{"id":18176,"date":"2024-04-03T07:59:42","date_gmt":"2024-04-03T07:59:42","guid":{"rendered":"https:\/\/soicau5010.minhngocxoso.com\/?p=18176"},"modified":"2024-04-03T07:59:42","modified_gmt":"2024-04-03T07:59:42","slug":"phuong-phap-soi-cau-tong-hieu-giai-dac-biet","status":"publish","type":"post","link":"https:\/\/ketqua886.com\/phuong-phap-soi-cau-tong-hieu-giai-dac-biet\/","title":{"rendered":"ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t"},"content":{"rendered":"
<\/p>\n\n\n
Soi c\u1ea7u l\u00f4 lu\u00f4n t\u1ed3n t\u1ea1i nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p, nhi\u1ec1u c\u00e1ch t\u00ednh kh\u00e1c nhau, phong ph\u00fa v\u00e0 \u0111a d\u1ea1ng gi\u00fap ng\u01b0\u1eddi ch\u01a1i c\u00f3 c\u00e1i nh\u00ecn t\u1ed5ng quan v\u1ec1 con s\u1ed1 s\u1ebd ra trong \u0111\u1ee3t soi c\u1ea7u. T\u1ef1u chung l\u1ea1i c\u1ee7a c\u00e1c ph\u01b0\u01a1ng ph\u00e1p v\u1eabn l\u00e0 t\u1ed5ng h\u1ee3p c\u00e1c c\u1eb7p s\u1ed1 \u0111\u00e3 tr\u00fang trong \u0111\u1ee3t soi c\u1ea7u g\u1ea7n nh\u1ea5t c\u0169ng nh\u01b0 l\u00e0 x\u00e1c su\u1ea5t xu\u1ea5t hi\u1ec7n c\u1ee7a c\u00e1c c\u1eb7p s\u1ed1 soi c\u1ea7u , l\u00f4 gan c\u1ee7a t\u1eebng nh\u00e0 \u0111\u00e0i trong kho\u1ea3ng th\u1eddi gian \u1ea5n \u0111\u1ecbnh.
H\u00f4m nay, xin chia s\u1ebb v\u1edbi c\u00e1c b\u1ea1n m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u m\u1edbi, \u0111\u01b0\u1ee3c c\u00e1c chuy\u00ean gia t\u1ed5ng h\u1ee3p trong nhi\u1ec1u \u0111\u1ee3t soi c\u1ea7u v\u00e0 t\u00ednh v\u1ec1 x\u00e1c xu\u1ea5t tr\u00fang gi\u1ea3i. \u0110\u00f3 l\u00e0 ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u 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. Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u mi\u1ec5n ph\u00ed n\u00e0y s\u1ebd l\u00e0 m\u1ed9t tr\u1ee3 th\u1ee7 \u0111\u1eafc l\u1ef1c \u0111\u1ec3 anh em tham kh\u1ea3o trong nh\u1eefng \u0111\u1ee3t soi c\u1ea7u s\u1eafp \u0111\u1ebfn \u0111\u00e2y. M\u1eddi anh em tham kh\u1ea3o nh\u00e9!<\/p>\n\n\n\n
\u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng ph\u00e1p m\u00e0 ch\u00fang ta s\u1ebd d\u1ef1a v\u00e0o 5 s\u1ed1 c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t \u0111\u1ec3 soi c\u1ea7u. C\u1ee5 th\u1ec3 l\u00e0 \u1edf d\u00e3y s\u1ed1 c\u1ee7a d\u00e3y \u0111\u1eb7c bi\u1ec7t, ta g\u1eafn theo th\u1ee9 t\u1ef1 t\u1eeb ABCDE. Khi \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y ta s\u1ebd l\u1ea5y A tr\u1eeb \u0111i C \u0111\u1ec3 ra \u1ea9n s\u1ed1 X, l\u1ea5y E tr\u1eeb \u0111i C \u0111\u1ec3 t\u00ecm ra \u1ea9n s\u1ed1 Y. K\u1ebft h\u1ee3p X v\u00e0 Y ta s\u1ebd \u0111\u01b0\u1ee3c c\u1eb7p s\u1ed1 c\u1ea7n t\u00ecm.
