{"id":16844,"date":"2024-03-21T01:01:51","date_gmt":"2024-03-21T01:01:51","guid":{"rendered":"http:\/\/dudoansongthulo.com\/?p=16844"},"modified":"2024-03-21T01:01:51","modified_gmt":"2024-03-21T01:01:51","slug":"du-doan-xo-mien-bac-lo-xien-2","status":"publish","type":"post","link":"https:\/\/dudoansongthulo.com\/du-doan-xo-mien-bac-lo-xien-2\/","title":{"rendered":"T\u00ecm hi\u1ec3u b\u00ed quy\u1ebft c\u1ee7a chuy\u00ean gia d\u1ef1 \u0111o\u00e1n x\u1ed5 s\u1ed1 mi\u1ec1n B\u1eafc l\u00f4 xi\u00ean"},"content":{"rendered":"
L\u00e0m sao \u0111\u1ec3 d\u1ef1 \u0111o\u00e1n x\u1ed5 s\u1ed1 mi\u1ec1n B\u1eafc l\u00f4 xi\u00ean<\/strong> ng\u00e0y h\u00f4m nay ch\u00ednh x\u00e1c? \u0110\u00f3 l\u00e0 b\u0103n kho\u0103n c\u1ee7a nh\u1eefng ng\u01b0\u1eddi \u0111am m\u00ea v\u1edbi nh\u1eefng con s\u1ed1. Sau \u0111\u00e2y, c\u00e1c b\u1ea1n s\u1ebd nh\u1eadn \u0111\u01b0\u1ee3c c\u00e1ch d\u1ef1 \u0111o\u00e1n l\u00f4 \u0111\u1ec1 s\u00e1t nh\u1ea5t hi\u1ec7n nay. Ch\u00fac c\u00e1c b\u1ea1n th\u00e0nh c\u00f4ng.<\/p>\n B\u00f3ng s\u1ed1 trong soi c\u1ea7u l\u00f4 \u0111\u1ec1<\/strong> c\u1ea7n bi\u1ebft:<\/p>\n C\u1ea7u s\u1ed1 6 l\u00e0 b\u00f3ng c\u1ee7a 1<\/p>\n C\u1ea7u s\u1ed1 7 l\u00e0 b\u00f3ng c\u1ee7a 2<\/p>\n C\u1ea7u s\u1ed1 8 l\u00e0 b\u00f3ng c\u1ee7a 3<\/p>\n C\u1ea7u s\u1ed1 9 l\u00e0 b\u00f3ng c\u1ee7a 4<\/p>\n C\u1ea7u s\u1ed1 0 l\u00e0 b\u00f3ng c\u1ee7a 5<\/p>\n D\u1ef1 \u0111o\u00e1n x\u1ed5 s\u1ed1 mi\u1ec1n B\u1eafc l\u00f4 xi\u00ean<\/strong> theo phong th\u1ee7y<\/p>\n Kim = 2. M\u1ed9c = 5. H\u1ecfa = 3. Th\u1ee7y =1. Th\u1ed5 = 4.<\/p>\n Chi\u1ebfu theo b\u00f3ng c\u1ee7a s\u1ed1 nh\u01b0 \u1edf tr\u00ean, ta c\u00f3 : Kim = 7. M\u1ed9c = 0. H\u1ecfa = 8. Th\u1ee7y = 6. Th\u1ed5 = 9.<\/p>\n Trong phong th\u1ee7y c\u00f3 d\u00e3y: Kim -> M\u1ed9c -> H\u1ecfa -> Th\u1ee7y -> Th\u1ed5 -> Kim.<\/p>\n T\u1eeb \u0111\u00f3 quy \u0111\u1ed5i \u0111\u01b0\u1ee3c : 2->5->3->1->4->2 v\u00e0 7->0->6->9->7<\/p>\n T\u1ed5ng s\u1ed1 trong gi\u1ea3i \u0111\u1eb7c bi\u1ec7t:<\/p>\n VD: Gi\u1ea3i \u0111\u1eb7c bi\u1ec7t l\u00e0 15326 -> t\u1ed5ng = 1+5+3+2+6 = 17<\/p>\nC\u01a1 b\u1ea3n \u0111\u1ec3 d\u1ef1 \u0111o\u00e1n x\u1ed5 s\u1ed1 mi\u1ec1n B\u1eafc l\u00f4 xi\u00ean<\/h2>\n