T\u1ed5ng \u0111\u1ec1 ch\u1eafc ch\u1eafn l\u00e0 c\u00e1i t\u00ean anh em \u0111\u00e3 nghe r\u1ea5t nhi\u1ec1u. Nh\u01b0ng v\u1eabn ch\u01b0a th\u1ef1c s\u1ef1 bi\u1ebft c\u00e1ch ch\u01a1i v\u00e0 t\u1ef7 l\u1ec7 d\u1ef1 \u0111o\u00e1n th\u1ebf n\u00e0o l\u00e0 ch\u00ednh x\u00e1c. V\u1eady th\u00ec nh\u1eefng th\u00f4ng tin sau \u0111\u00e2y ch\u00ednh l\u00e0 gi\u1ea3i ph\u00e1p t\u1ed1t nh\u1ea5t. Cho t\u1ea5t c\u1ea3 nh\u1eefng ai mu\u1ed1n nh\u1edb v\u00e0 l\u00e0m ch\u1ee7 \u0111\u01b0\u1ee3c t\u1ed5ng \u0111\u1ec1 mi\u1ec1n B\u1eafc. V\u1edbi c\u00e1ch ch\u01a1i n\u00e0y, ch\u1eafc ch\u1eafn x\u00e1c su\u1ea5t tr\u00fang s\u1ebd l\u00ean \u0111\u1ebfn 90%<\/em><\/p>\n <\/p>\n \u0110\u1ec1 t\u1ed5ng mi\u1ec1n B\u1eafc \u0111ang l\u00e0 c\u00e1ch ch\u01a1i \u0111\u01b0\u1ee3c r\u1ea5t nhi\u1ec1u ng\u01b0\u1eddi ch\u01a1i \u00e1p d\u1ee5ng hi\u1ec7n nay. V\u1eady b\u1ea1n \u0111\u00e3 th\u1ef1c s\u1ef1 hi\u1ec3u \u0111\u01b0\u1ee3c \u0111\u1ec1 t\u1ed5ng l\u00e0 g\u00ec ch\u01b0a? \u0110\u1ec3 c\u00f3 th\u1ec3 n\u1eafm r\u00f5 v\u00e0 \u00e1p d\u1ee5ng \u0111\u01b0\u1ee3c c\u00e1ch ch\u01a1i n\u00e0y. th\u00ec vi\u1ec7c \u0111\u1ea7u ti\u00ean l\u00e0 anh em c\u1ea7n hi\u1ec3u \u0111\u01b0\u1ee3c th\u1ebf n\u00e0o l\u00e0 \u0111\u1ec1 t\u1ed5ng<\/p>\n T\u1ed5ng \u0111\u1ec1 l\u00e0 t\u1eadp h\u1ee3p c\u00e1c con s\u1ed1 m\u00e0 khi ta c\u1ed9ng 2 ch\u1eef s\u1ed1 c\u1ee7a c\u00e1c s\u1ed1 th\u00ec \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 nh\u01b0 nhau. K\u1ebft qu\u1ea3 c\u1ee7a t\u1ed5ng 2 ch\u1eef s\u1ed1 n\u00e0y s\u1ebd t\u00ednh theo c\u00e1ch t\u00ednh c\u1ee7a b\u00e0i 3 c\u00e2y (b\u00e0i c\u00e0o). \u0110\u1ec3 d\u1ec5 hi\u1ec3u h\u01a1n v\u1ec1 kh\u00e1i ni\u1ec7m n\u00e0y, ta c\u00f9ng xem v\u00ed d\u1ee5 sau \u0111\u00e2y<\/p>\n V\u00ed d\u1ee5:<\/p>\n + S\u1ed1 20 s\u1ebd n\u1eb1m trong T\u1ed5ng 2 v\u00ec c\u00f3 t\u1ed5ng 2 ch\u1eef s\u1ed1 2 + 0 = 2<\/p>\n + T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 11 c\u0169ng n\u1eb1m trong T\u1ed5ng 2 v\u00ec 1 + 1 = 2<\/p>\n + S\u1ed1 57 c\u00f3 t\u1ed5ng 2 ch\u1eef s\u1ed1 l\u00e0 5 + 7 = 12 tuy nhi\u00ean theo c\u00e1ch t\u00ednh c\u1ee7a b\u00e0i 3 c\u00e2y th\u00ec 5 + 7 = 12 t\u00ednh l\u00e0 2 \u0111i\u1ec3m. V\u00ec v\u1eady s\u1ed1 57 c\u0169ng n\u1eb1m trong T\u1ed5ng 2.<\/p>\n + T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 s\u1ed1 84 c\u00f3 8 + 4 = 12 c\u0169ng n\u1eb1m trong T\u1ed5ng 2.<\/p>\nC\u00e1ch ch\u01a1i \u0111\u1ec1 t\u1ed5ng mi\u1ec1n B\u1eafc m\u1edbi nh\u1ea5t<\/strong><\/span><\/h2>\n
1\/ T\u1ed5ng \u0111\u1ec1 l\u00e0 g\u00ec?<\/span><\/strong><\/h3>\n