V\u00ed d\u1ee5 nh\u00e9! Ta c\u00f3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 60327. G\u1eafn c\u00e1c s\u1ed1 theo ph\u01b0\u01a1ng ph\u00e1p, ta \u0111\u01b0\u1ee3c quy tr\u00ecnh sau:<\/p>\n\n\n\n
C\u00f3 m\u1ed9t v\u00e0i tr\u01b0\u1eddng h\u1ee3p c\u1ea7n ch\u00fa \u00fd khi s\u1eed d\u1ee5ng phu\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u \u0111\u1eb7c bi\u1ec7t n\u00e0y l\u00e0:<\/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 ta kh\u00f4ng d\u00f9ng ph\u00e9p tr\u1eeb m\u00e0 ta s\u1ebd d\u00f9ng ph\u00e9p c\u1ed9ng.
V\u00ed d\u1ee5 nh\u00e9: Gi\u1ea3 s\u1eed ta c\u00f3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 82752.
Ta c\u00f3 X = 8-7 = 1.
Khi t\u00ecm Y ta th\u1ea5y C = 7 c\u00f2n E = 2 (7 v\u00e0 2 l\u00e0 b\u00f3ng c\u1ee7a nhau) n\u00ean Y = 7 + 2 = 9
Nh\u01b0 v\u1eady ta t\u00ecm \u0111\u01b0\u1ee3c c\u1eb7p 19, 91 \u0111\u00e1nh v\u00e0o ng\u00e0y h\u00f4m sau. N\u1ebfu tr\u01b0\u1eddng h\u1ee3p 9 v\u00e0 4 l\u00e0 b\u00f3ng c\u1ee7a nhau th\u00ec Y = 9 + 4 = 13 th\u00ec Y = 3.<\/p>\n\n\n\n
N\u1ebfu xu\u1ea5t hi\u1ec7n m\u1ed9t c\u1eb7p t\u1ed5ng hi\u1ec7u l\u00e0 b\u00f3ng c\u1ee7a nhau m\u00e0 ta l\u1ea1i c\u00f3 C = A+ E th\u00ec kh\u00f4ng n\u00ean \u0111\u00e1nh tr\u01b0\u1eddng h\u1ee3p n\u00e0y.
V\u00ed d\u1ee5 : gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 54621
Ta c\u00f3 A =5, C=6, E=1. Ta th\u1ea5y 6 v\u00e0 1 l\u00e0 b\u00f3ng c\u1ee7a nhau v\u00e0 c\u00f3 6 = 5 +1 (C= A +E) n\u00ean ta kh\u00f4ng s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u mi\u1ec5n ph\u00ed n\u00e0y. L\u00fd gi\u1ea3i v\u1ec1 t\u00ecnh hu\u1ed1ng nayfn h\u01b0u sau: Do tr\u00e1nh b\u00f3ng l\u00e0 ch\u00fang ta d\u00f9ng ph\u00e9p c\u1ed9ng, m\u00e0 xu\u1ea5t hi\u1ec7n hi\u1ec7u b\u00f3ng xu\u1ea5t hi\u1ec7n r\u1ea5t kh\u00f3 \u0111\u00e1nh v\u00ec n\u00f3 kh\u00f4ng c\u00f2n l\u00e0 s\u1ed1 ng\u1eabu nhi\u00ean.<\/p>\n\n\n\n
N\u1ebfu A ho\u1eb7c C ho\u1eb7c E = 0 th\u00ec ta s\u1ebd coi n\u00f3 l\u00e0 1. V\u00e0 ta l\u00e0m nh\u01b0 b\u00ecnh th\u01b0\u1eddng .
V\u00ed d\u1ee5: Ta c\u00f3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t 89370.
