nhau. Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/span><\/p>\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) => <\/span><\/p>\n Kh\u00f4ng \u0110\u00e1nh<\/span><\/p>\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.<\/span><\/p>\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.<\/span><\/p>\n 4. T\u1ed5ng \u0110\u1eb7c bi\u1ec7t = 27 => Kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/span><\/p>\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.<\/span><\/p>\n
\nt\u1ee9c l\u00e0 : 2->5->3->1->4->2 v\u00e0 7->0->alt6->9->7<\/span><\/p>\nII. C\u00e1ch t\u00ednh<\/span><\/h3>\n
<\/p>\n
\ntrong 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<\/span>
\nvd : MN = 35 ( 3#5) CD = 39 ( 3#9, C=M) , xin nh\u1edb \u1edf \u0111\u00e2y l\u00e0 CD ch\u01b0a c\u00f3 quan h\u1ec7 g\u00ec v\u1edbi nhau.<\/span><\/p>\n
\nN\u1ebfu k\u1ebft qu\u1ea3 ng\u00e0y 1 c\u00f3 CD 2 nh\u00e1y, CE 1 nh\u00e1y, v\u1edbi E l\u00e0 b\u00f3ng c\u1ee7a D => l\u00f4 ng\u00e0y th\u1ee9 3 c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c suy ra t\u1eeb CE.<\/span><\/p>\n<\/p>\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)<\/span><\/h3>\n
c\u1ea7u vip \u0103n ngay<\/span><\/h3>\n