C\u00e1ch t\u00ednh l\u00f4 xi\u00ean quay<\/strong>\u00a0l\u00e0 nh\u01b0 th\u1ebf n\u00e0o? Kh\u00e1i ni\u1ec7m v\u00e0 c\u00e1ch \u0111\u00e1nh ra l\u00e0m sao cho d\u1ec5 tr\u00fang v\u00e0 chu\u1ea9n v\u1edbi c\u00f4ng th\u1ee9c m\u00e0 c\u00e1c b\u1eadc ti\u1ec1n b\u1ed1i \u0111\u00e3 truy\u1ec1n d\u1ea1y t\u1eeb c\u00e1ch \u0111\u00e2y h\u00e0ng ch\u1ee5c n\u0103m?<\/p>\nN\u1ebfu anh em \u0111ang th\u1eafc m\u1eafc v\u1ec1 ch\u1ee7 \u0111\u1ec1 n\u00e0y th\u00ec t\u00f4i xin \u0111\u01b0\u1ee3c t\u1ed5ng h\u1ee3p \u0111\u1ebfn cho anh em d\u00e2n ch\u01a1i nh\u1eefng ki\u1ebfn th\u1ee9c v\u00f4 c\u00f9ng quan tr\u1ecdng v\u1ec1\u00a0c\u00e1ch ch\u01a1i l\u00f4 \u0111\u1ec1<\/strong>\u00a01 \u0111\u1ed3ng v\u1ed1n 100 \u0111\u1ed3ng l\u1eddi n\u00e0y, m\u1eddi anh em tham kh\u1ea3o ngay nh\u00e9.<\/p>\n<\/span>L\u00f4 xi\u00ean quay l\u00e0 g\u00ec?<\/h2>\nL\u00f4 xi\u00ean quay<\/strong>\u00a0l\u00e0 m\u1ed9t c\u00e1ch ch\u01a1i l\u00f4 \u0111\u1ec1 \u0111\u01a1n gi\u1ea3n, \u0111\u00f3 l\u00e0 t\u1eadp h\u1ee3p c\u00e1c t\u1ed5 h\u1ee3p c\u1eb7p l\u00f4 xi\u00ean 2, xi\u00ean 3 v\u00e0 xi\u00ean 4, ch\u01a1i theo c\u00e1ch n\u00e0y anh em kh\u00f4ng c\u1ea7n ph\u1ea3i k\u1ec3 c\u00e1c c\u1eb7p l\u00f4 xi\u00ean m\u00e0 ch\u1ec9 c\u1ea7n ghi t\u00ean c\u00e1c con l\u00f4 \u0111\u1ec3 tr\u00e1nh nh\u1ea7m l\u1eabn, m\u1ea5t th\u1eddi gian m\u00e0 t\u1ea5t c\u1ea3 \u0111\u1ec1u c\u00f3 th\u1ec3 ng\u1ea7m hi\u1ec3u \u0111\u01b0\u1ee3c.<\/p>\nV\u00ed d\u1ee5: Anh em mu\u1ed1n \u0111\u00e1nh l\u00f4 xi\u00ean quay 01, 02, 03, 04 th\u00ec ta c\u00f3:<\/p>\n
\n- C\u00e1c c\u1eb7p xi\u00ean 2: 01 \u2013 02, 01 \u2013 03<\/li>\n
- C\u00e1c c\u1eb7p xi\u00ean 3: 01 \u2013 02 \u2013 03, 01 \u2013 02 \u2013 04, 01 \u2013 03 \u2013 04, 02 \u2013 03 \u2013 04.<\/li>\n
- C\u00e1c c\u1eb7p xi\u00ean 4: 01 \u2013 02 \u2013 03 \u2013 04.<\/li>\n<\/ul>\n
