{"id":17397,"date":"2024-05-08T22:17:25","date_gmt":"2024-05-08T22:17:25","guid":{"rendered":"http:\/\/soicau5018.minhngocxoso.com\/cong-thuc-tinh-de-co-ban\/"},"modified":"2024-05-08T22:17:25","modified_gmt":"2024-05-08T22:17:25","slug":"cong-thuc-tinh-de-co-ban","status":"publish","type":"post","link":"https:\/\/soicaumienbac8.com\/cong-thuc-tinh-de-co-ban\/","title":{"rendered":"c\u00f4ng th\u1ee9c t\u00ednh \u0111\u1ec1 c\u01a1 b\u1ea3n"},"content":{"rendered":"

C\u00d4NG TH\u1ee8C T\u00cdNH \u0110\u1ec0 C\u01a0 B\u1ea2N<\/span><\/p>\n

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C\u00d4NG TH\u1ee8C T\u00cdNH L\u00d4 \u0110\u1ec0 C\u01a0 B\u1ea2N (SOI C\u1ea6U L\u00d4 \u0110\u1ec0)<\/h4>\n

I. M\u1ed9t s\u1ed1 kh\u00e1i ni\u1ec7m c\u01a1 b\u1ea3n<\/h3>\n

1. B\u00f3ng l\u00e0 g\u00ec (kh\u00e1i ni\u1ec7m n\u00e0y ch\u1eafc th\u1eb1ng n\u00e0o c\u0169ng bi\u1ebft)
\nPhi\u1ebfu \u0111\u00e1nh l\u00f4 \u0111\u1ec1 c\u1ee7a m\u1ed9t sinh vi\u00ean
\n1 c\u00f3 b\u00f3ng l\u00e0 6
\n2 c\u00f3 b\u00f3ng l\u00e0 7
\n3 c\u00f3 b\u00f3ng l\u00e0 8
\n4 c\u00f3 b\u00f3ng l\u00e0 9
\n5 c\u00f3 b\u00f3ng l\u00e0 0
\n2. 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):
\n\u1edf \u0111\u00e2y ch\u1ec9 c\u1ea7n quan t\u00e2m \u0111\u1ebfn s\u1ef1 T\u01b0\u01a1ng sinh:
\nKim = 2 M\u1ed9c = 5 H\u1ecfa = 3 Th\u1ee7y =1 Th\u1ed5 = 4
\nNh\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
\nKim -> M\u1ed9c -> H\u1ecfa -> Th\u1ee7y -> Th\u1ed5 -> Kim
\nt\u1ee9c l\u00e0 : 2->5->3->1->4->2 v\u00e0 7->0->alt6->9->7
\n3. 2 s\u1ed1 cu\u1ed1i c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t:
\nVD: gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 09846 -> 2 s\u1ed1 cu\u1ed1i l\u00e0 46
\n4. T\u1ed5ng c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t :
\nVn: gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 12536 -> t\u1ed5ng = 1+2+5+3+6 = 17<\/p>\n

II. C\u00e1ch t\u00ednh<\/h3>\n

Ta x\u00e9t k\u1ebft qu\u1ea3 trong 2 h\u00f4m \u0111\u1ec3 \u0111\u00e1nh h\u00f4m th\u1ee9 3.
\nGi\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 )
\nVD 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)
\nTuy 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.
\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:
\ng\u1ecdi \u0110\u1ec1 ng\u00e0y 1 = MN
\n\u0110\u1ec1 ng\u00e0y 2 = CD
\nL\u00f4 ng\u00e0y 3 (d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c) = AB
\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
\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.
\nSau \u0111\u00e2y ta s\u1ebd \u0111i v\u00e0o t\u1eebng ng\u00e0y c\u1ee5 th\u1ec3 :
\n1. T\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t = 20 ; 25
\nNg\u00e0y 1 c\u00f3 \u0110\u1ec1 = MN -> L\u00f4 ng\u00e0y 3 = AB. \u0110\u00e1nh BA
\n2. T\u1ed5ng \u0111b= s\u1ed1 ch\u1eb5n ( 22, 24, 26,28)
\nng\u00e0y 1 \u0111\u1ec1= MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3= AB
\nl\u1ea5y 10-B= I , gi\u1eef nguy\u00ean A. \u0110\u00e1nh AI
\n3. T\u1ed5ng = s\u1ed1 l\u1ebb v\u00e0 30 (21, 23, 29, 30)
\nng\u00e0y 1 \u0111\u1ec3 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3 = AB. \u0110\u00e1nh AB<\/p>\n

M\u1ed9t s\u1ed1 tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t :<\/h4>\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)
\nt\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)
\n\u0110\u00e1nh BA
\n2. \u0110\u1ec1 ng\u00e0y 1 -> l\u00f4 ng\u00e0y 3 l\u00e0 AB. T\u1ed5ng ng\u00e0y 2 = xy -> AB
\n\u0110\u00e1nh BA
\nvd: \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
\n3. \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)
\n\u0110\u00e1nh BA
\n4. M\u1ed9t c\u00e1ch t\u00ecm kh\u00e1c c\u00f3 \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y :
\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.
\nVD: 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)
\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
\nK\u1ebft qu\u1ea3 \u0111\u00ea!<\/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)<\/h3>\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)
\n\u0110\u1ec1 ng\u00e0y 1 v\u00e0 2 l\u00e0 b\u00f3ng c\u1ee7a nhau. Kh\u00f4ng n\u00ean \u0111\u00e1nh.
\nVD: \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 \u0110\u00e1nh
\n2. 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.
\n3. 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.
\n4. T\u1ed5ng \u0110\u1eb7c bi\u1ec7t = 27 => Kh\u00f4ng n\u00ean \u0111\u00e1nh.
\n5. 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<\/div>\n<\/div>\n<\/article>\n

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