C\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean nh\u01b0 th\u1ebf n\u00e0o?<\/strong><\/h2>\r\n\r\n\r\n\r\nC\u00f3 nh\u1eefng c\u00e1ch t\u00ednh l\u00f4 xi\u00ean nh\u00e1y quay 3, 4 n\u00e0o? C\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean ch\u00ednh x\u00e1c cao l\u00e0 g\u00ec? C\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean ngh\u0129a l\u00e0 ch\u00fang ta c\u01b0\u1ee3c v\u00e0o c\u00e1c b\u1ed9 s\u1ed1 con l\u00f4 \u0111\u01a1n l\u1ebb. \u0110\u01b0\u1ee3c t\u00ednh l\u00e0 tr\u00fang khi t\u1ea5t c\u1ea3 c\u00e1c con trong b\u1ed9 l\u00f4 m\u00e0 b\u1ea1n \u0111\u00e3 ch\u1ecdn ph\u1ea3i c\u00f9ng v\u1ec1 trong 1 ng\u00e0y. Ch\u1ec9 c\u1ea7n 1 trong s\u1ed1 \u0111\u00f3 kh\u00f4ng ra th\u00ec coi nh\u01b0 th\u1ea5t b\u1ea1i.\u00a0<\/p>\r\n\r\n\r\n\r\n
V\u00ed d\u1ee5:<\/strong><\/p>\r\n\r\n\r\n\r\n\r\n- Ch\u01a1i xi\u00ean 2 l\u00e0 th\u00ec 2 con c\u00f9ng v\u1ec1 l\u00e0 th\u1eafng.<\/li>\r\n
- Xi\u00ean 3 ph\u1ea3i 3 con c\u00f9ng n\u1ed5.<\/li>\r\n
- Xi\u00ean 4 th\u00ec 4 con c\u00f9ng xu\u1ea5t hi\u1ec7n.<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n
Xi\u00ean 2 kh\u1ea3 n\u0103ng tr\u00fang l\u00e0 cao nh\u1ea5t, c\u00e0ng xi\u00ean nhi\u1ec1u s\u1ed1 th\u00ec c\u00e0ng kh\u00f3 th\u1eafng. Mu\u1ed1n \u0103n \u0111\u01b0\u1ee3c c\u00e1ch ch\u01a1i n\u00e0y anh em ph\u1ea3i b\u1ecf ra nhi\u1ec1u th\u1eddi gian t\u00ecm hi\u1ec3u nghi\u00ean c\u1ee9u.<\/p>\r\n\r\n\r\n\r\n
\u0110\u00e1nh l\u00f4 xi\u00ean \u0103n bao nhi\u00eau?<\/strong><\/h2>\r\n\r\n\r\n\r\nH\u01b0\u1edbng d\u1eabn t\u00ednh l\u00f4 xi\u00ean 2 mi\u1ec1n b\u1eafc v\u00e0 c\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean 2 chu\u1ea9n t\u1eeb cao th\u1ee7:<\/p>\r\n\r\n\r\n\r\n
V\u1edbi l\u00f4 xi\u00ean 2, 1 \u0111i\u1ec3m c\u00f3 gi\u00e1 tr\u1ecb 10.000 VN\u0110. T\u1ef7 l\u1ec7 \u0103n c\u1ee7a xi\u00ean 2 theo c\u00e1ch ch\u01a1i truy\u1ec1n th\u1ed1ng l\u00e0 \u0111\u00e1nh 1 \u0111\u01b0\u1ee3c 10. Ngh\u0129a l\u00e0 v\u1edbi 1 \u0111i\u1ec3m tr\u00fang s\u1ebd \u0111\u01b0\u1ee3c 100.000 VN\u0110.<\/p>\r\n\r\n\r\n\r\n