<\/span>C\u00e1ch t\u00ednh l\u00f4 xi\u00ean quay: C\u00e1c d\u1ea1ng \u0111\u00e1nh c\u01a1 b\u1ea3n<\/h2>\nXi\u00ean quay c\u00f3 nhi\u1ec1u h\u00ecnh th\u1ee9c \u0111\u00e1nh nh\u01b0 xi\u00ean quay 2, 3 v\u00e0 4. V\u1eady c\u1ee5 th\u1ec3 c\u1ee7a t\u1eebng c\u00e1ch \u0111\u00e1nh l\u00e0 nh\u01b0 th\u1ebf n\u00e0o? C\u00e1ch t\u00ednh ti\u1ec1n th\u1eafng ra l\u00e0m sao? M\u1eddi anh em theo d\u00f5i<\/p>\n
<\/span>C\u00e1ch \u0111\u00e1nh xi\u00ean quay 2<\/h3>\nL\u00f4 xi\u00ean quay 2\u00a0<\/strong>l\u00e0 t\u1ed5 h\u1ee3p c\u00e1c c\u1eb7p xi\u00ean 2 c\u00f3 trong b\u1ed9 s\u1ed1 m\u00e0 anh em mu\u1ed1n \u0111\u00e1nh. \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 th\u1eafng l\u00e0 anh em ph\u1ea3i tr\u00fang c\u1ea3 2 con trong b\u1ed9 xi\u00ean th\u00ec m\u1edbi \u0111\u01b0\u1ee3c nh\u1eadn th\u01b0\u1edfng.<\/p>\nV\u00ed d\u1ee5:<\/p>\n
\n- \u0110\u00e1nh quay 2 v\u1edbi b\u1ed9 s\u1ed1 01, 02, 03 s\u1ebd bao g\u1ed3m \u0111\u00e1nh 3 c\u1eb7p xi\u00ean 2 l\u00e0 01 \u2013 02, 01 \u2013 03, 02- 03.<\/li>\n
- Ti\u1ec1n v\u1ed1n: Anh em ch\u01a1i 10.000\u0111 th\u00ec ph\u1ea3i tr\u1ea3 cho c\u1ea3 3 con xi\u00ean 2 t\u1ed5ng c\u1ed9ng l\u00e0 30.000\u0111.<\/li>\n
- Tr\u00fang th\u01b0\u1edfng: Tr\u00fang 1 c\u1eb7p xi\u00ean 2 th\u00ec \u0103n, 2 c\u1eb7p xi\u00ean 2 c\u00f9ng v\u1ec1 th\u00ec nh\u00e2n \u0111\u00f4i ti\u1ec1n th\u01b0\u1edfng,\u2026<\/li>\n<\/ul>\n
<\/span>\u0110\u00e1nh l\u00f4 xi\u00ean quay 3 nh\u01b0 th\u1ebf n\u00e0o?<\/h3>\nL\u00f4 xi\u00ean quay 3<\/strong>\u00a0l\u00e0 t\u1ed5 h\u1ee3p c\u00e1c c\u1eb7p xi\u00ean 3 c\u00f3 trong b\u1ed9 s\u1ed1 m\u00e0 anh em mu\u1ed1n \u0111\u00e1nh. N\u1ebfu anh em tr\u00fang c\u1ea3 3 con th\u00ec \u0111\u01b0\u1ee3c t\u00ednh l\u00e0 \u0103n xi\u00ean 3, tr\u01b0\u1ee3t 1 con tr\u00fang 2 con th\u00ec \u0111\u01b0\u1ee3c t\u00ednh l\u00e0 \u0103n xi\u00ean 2, tr\u01b0\u1ee3t 2 con th\u00ec anh em thua.<\/p>\nV\u00ed d\u1ee5:<\/p>\n
\n- \u0110\u00e1nh xi\u00ean quay 3 v\u1edbi b\u1ed9 s\u1ed1 01, 02, 03, 04 s\u1ebd bao g\u1ed3m \u0111\u00e1nh 4 c\u1eb7p l\u00f4 xi\u00ean 3 l\u00e0 01 \u2013 02 \u2013 03, 01 \u2013 02 \u2013 04, 02 \u2013 03 \u2013 04, 01 \u2013 03 \u2013 04.<\/li>\n
