{"id":18158,"date":"2024-03-07T09:33:49","date_gmt":"2024-03-07T09:33:49","guid":{"rendered":"https:\/\/soicau4013.minhngocxoso.com\/?p=18158"},"modified":"2024-03-07T09:33:49","modified_gmt":"2024-03-07T09:33:49","slug":"soi-cau-xsmb-theo-phuong-phap-tong-hieu-giai-dac-biet","status":"publish","type":"post","link":"https:\/\/socauxsmbmienphi.com\/soi-cau-xsmb-theo-phuong-phap-tong-hieu-giai-dac-biet\/","title":{"rendered":"soi c\u1ea7u xsmb theo ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t"},"content":{"rendered":"\n
Soi c\u1ea7u XSMB theo ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 m\u1ed9t trong nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 c\u00f3 \u0111\u1ed9 ch\u00ednh x\u00e1c cao. Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 n\u00e0y th\u01b0\u1eddng \u0111\u01b0\u1ee3c m\u00ecnh \u00e1p d\u1ee5ng \u0111\u1ec3 c\u1eadp nh\u1eadt trong c\u00e1c b\u00e0i d\u1ef1 \u0111o\u00e1n XSMB h\u00e0ng ng\u00e0y \u0111\u1ec3 anh em tham kh\u1ea3o. Tuy n\u00f3 kh\u00f4ng d\u1ef1a v\u00e0o th\u1ed1ng k\u00ea x\u00e1c su\u1ea5t khoa h\u1ecdc nh\u01b0ng l\u1ea1i c\u00f3 \u0111\u1ed9 ch\u00ednh x\u00e1c cao. C\u00e1ch soi c\u1ea7u n\u00e0y d\u1ef1a v\u00e0o t\u1ed5ng hi\u1ec7u gi\u1eefa c\u00e1c v\u1ecb tr\u00ed c\u1ee7a c\u00e1c s\u1ed1 trong gi\u1ea3i \u0111\u1eb7c bi\u1ec7t. \u1ede b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y m\u00ecnh s\u1ebd h\u01b0\u1edbng d\u1eabn c\u00e1c b\u1ea1n c\u00e1ch soi c\u1ea7u theo ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u n\u00e0y.<\/p>\n\n\n\n
\u0110\u1ec3 soi c\u1ea7u l\u00f4 MB mi\u1ec5n ph\u00ed theo ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ch\u00fang ta l\u00e0m theo c\u00e1c b\u01b0\u1edbc d\u01b0\u1edbi \u0111\u00e2y.<\/p>\n\n\n\n
B\u01b0\u1edbc 1: Ch\u00fang ta x\u00e9t gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u1ec1 trong ng\u00e0y h\u00f4m nay: ABCDE.<\/p>\n\n\n\n
L\u1ea5y v\u00ed d\u1ee5 ng\u00e0y 2\/10\/2024 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0: 23881.<\/p>\n\n\n\n
B\u01b0\u1edbc 2: Ch\u00fang ta l\u1ea5y hi\u1ec7u A \u2013 C = X v\u00e0 E \u2013 C = Y. Ta s\u1ebd \u0111\u01b0\u1ee3c l\u00f4 XY.<\/p>\n\n\n\n
