{"id":18300,"date":"2024-04-09T09:55:36","date_gmt":"2024-04-09T09:55:36","guid":{"rendered":"https:\/\/soicau2089.minhngocxoso.com\/?p=18300"},"modified":"2024-04-09T09:55:36","modified_gmt":"2024-04-09T09:55:36","slug":"phuong-phap-soi-cau-lo-3-cay-la-gi","status":"publish","type":"post","link":"https:\/\/soicauxienmb.com\/phuong-phap-soi-cau-lo-3-cay-la-gi\/","title":{"rendered":"Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 c\u00e2y l\u00e0 g\u00ec?"},"content":{"rendered":"
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ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 c\u00e2y v\u1edbi \u0111\u1ed9 ch\u00ednh x\u00e1c c\u1ef1c cao<\/b><\/h3>\n<\/div>\n<\/div>\n<\/div>\n
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Hi\u1ec7n nay, c\u00f3 nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 \u0111\u1ec1 \u0111\u01b0\u1ee3c anh em ch\u00fang ta \u00e1p d\u1ee5ng \u0111\u1ec3 d\u1ef1 \u0111o\u00e1n kqxs \u1edf c\u1ea3 ba mi\u1ec1n. M\u1ed7i ph\u01b0\u01a1ng ph\u00e1p s\u1ebd c\u00f3 nh\u1eefng \u0111\u1eb7c \u0111i\u1ec3m, c\u00e1ch t\u00ednh c\u0169ng nh\u01b0 c\u00e1ch \u00e1p d\u1ee5ng kh\u00e1c nhau \u0111\u1ec3 c\u00f3 th\u1ec3 \u0111\u01b0a ra c\u1eb7p l\u00f4 \u0111\u1ec1 chu\u1ea9n nh\u1ea5t cho anh em tr\u01b0\u1edbc m\u1ed7i \u0111\u1ee3t soi c\u1ea7u l\u00f4.<\/p>\n

H\u00f4m nay c\u00e1c chuy\u00ean gia s\u1ebd b\u1eadt m\u00ed cho anh em m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p \u0111ang n\u1ed5i b\u1eadt hi\u1ec7n nay, \u0111\u00f3 l\u00e0  ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u 3 c\u00e2y. C\u00f9ng t\u00ecm hi\u1ec3u xem ph\u01b0\u01a1ng ph\u00e1p n\u00e0y c\u00f3 \u0111i\u1ec1u g\u00ec \u0111\u1eb7c bi\u1ec7t nh\u00e9!<\/p>\n

Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 c\u00e2y l\u00e0 g\u00ec?<\/h2>\n

Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 c\u00e2y l\u00e0 trong nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u kh\u00e1 \u0111\u01a1n gi\u1ea3n. Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y \u0111\u01b0\u1ee3c x\u00e2y d\u1ef1ng d\u1ef1a v\u00e0o gi\u1ea3i \u0111\u1eb7c bi\u1ec7t c\u1ee7a k\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1 2 ng\u00e0y g\u1ea7n nh\u1ea5t \u0111\u1ec3 quy\u1ebft \u0111\u1ecbnh con s\u1ed1 s\u1ebd \u0111\u00e1nh v\u00e0o ng\u00e0y th\u1ee9 3<\/p>\n

Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 ch\u01a1i nh\u01b0 th\u1ebf n\u00e0o?<\/h2>\n

Ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 c\u00e2y n\u00e0y d\u1ef1a v\u00e0o gi\u1ea3i \u0111\u1eb7c bi\u1ec7t c\u1ee7a 2 h\u00f4m li\u00ean ti\u1ebfp \u0111\u1ec3 \u0111\u00e1nh h\u00f4m th\u1ee9 3. Ta c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p \u0111\u00e1nh nh\u01b0 sau:<\/p>\n

X\u1ebfp 2 s\u1ed1 \u0111\u1ec1 ra trong hai ng\u00e0y g\u1ea7n nh\u1ea5t<\/h3>\n

\u1ede ph\u01b0\u01a1ng ph\u00e1p n\u00e0y gi\u00fap b\u1ea1n soi c\u1ea7u s\u1ed1 \u0111\u1ec1 ra ng\u00e0y ti\u1ebfp theo \u0111\u00f3 l\u00e0 ng\u00e0y 3 \u0111\u1ec1 ra XX. \u0110\u1ec3 t\u00ecm ra \u0111\u01b0\u1ee3c s\u1ed1 X, ch\u00fang ta c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng b\u00f9: Ng\u01b0\u1eddi ch\u01a1i l\u1ea5y m\u1ed9t ch\u1eef s\u1ed1 c\u1ee7a ng\u00e0y th\u1ee9 1 c\u1ed9ng v\u1edbi m\u1ed9t ch\u1eef s\u1ed1 c\u1ee7a ng\u00e0y th\u1ee9 2 v\u00e0 c\u1ed9ng v\u1edbi X sao cho k\u1ebft qu\u1ea3 b\u1eb1ng 10 ho\u1eb7c b\u1eb1ng 20.<\/p>\n