- Ti\u1ec1n v\u1ed1n: Anh em ch\u01a1i 10.000\u0111 th\u00ec ph\u1ea3i tr\u1ea3 cho c\u1ea3 4 con xi\u00ean 3 t\u1ed5ng c\u1ed9ng l\u00e0 40.000\u0111.<\/li>\n
- Tr\u00fang th\u01b0\u1edfng: Tr\u00fang 1 c\u1eb7p xi\u00ean 3 th\u00ec \u0103n, 2 c\u1eb7p xi\u00ean 3 th\u00ec nh\u00e2n \u0111\u00f4i ti\u1ec1n th\u01b0\u1edfng,\u2026 V\u00ed d\u1ee5 trong b\u1ed9 xi\u00ean 3: 01 \u2013 02 \u2013 03 m\u00e0 ch\u1ec9 tr\u00fang 01 \u2013 02 th\u00ec anh em \u0111\u01b0\u1ee3c t\u00ednh l\u00e0 \u0103n m\u1ed9t con xi\u00ean 2.<\/li>\n
- X\u00e1c su\u1ea5t tr\u00fang xi\u00ean quay 3 \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c nh\u01b0 sau:\u00a021,5% x 21,5% x 21,5% = 0.0134 = 1.34%.<\/li>\n
- Bao g\u1ed3m: 1 c\u1eb7p xi\u00ean 3 v\u00e0 2 c\u1eb7p xi\u00ean 2.<\/li>\n<\/ul>\n
Ngo\u00e0i c\u00e1ch \u0111\u00e1nh xi\u00ean, th\u00ec\u00a0c\u00e1ch<\/strong>\u00a0\u0111\u00e1nh \u0111\u1ec1 \u0111\u1ea7u \u0111u\u00f4i<\/strong>\u00a0c\u0169ng l\u00e0 m\u1ed9t trong nh\u1eefng c\u00e1ch \u0111\u00e1nh \u0111\u01b0\u1ee3c d\u00e2n ch\u01a1i l\u00f4 \u0111\u1ec1 c\u1ef1c k\u1ef3 \u01b0a th\u00edch. Xem c\u1ee5 th\u1ec3\u00a0t\u1ea1i \u0111\u00e2y<\/strong>.<\/p>\n<\/span>H\u01b0\u1edbng d\u1eabn c\u00e1ch \u0111\u00e1nh l\u00f4 xi\u00ean quay 4<\/h3>\nL\u00f4 xi\u00ean quay 4<\/strong>\u00a0l\u00e0 t\u1ed5 h\u1ee3p c\u00e1c c\u1eb7p xi\u00ean 4 c\u00f3 trong b\u1ed9 s\u1ed1 m\u00e0 anh em mu\u1ed1n \u0111\u00e1nh. N\u1ebfu tr\u00fang c\u1ea3 4 con th\u00ec \u0111\u01b0\u1ee3c t\u00ednh l\u00e0 \u0103n xi\u00ean 4, tr\u01b0\u1ee3t 2 con tr\u00fang 2 con th\u00ec \u0111\u01b0\u1ee3c t\u00ednh l\u00e0 \u0103n xi\u00ean 2, tr\u01b0\u1ee3t 1 con tr\u00fang 3 con th\u00ec \u0111\u01b0\u1ee3c t\u00ednh l\u00e0 \u0103n xi\u00ean 3, tr\u01b0\u1ee3t 3 con th\u00ec anh em thua.<\/p>\nV\u00ed d\u1ee5:<\/p>\n
\n- \u0110\u00e1nh xi\u00ean quay 4 v\u1edbi b\u1ed9 s\u1ed1 01, 02, 03, 04, 05 s\u1ebd bao g\u1ed3m \u0111\u00e1nh 5 c\u1eb7p l\u00f4 xi\u00ean 4 l\u00e0 01 \u2013 02 \u2013 03 \u2013 04, 01 \u2013 02 \u2013 03 \u2013 05, 01 \u2013 02 \u2013 04 \u2013 05, 01 \u2013 03 \u2013 04 \u2013 05, 02 \u2013 03 \u2013 04 \u2013 05.<\/li>\n