V\u00ed d\u1ee5:<\/em> \u1ede ng\u00e0y 2\/10\/2024 ch\u00fang ta c\u00f3 A = 2, C = 8, E = 1<\/p>\n\n\n\n L\u01b0u \u00fd:<\/em> N\u1ebfu A l\u1edbn h\u01a1n C th\u00ec X = A \u2013 C. C\u00f2n n\u1ebfu A b\u00e9 h\u01a1n C th\u00ec X = C \u2013 A. T\u00ecm Y c\u0169ng t\u01b0\u01a1ng t\u1ef1.<\/p>\n\n\n\n B\u01b0\u1edbc 3: Ch\u00fang ta \u0111\u00e1nh c\u1eb7p XY \u2013 YX khung 3 ng\u00e0y v\u1edbi t\u1ef7 l\u1ec7 1:3:10 (ng\u00e0y 1 \u0111\u00e1nh 10 \u0111i\u1ec3m m\u1ed9t con kh\u00f4ng tr\u00fang th\u00ec ng\u00e0y 2 \u0111\u00e1nh 30 \u0111i\u1ec3m m\u1ed9t con, ng\u00e0y cu\u1ed1i 3 \u0111\u00e1nh 100 \u0111i\u1ec3m 1 con). C\u00e1c b\u1ea1n \u0111\u00e1nh c\u1eb7p s\u1ed1 soi \u0111\u01b0\u1ee3c trong 3 ng\u00e0y khi n\u00e0o tr\u00fang th\u00ed b\u1ecf, n\u1ebfu kh\u00f4ng v\u1ec1 c\u0169ng b\u1ecf.<\/p>\n\n\n\n V\u00ed d\u1ee5: Ng\u00e0y 2\/10\/2024 ta t\u00ednh \u0111\u01b0\u1ee3c XY = 67, th\u00ec ng\u00e0y 3\/10\/2024 ch\u00fang ta \u0111\u00e1nh c\u1eb7p 676. K\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1 MB 3\/10\/2024 \u0111\u00e3 v\u1ec1 l\u00f4 67.<\/p>\n\n\n\n \u0110\u1ec3 c\u00f3 \u0111\u01b0\u1ee3c \u0111\u1ed9 ch\u00ednh x\u00e1c cao khi \u00e1p d\u1ee5ng soi c\u1ea7u XSMB theo ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t, c\u00e1c b\u1ea1n c\u1ea7n l\u01b0u \u00fd m\u1ed9t s\u1ed1 tr\u01b0\u1eddng h\u1ee3p sau:<\/p>\n\n\n\n N\u1ebfu A l\u00e0 b\u00f3ng d\u01b0\u01a1ng c\u1ee7a C ho\u1eb7c E l\u00e0 b\u00f3ng d\u01b0\u01a1ng c\u1ee7a C th\u00ec khi t\u00ecm X v\u00e0 Y ch\u00fang ta kh\u00f4ng d\u00f9ng ph\u00e9p tr\u1eeb m\u00e0 ta s\u1ebd d\u00f9ng ph\u00e9p c\u1ed9ng.<\/p>\n\n\n\n L\u1ea5y v\u00ed d\u1ee5:<\/em> K\u1ebft qu\u1ea3 XSMB ng\u00e0y 3\/10\/2024 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u1ec1 17227.<\/p>\n\n\n\n Ta c\u00f3 A = 1, C = 2 v\u00e0 E =7. (2 v\u00e0 7 l\u00e0 b\u00f3ng c\u1ee7a nhau)<\/p>\n\n\n\n Theo nh\u01b0 tr\u00ean ta t\u00ecm \u0111\u01b0\u1ee3c c\u1eb7p 191 \u0111\u00e1nh cho ng\u00e0y 4\/10\/2024. K\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1 MB 4\/10\/2024 ta c\u00f3 c\u1ea3 l\u00f4 19 \u2013 91.<\/p>\n\n\n\n N\u1ebfu t\u1ed3n t\u1ea1i 0 v\u00e0 6 l\u00e0 hi\u1ec7u c\u1ee7a nhau th\u00ec khi 0 bi\u1ebfn th\u00e0nh 1 l\u00e0 b\u00f3ng c\u1ee7a 6. Nh\u01b0ng ch\u00fang ta v\u1eabn kh\u00f4ng d\u00f9ng c\u1ed9ng nh\u01b0 tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t 1 (1 l\u00e0 b\u00f3ng c\u1ee7a 6) m\u00e0 v\u1eabn d\u00f9ng ph\u00e9p tr\u1eeb l\u00e0 6 \u2013 1 = 5.