V\u1edbi v\u00ed d\u1ee5 tr\u00ean ta c\u00f3 3+2+X=10 ho\u1eb7c 20 -> X=5V\u00ed d\u1ee5: X\u1ebfp 2 s\u1ed1 \u0111\u1ec1 ra trong hai ng\u00e0y 1 v\u00e0 2. Ng\u00e0y 1 \u0111\u1ec1 ra 32. Ng\u00e0y 2 \u0111\u1ec1 ra 22<\/p>\n

2+2+X = 10 ho\u1eb7c 20 -> X=6<\/p>\n

V\u1eady 56 \u2013 65 l\u00e0 s\u1ed1 c\u1ea7n t\u00ecm \u0111\u1ec3 ch\u01a1i \u0111\u1ec1.<\/p>\n

\"\"<\/p>\n

L\u1ea5y 2 s\u1ed1 \u0111\u1ea7u v\u00e0 2 s\u1ed1 cu\u1ed1i c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t hai ng\u00e0y g\u1ea7n nh\u1ea5t<\/h3>\n

V\u00ed d\u1ee5: Ng\u00e0y 1 k\u1ebft qu\u1ea3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t: 58432 ta l\u1ea5y 52<\/p>\n

Ng\u00e0y 2 k\u1ebft qu\u1ea3 gi\u1ea3i \u0111\u1eb7c bi\u1ec7t: 73122 ta l\u1ea5y 72<\/p>\n

Ph\u01b0\u01a1ng ph\u00e1p 3 c\u00e2y \u0111\u00e1nh l\u00f4 \u0111\u1ec1 trong tr\u01b0\u1eddng h\u1ee3p n\u00e0y nh\u01b0 sau:<\/p>\n

5+7+X = 10 ho\u1eb7c 20 -> X=8<\/p>\n

5+2+X = 10 ho\u1eb7c 20 -> X=3<\/p>\n

2+7+X = 10 ho\u1eb7c 20 -> X=1<\/p>\n

2+2+X = 10 ho\u1eb7c 20 -> X=6<\/p>\n

Nh\u01b0 v\u1eady v\u1edbi ph\u01b0\u01a1ng ph\u00e1p t\u00ednh nh\u01b0 tr\u00ean ch\u00fang ta c\u00f3 th\u1ec3 d\u1ec5 d\u00e0ng t\u00ecm ra c\u00e1c k\u1ebft qu\u1ea3 X t\u1eeb v\u00ed d\u1ee5 l\u00e0 nh\u1eefng s\u1ed1 1, 3, 5, 6, 8 \u0111\u1ec3 \u1ee9ng d\u1ee5ng v\u00e0o b\u1eaft con \u0111\u1ec1 s\u1ebd v\u1ec1 trong ng\u00e0y 17\/09. L\u01b0u \u00fd \u0111\u1ebfn nh\u1eefng con s\u1ed1 tr\u00f9ng nhau v\u00e0 th\u01b0\u1eddng s\u1ebd c\u00f3 1-3 con s\u1ed1 \u0111\u00f3, k\u1ebft qu\u1ea3 v\u1ec1 nh\u01b0 th\u1ebf n\u00e0o b\u1ea1n c\u0169ng bi\u1ebft r\u1ed3i \u0111\u00f3.<\/p>\n

Soi c\u1ea7u l\u00f4 ph\u01b0\u01a1ng ph\u00e1p 3 c\u00e2y c\u1ea7n ch\u00fa \u00fd nh\u1eefng \u0111i\u1ec1u g\u00ec?<\/h2>\n

M\u1eb7c d\u00f9 ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 3 c\u00e2y n\u00e0y kh\u00e1 \u0111\u01a1n gi\u1ea3n v\u00e0 hi\u1ec7u qu\u1ea3 mang l\u1ea1i \u0111\u01b0\u1ee3c \u0111\u00e1nh gi\u00e1 kh\u00e1 cao, tuy nhi\u00ean, v\u1eabn c\u00f3 m\u1ed9t v\u00e0i l\u01b0u \u00fd nho nh\u1ecf cho anh em khi s\u1eed d\u1ee5ng \u0111\u1ec3 c\u00f3 \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c nh\u1ea5t.<\/p>\n