- Ti\u1ec1n v\u1ed1n: Anh em ch\u01a1i 10.000\u0111 th\u00ec ph\u1ea3i tr\u1ea3 cho c\u1ea3 5 con xi\u00ean 5 t\u1ed5ng c\u1ed9ng l\u00e0 50.000\u0111.<\/li>\n
- Tr\u00fang th\u01b0\u1edfng: Tr\u00fang 1 c\u1eb7p l\u00e0 \u0103n, 2 c\u1eb7p th\u00ec nh\u00e2n \u0111\u00f4i ti\u1ec1n th\u01b0\u1edfng,\u2026. V\u00ed d\u1ee5 trong b\u1ed9 01 \u2013 02 \u2013 03 \u2013 04 ch\u1ec9 tr\u00fang 3 con th\u00ec tr\u00fang xi\u00ean 3, 2 con th\u00ec tr\u00fang xi\u00ean 2.<\/li>\n
- X\u00e1c su\u1ea5t tr\u00fang \u0111\u01b0\u1ee3c quay 4 \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c nh\u01b0 sau:\u00a021.5% x 21,5% x 21,5% x 21,5% = 0.00214.<\/li>\n
- Bao g\u1ed3m 1 c\u1eb7p xi\u00ean 4, 4 c\u1eb7p xi\u00ean 3 v\u00e0 6 c\u1eb7p xi\u00ean 2.<\/li>\n<\/ul>\n
<\/span>C\u00e1ch t\u00ednh ti\u1ec1n l\u00f4 xi\u00ean quay (c\u1eadp nh\u1eadt 2024)<\/h2>\nKh\u00e1c v\u1edbi l\u00f4 xi\u00ean th\u00f4ng th\u01b0\u1eddng, ch\u01a1i xi\u00ean quay s\u1ebd c\u00f3 c\u00e1ch t\u00ednh ti\u1ec1n kh\u00e1c m\u1ed9t ch\u00fat do m\u1ed7i con xi\u00ean \u0111\u01b0\u1ee3c quy\u1ec1n tr\u01b0\u1ee3t 1 ho\u1eb7c 2 con s\u1ed1 th\u00e0nh \u0103n xi\u00ean 2, xi\u00ean 3.<\/p>\n
T\u00f4i s\u1ebd t\u1ed5ng h\u1ee3p cho anh em c\u00e1ch t\u00ednh ti\u1ec1n h\u00ecnh th\u1ee9c c\u00e1 c\u01b0\u1ee3c n\u00e0y m\u1ed9t c\u00e1ch d\u1ec5 d\u00e0ng nh\u1ea5t \u0111\u1ec3 mang l\u1ea1i l\u1ee3i nhu\u1eadn c\u1ef1c cao cho anh em, tr\u00e1nh thi\u1ebfu hi\u1ec3u bi\u1ebft m\u00e0 m\u1ea5t ti\u1ec1n oan.<\/p>\n
<\/p>\n
S\u1ed1 ti\u1ec1n th\u01b0\u1edfng m\u00e0 anh em s\u1ebd nh\u1eadn \u0111\u01b0\u1ee3c n\u1ebfu chi\u1ebfn th\u1eafng.<\/p>\n<\/div>\n
<\/span>C\u00e1ch t\u00ednh ti\u1ec1n l\u00f4 xi\u00ean quay 2<\/h3>\nL\u00f4 xi\u00ean 2 th\u00f4ng th\u01b0\u1eddng t\u1ef7 l\u1ec7 anh em tr\u00fang s\u1ebd l\u00e0 1 \u0103n 10.<\/p>\n