Ta c\u00f3 X = 8+3 = 11 khi \u0111\u00f3 X = 1 (8 v\u00e0 3 l\u00e0 b\u00f3ng c\u1ee7a nhau)
Khi t\u00ecm Y ta th\u1ea5y E = 0 n\u00ean ta coi l\u00e0 E = 1 n\u00ean Y = 3-1 = 2.
Nh\u01b0 v\u1eady ta t\u00ecm \u0111\u01b0\u1ee3c c\u1eb7p 12, 21 \u0111\u00e1nh v\u00e0o ng\u00e0y h\u00f4m sau.<\/p>\n\n\n\n
1. 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 6-1 = 5.
2. 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
N\u1ebfu X = Y th\u00ec ta s\u1ebd s\u1ebd \u0111\u00e1nh c\u1eb7p CX, XC khung 2 ng\u00e0y nh\u00e9.
V\u00ed d\u1ee5: gi\u1ea3 s\u1eed gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u1ec1 58255
Ta c\u00f3 X = 5-2 = 3
Ta c\u00f3 Y = 5-2 = 3
Ta c\u00f3 C = 2 (s\u1ed1 \u1edf gi\u1eefa)
V\u00ec X=Y n\u00ean ng\u00e0y 14-05 ta \u0111\u00e1nh 23, 32 tr\u00fang 32 ng\u00e0y th\u1ee9 2.<\/p>\n\n\n
<\/p>\n\n\n
M\u1eb7c d\u00f9 ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u \u0111\u1eb7c bi\u1ec7t \u0111\u01b0\u1ee3c \u0111\u00e1nh gi\u00e1 l\u00e0 kh\u00e1 hi\u1ec7u qu\u1ea3 v\u00e0 con s\u1ed1 d\u1ef1 \u0111o\u00e1n ch\u00ednh x\u00e1c, th\u1ebf nh\u01b0ng khi s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y b\u1ea1n c\u0169ng c\u1ea7n ch\u00fa \u00fd nh\u1eefng \u0111i\u1ec1u sau:<\/p>\n\n\n\n
T\u00ecm ra c\u1eb7p XY, nh\u01b0ng ta th\u1ea5y c\u1eb7p XY, YX v\u1ec1 2 nh\u00e1y ho\u1eb7c v\u1ec1 c\u1ea3 c\u1eb7p th\u00ec kh\u00f4ng n\u00ean \u0111\u00e1nh, v\u00ec n\u1ebfu ra nh\u01b0 th\u1ebf th\u00ec x\u00e1c su\u1ea5t 2 ng\u00e0y h\u00f4m sau ra XYX s\u1ebd kh\u00f3 h\u01a1n.
V\u00ed d\u1ee5:Gi\u1ea3 s\u1eed c\u00f3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t 20459.
Ta c\u00f3 X = 4-2 = 2, Y = 9+4 = 13 (3) (9 v\u00e0 4 l\u00e0 b\u00f3ng c\u1ee7a nhau).
Nh\u01b0 v\u1eady ta s\u1ebd t\u00ecm \u0111\u01b0\u1ee3c c\u1eb7p 23,32. Nh\u01b0 v\u1eady l\u00f4 23 \u0111\u00e3 ra 2 nh\u00e1y n\u00ean c\u1ea7u n\u00e0y ng\u00e0y h\u00f4m sau kh\u00f4ng n\u00ean ch\u01a1i.<\/p>\n\n\n\n
N\u1ebfu x\u1ea3y ra c\u00f9ng l\u00fac 2 tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t 2 v\u00e0 3 th\u00ec kh\u00f4ng n\u00ean \u0111\u00e1nh.
V\u00ed d\u1ee5:Gi\u1ea3 s\u1eed ta c\u00f3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 31270.
Ta c\u00f3 X = 3-2= 1.
Khi t\u00ecm Y ta th\u1ea5y C = 0 => C = 1 (tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t 2). N\u00ean Y = 2 -1 = 1.