<\/p>\n\n\n\n V\u00ed d\u1ee5:<\/em> Ng\u00e0y 19\/9\/2024 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u1ec1 80620.<\/p>\n\n\n\n Ch\u00fang ta c\u00f3 X = 8 \u2013 6 = 2.<\/p>\n\n\n\n Khi t\u00ecm Y ta th\u1ea5y E = 0 => E = 1. \u0110\u00fang ra ph\u1ea3i l\u1ea5y Y \u2013 6 + 1 (v\u00ec 6 l\u00e0 b\u00f3ng c\u1ee7a 1), nh\u01b0ng ta l\u1ea5y Y = 6 \u2013 1 (coi 1 l\u00e0 0 n\u00ean kh\u00f4ng ph\u1ea3i b\u00f3ng) => Y = 5.<\/p>\n\n\n\n Ta t\u00ecm \u0111\u01b0\u1ee3c c\u1eb7p 252 \u0111\u00e1nh cho ng\u00e0y 20\/9\/2024 v\u00e0 \u0111\u01b0\u01a1ng nhi\u00ean l\u00e0 ch\u00fang ta \u0103n r\u1ed3i.<\/p>\n\n\n\n Khi x\u1ea3y ra tr\u01b0\u1eddng h\u1ee3p n\u1ebfu t\u1ed3n t\u1ea1i 0 v\u00e0 5 l\u00e0 hi\u1ec7u c\u1ee7a nhau th\u00ec 0 s\u1ebd bi\u1ebfn th\u00e0nh 1. V\u00e0 khi t\u00ecm X,Y ta s\u1ebd d\u00f9ng ph\u00e9p c\u1ed9ng 5+1 = 6 (do ta x\u00e9t 0 b\u00f3ng 5 ch\u1ee9 kh\u00f4ng x\u00e9t khi 0 bi\u1ebfn th\u00e0nh 1).<\/p>\n\n\n\n V\u00ed d\u1ee5:<\/em> Ng\u00e0y 31\/8\/2024 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u1ec1 70560.<\/p>\n\n\n\n Ta c\u00f3 X = 7 \u2013 5 = 2.<\/p>\n\n\n\n Khi t\u00ecm Y ch\u00fang ta th\u1ea5y E = 0 => E = 1. Y = 5 + 1 = 6. (V\u00ec 5 v\u00e0 0 l\u00e0 b\u00f3ng c\u1ee7a nhau).<\/p>\n\n\n\n Ng\u00e0y 1\/9\/2024 ta \u0111\u00e1nh 262 v\u00e0 \u0103n 62 2 nh\u00e1y.<\/p>\n\n\n\n Trong tr\u01b0\u1eddng h\u1ee3p n\u1ebfu X = Y th\u00ec ta s\u1ebd s\u1ebd \u0111\u00e1nh c\u1eb7p CX, XC khung 3 ng\u00e0y nh\u00e9.<\/p>\n\n\n\n V\u00ed d\u1ee5:<\/em> Ng\u00e0y 23\/9\/2024 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u1ec1 92329.<\/p>\n\n\n\n Ta c\u00f3 X = 9 \u2013 3 = 6<\/p>\n\n\n\n Ta c\u00f3 Y = 9 \u2013 3 = 6<\/p>\n\n\n\n Ta c\u00f3 C = 3 (s\u1ed1 \u1edf gi\u1eefa).<\/p>\n\n\n\n V\u00ec X = Y n\u00ean ng\u00e0y 24\/9\/2024 ta \u0111\u00e1nh 363 v\u00e0 k\u1ebft qu\u1ea3 tr\u00fang 36.<\/p>\n\n\n\nTr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t khi soi c\u1ea7u XSMB theo ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t<\/h3>\n\n\n\n
A l\u00e0 b\u00f3ng c\u1ee7a C ho\u1eb7c E l\u00e0 b\u00f3ng c\u1ee7a C<\/h4>\n\n\n\n
0 v\u00e0 6 l\u00e0 hi\u1ec7u c\u1ee7a nhau<\/h4>\n\n\n\n
0 v\u00e0 5 l\u00e0 hi\u1ec7u c\u1ee7a nhau<\/h4>\n\n\n\n
N\u1ebfu X = Y<\/h4>\n\n\n\n