V\u00ed d\u1ee5: Xi\u00ean quay 2 (01, 02, 03) s\u1ebd c\u00f3 3 c\u1eb7p xi\u00ean 2, anh em c\u01b0\u1ee3c 10.000\u0111 th\u00ec t\u1ed5ng ti\u1ec1n anh em ph\u1ea3i tr\u1ea3 l\u00e0 30.000\u0111. Tr\u00fang 1 c\u1eb7p xi\u00ean 2 th\u00ec anh em \u0103n 100.000\u0111, 2 c\u1eb7p xi\u00ean 2 th\u00ec \u0103n 200.000\u0111.<\/p>\n
Th\u1eafc m\u1eafc v\u1eady ch\u01a1i xi\u00ean c\u00f3 \u0103n \u0111\u01b0\u1ee3c nhi\u1ec1u ti\u1ec1n h\u01a1n xi\u00ean quay kh\u00f4ng? Xem ngay\u00a0c\u00e1ch t\u00ednh ti\u1ec1n l\u00f4 xi\u00ean<\/strong>\u00a0t\u1ea1i \u0111\u00e2y.<\/p>\n<\/span>L\u00f4 xi\u00ean quay 3 \u0103n \u0111\u01b0\u1ee3c bao nhi\u00eau ti\u1ec1n?<\/h3>\nL\u00f4 xi\u00ean 3 th\u00f4ng th\u01b0\u1eddng th\u00ec t\u1ef7 l\u1ec7 anh em tr\u00fang s\u1ebd l\u00e0 1 \u0103n 40 (t\u00f9y nh\u00e0 c\u00e1i).<\/p>\n
V\u00ed d\u1ee5: Anh em \u0111\u00e1nh xi\u00ean quay 3 c\u1ee7a b\u1ed9 s\u1ed1 (01, 02, 03, 04) s\u1ebd c\u00f3 4 c\u1eb7p xi\u00ean 3, \u0111\u00e1nh 10.000\u0111 th\u00ec t\u1ed5ng ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0 40.000\u0111.<\/p>\n
Tr\u00fang 1 c\u1eb7p xi\u00ean 3 th\u00ec \u0103n 400.000\u0111, 2 c\u1eb7p xi\u00ean 3 th\u00ec \u0103n 800.000\u0111. Tr\u01b0\u1eddng h\u1ee3p tr\u01b0\u1ee3t 1 con tr\u00fang 2 con l\u00f4 th\u00ec \u0103n con xi\u00ean 2 (t\u00ednh ti\u1ec1n nh\u01b0 tr\u00ean).<\/p>\n
<\/span>Xi\u00ean quay 4 \u0103n bao nhi\u00eau?<\/h3>\nL\u00f4 xi\u00ean 4 th\u00f4ng th\u01b0\u1eddng th\u00ec t\u1ef7 l\u1ec7 anh em tr\u00fang s\u1ebd l\u00e0 1 \u0103n 100 (t\u00f9y nh\u00e0 c\u00e1i).<\/p>\n
V\u00ed d\u1ee5: Anh em \u0111\u00e1nh xi\u00ean quay 4 c\u1ee7a b\u1ed9 s\u1ed1 (01, 02, 03, 04, 05) s\u1ebd c\u00f3 5 c\u1eb7p xi\u00ean 4, \u0111\u00e1nh 10.000\u0111 th\u00ec t\u1ed5ng v\u1ed1n l\u00e0 50.000\u0111.<\/p>\n
Tr\u00fang 1 c\u1eb7p xi\u00ean 4 th\u00ec \u0103n 1.000.000\u0111, 2 c\u1eb7p xi\u00ean 4 th\u00ec \u0103n 2.000.000\u0111. Tr\u01b0\u1eddng h\u1ee3p tr\u01b0\u1ee3t 1 con tr\u00fang 3 con l\u00f4 th\u00ec \u0103n con xi\u00ean 3 (t\u00ednh ti\u1ec1n nh\u01b0 xi\u00ean 3), tr\u01b0\u1ee3t 2 con tr\u00fang 2 con l\u00f4 th\u00ec \u0103n con xi\u00ean 2 (t\u00ednh ti\u1ec1n nh\u01b0 xi\u00ean 2).<\/p>\n