{"id":18147,"date":"2024-05-14T08:54:40","date_gmt":"2024-05-14T08:54:40","guid":{"rendered":"https:\/\/soicau4022.minhngocxoso.com\/?p=18147"},"modified":"2024-05-14T08:54:40","modified_gmt":"2024-05-14T08:54:40","slug":"tong-hop-cac-cong-thuc-soi-cau-lo-de","status":"publish","type":"post","link":"https:\/\/xsmbtailoc.com\/tong-hop-cac-cong-thuc-soi-cau-lo-de\/","title":{"rendered":"t\u1ed5ng h\u1ee3p c\u00e1c c\u00f4ng th\u1ee9c soi c\u1ea7u l\u00f4 \u0111\u1ec1"},"content":{"rendered":"
<\/p>\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 ; alt6 )
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)
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.
# \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:
g\u1ecdi \u0110\u1ec1 ng\u00e0y 1 = MN
\u0110\u1ec1 ng\u00e0y 2 = CD
L\u00f4 ng\u00e0y 3 (d\u1ef1 \u0111o\u00e1n \u0111\u01b0\u1ee3c) = AB
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
vd : 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.<\/p>\n\n\n\n
1. T\u1ed5ng gi\u1ea3i \u0111\u1eb7c bi\u1ec7t = 20 ; 25
Ng\u00e0y 1 c\u00f3 \u0110\u1ec1 = MN -> L\u00f4 ng\u00e0y 3 = AB. \u0110\u00e1nh BA
2. T\u1ed5ng \u0111b= s\u1ed1 ch\u1eb5n ( 22, 24, 26,28)
ng\u00e0y 1 \u0111\u1ec1= MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3= AB
l\u1ea5y 10-B= I , gi\u1eef nguy\u00ean A. \u0110\u00e1nh AI
3. T\u1ed5ng = s\u1ed1 l\u1ebb v\u00e0 30 (21, 23, 29, 30)
ng\u00e0y 1 \u0111\u1ec3 = MN -> d\u1ef1 \u0111o\u00e1n L\u00f4 ng\u00e0y 3 = AB. \u0110\u00e1nh AB<\/p>\n\n\n\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)
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)
\u0110\u00e1nh BA<\/p>\n\n\n\n
\u0110\u00e1nh BA
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
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)
\u0110\u00e1nh BA
4. M\u1ed9t c\u00e1ch t\u00ecm kh\u00e1c c\u00f3 \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y :
N\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.
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)
\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
K\u1ebft qu\u1ea3 \u0111\u00ea!<\/p>\n\n\n\n
<\/p>\r\n\r\n
M\u1ed9t s\u1ed1 kh\u00e1i ni\u1ec7m c\u01a1 b\u1ea3n C\u00d4NG TH\u1ee8C \u0110\u00c1NH L\u00d4 \u0110\u1ec0 B\u00f3ng l\u00e0 g\u00ec (kh\u00e1i ni\u1ec7m n\u00e0y ch\u1eafc th\u1eb1ng n\u00e0o c\u0169ng bi\u1ebft)Phi\u1ebfu \u0111\u00e1nh l\u00f4 \u0111\u1ec1 c\u1ee7a m\u1ed9t sinh vi\u00ean1 c\u00f3 b\u00f3ng l\u00e0 62 c\u00f3 b\u00f3ng l\u00e0 73 c\u00f3 b\u00f3ng l\u00e0 84 c\u00f3 b\u00f3ng l\u00e0 95 c\u00f3 b\u00f3ng l\u00e0 0 2. V\u00f2ng … <\/p>\n","protected":false},"author":1,"featured_media":18148,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"yoast_head":"\n