Nh\u01b0 v\u1eady X=Y= 1. T\u00ecnh hu\u1ed1ng n\u00e0y \u0111\u00e3 t\u1ed3n t\u1ea1i c\u1ea3 tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t 2 v\u00e0 3 n\u00ean chngs ta c\u0169ng kh\u00f4ng n\u00ean \u0111\u00e1nh.<\/p>\n\n\n\n
1. N\u00eau c\u1ea7u n\u00e0y \u0111\u00e1nh tr\u00fang c\u1ea3 c\u1eb7p th\u00ec ngh\u1ec9 2 ng\u00e0y.
V\u00ed d\u1ee5:h\u00f4m nay ta t\u00ednh \u0111\u01b0\u1ee3c \u0111\u00e1nh c\u1eb7p 15, 51 th\u00ec tr\u00fang c\u1ea3 15, 51 n\u00ean hai ng\u00e0y ti\u1ebfp theo ta kh\u00f4ng d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0y n\u1eefa. \u0110\u1ebfn ng\u00e0yth\u1ee9 ba ta m\u1edbi d\u00f9ng k\u1ebft qu\u1ea3 ng\u00e0y \u0111\u1ea7u \u0111\u1ec3 soi.
2. N\u1ebfu hai ng\u00e0y li\u00ean ti\u1ebfp c\u1ea7u \u0111\u1ec1u tr\u00fang c\u1ea3 2 c\u1eb7p th\u00ec ta ngh\u1ec9 4 ng\u00e0y t\u00ednh t\u1eeb ng\u00e0y v\u1ec1 g\u1ea7n nh\u1ea5t. (3 ng\u00e0y li\u00ean ti\u1ebfp th\u00ec ngh\u1ec9 6 ng\u00e0y)
3. Trong nh\u1eefng ng\u00e0y ngh\u1ec9 ta v\u1eabn soi c\u1ea7u b\u00ecnh th\u01b0\u1eddng nh\u01b0ng kh\u00f4ng \u0111\u00e1nh, n\u1ebfu c\u1ea7u \u0111\u00f3 v\u1ec1 c\u1ea3 c\u1eb7p th\u00ec ta l\u1ea1i ngh\u1ec9 2 ng\u00e0y ti\u1ebfp theo.<\/p>\n\n\n\n
Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u t\u1ed5ng hi\u1ec7u \u0111\u1eb7c bi\u1ec7t \u0111\u01b0\u1ee3c nhi\u1ec1u ng\u01b0\u1eddi tin d\u00f9ng v\u00ec s\u1ef1 \u0111\u01a1n gi\u1ea3n, ch\u00ednh x\u00e1c m\u00e0 n\u00f3 mang l\u1ea1i. V\u1edbi c\u00e1ch soi c\u1ea7u mi\u1ec5n ph\u00ed v\u1edbi ph\u01b0\u01a1ng ph\u00e1p t\u1ed5ng hi\u1ec7u gi\u1ea3i \u0111\u1eb7c bi\u1ec7t n\u00e0y ch\u1ec9 v\u1edbi m\u1eb9o c\u1ed9ng tr\u1eeb c\u00e1c ch\u1eef s\u1ed1 trong gi\u1ea3i \u0111\u1eb7c bi\u1ec7t v\u00e0 m\u1ed9t s\u1ed1 quy t\u1eafc \u0111\u1ec3 tr\u00e1nh c\u00e1c l\u00f4 k\u00e9p th\u00ec \u0111\u00e2y \u0111\u00e3 l\u00e0 m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p hay mang l\u1ea1i nhi\u1ec1u l\u1ee3i nhu\u1eadn cho anh em. <\/p>\n\n\n
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Soi c\u1ea7u l\u00f4 lu\u00f4n t\u1ed3n t\u1ea1i nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p, nhi\u1ec1u c\u00e1ch t\u00ednh kh\u00e1c nhau, phong ph\u00fa v\u00e0 \u0111a d\u1ea1ng gi\u00fap […]<\/p>\n","protected":false},"author":1,"featured_media":18178,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[1265,3308,3309,3311,3312,956,107,3061,3310,3306,3304,3305,1003,1192,124,1193,3313,2374,3231,3307,490,431,1343,2767],"yoast_head":"\n