C\u00f2n ch\u01a1i tr\u1ef1c tuy\u1ebfn th\u00ec t\u1ef7 l\u1ec7 th\u01b0\u1edfng cao h\u01a1n, 1 \u0103n 16.5. C\u01b0\u1ee3c 1 \u0111i\u1ec3m \u0111\u00fang th\u1eafng \u0111\u01b0\u1ee3c 165.000 VN\u0110.<\/p>\r\n\r\n\r\n\r\n
C\u00e1ch t\u00ednh ti\u1ec1n l\u00f4 xi\u00ean 3<\/strong><\/h2>\r\n\r\n\r\n\r\nC\u00f3 nh\u1eefng c\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean 3 mi\u1ec1n b\u1eafc d\u1ec5 tr\u00fang n\u00e0o? 1 \u0111i\u1ec3m xi\u00ean 3 c\u0169ng c\u00f3 gi\u00e1 l\u00e0 10.000 VN\u0110 nh\u01b0ng t\u1ef7 l\u1ec7 th\u01b0\u1edfng cao h\u01a1n 1 \u0103n 45 \u0111\u1ed1i v\u1edbi c\u00e1ch b\u00ecnh th\u01b0\u1eddng, c\u00f2n \u0111\u00e1nh online th\u00ec 1 \u0103n t\u1eadn 65.<\/p>\r\n\r\n\r\n\r\n
C\u1ee5 th\u1ec3 v\u1edbi 1 \u0111i\u1ec3m ch\u01a1i truy\u1ec1n th\u1ed1ng b\u1ea1n nh\u1eadn v\u1ec1 450.000 VN\u0110, c\u00f2n online th\u00ec \u0111\u01b0\u1ee3c 650.000 VN\u0110.<\/p>\r\n\r\n\r\n\r\n
C\u00e1ch t\u00ednh ti\u1ec1n l\u00f4 xi\u00ean 4<\/strong><\/h2>\r\n\r\n\r\n\r\nT\u01b0\u01a1ng t\u1ef1 xi\u00ean 2, 3, \u0111i\u1ec3m xi\u00ean 4 gi\u00e1 c\u0169ng l\u00e0 10.000 VN\u0110. V\u1edbi c\u00e1ch truy\u1ec1n th\u1ed1ng th\u00ec 1 \u0103n 100. T\u1ee9c 1 \u0111i\u1ec3m tr\u00fang th\u01b0\u1edfng 1.000.000 VN\u0110. C\u00f2n h\u00ecnh th\u1ee9c tr\u1ef1c tuy\u1ebfn th\u00ec cao h\u01a1n r\u1ea5t nhi\u1ec1u, 1 \u0103n 170. Tr\u00fang 1 \u0111i\u1ec3m \u0103n 1.700.000 VN\u0110.<\/p>\r\n\r\n\r\n\r\n
T\u00ecm hi\u1ec3u v\u1ec1 l\u00f4 xi\u00ean quay<\/strong><\/h2>\r\n\r\n\r\n\r\nM\u1ed9t bi\u1ebfn th\u1ec3 kh\u00e1c c\u1ee7a l\u00f4 xi\u00ean g\u00e2y s\u1ed1t trong gi\u1edbi \u0111am m\u00ea s\u1ed1 h\u1ecdc th\u1eddi gian g\u1ea7n \u0111\u00e2y l\u00e0 l\u00f4 xi\u00ean quay. T\u1eeb quay \u1edf \u0111\u00e2y c\u00f3 ngh\u0129a l\u00e0 quay \u0111\u1ea7u l\u00f9i l\u1ea1i. B\u1ea1n \u0111\u00e1nh xi\u00ean quay 3 nh\u01b0ng v\u1ec1 2 con 1 con tr\u01b0\u1ee3t th\u00ec t\u00ednh th\u01b0\u1edfng th\u00e0nh l\u00f4 xi\u00ean 2. Ch\u01a1i xi\u00ean quay 4, \u0111o\u00e1n sai con \u0111\u01b0\u1ee3c \u0103n ti\u1ec1n t\u01b0\u01a1ng \u0111\u01b0\u01a1ng gi\u00e1 tr\u1ecb xi\u00ean 3; sai 2 con th\u00ec coi l\u00e0 xi\u00ean 2.<\/p>\r\n\r\n\r\n\r\n
L\u00f4 xi\u00ean quay t\u00ednh nh\u01b0 th\u1ebf n\u00e0o?<\/strong><\/h2>\r\n\r\n\r\n\r\n\r\n