Trong 2 tu\u1ea7n \u0111\u1ea7u ti\u00ean (v\u00f9ng m\u00e0u \u0111\u1ecf), ch\u1ecdn ra 14 b\u1ed9 s\u1ed1 c\u00f3 s\u1ed1 l\u1ea7n ra nhi\u1ec1u nh\u1ea5t.<\/li>\n
Trong 16 ng\u00e0y c\u00f2n l\u1ea1i (v\u00f9ng m\u00e0u xanh), ch\u1ecdn ra 2 con s\u1ed1 c\u00f3 s\u1ed1 l\u1ea7n ra nhi\u1ec1u nh\u1ea5t.<\/li>\n<\/ol>\n
\u0110\u1ec3 soi d\u00e0n \u0111\u1ec1 20 s\u1ed1, anh em c\u1ea7n ti\u1ebfn h\u00e0nh l\u1ecdc ra 5 con s\u1ed1 \u0111\u1ea7u v\u00e0 4 con s\u1ed1 \u0111u\u00f4i hay v\u1ec1 nh\u1ea5t trong th\u1eddi gian 30 l\u1ea7n quay tr\u1edf l\u1ea1i \u0111\u00e2y. Ch\u00fang ta s\u1ebd ti\u1ebfn h\u00e0nh gh\u00e9p c\u00e1c con s\u1ed1 \u0111\u00f3 th\u00e0nh c\u1eb7p theo quy t\u1eafc nh\u1ea5t \u0111\u1ecbnh. T\u1eeb \u0111\u00f3 t\u1ea1o ra \u0111\u01b0\u1ee3c d\u00e0n \u0111\u1ec1 20 s\u1ed1 b\u1ea5t t\u1eed \u0111\u00e1nh trong v\u00f2ng 3 ng\u00e0y s\u1eafp t\u1edbi.<\/span><\/p>\n
Ch\u1ecdn ch\u1ebf \u0111\u1ed9 xem th\u1ed1ng k\u00ea \u0111\u1ea7u \u0111u\u00f4i c\u1ee7a KQXS \u0111\u00e0i anh em mu\u1ed1n ch\u01a1i.<\/span><\/li>\n
Ch\u1ecdn th\u1eddi gian th\u1ed1ng k\u00ea 30 ng\u00e0y cho t\u1edbi ng\u00e0y soi c\u1ea7u.<\/span><\/li>\n
Ch\u1ecdn ra 5 con s\u1ed1 \u0111\u1ea7u c\u00f9ng 4 con s\u1ed1 \u0111u\u00f4i xu\u1ea5t hi\u1ec7n nhi\u1ec1u nh\u1ea5t trong v\u00f2ng 30 ng\u00e0y \u0111\u00f3.<\/span><\/li>\n
D\u00e0n 34 s\u1ed1 mi\u1ec1n B\u1eafc hay ra nh\u1ea5t l\u00e0 d\u00e0n \u0111\u1ec1 c\u00f3 d\u1ea1ng chia h\u1ebft cho 3. D\u00e0n \u0111\u1ec1 n\u00e0y anh em c\u00f3 th\u1ec3 xem chi ti\u1ebft c\u00e1ch t\u00ednh trong v\u00e0i vi\u1ebft v\u1ec1 T\u1ed5ng \u0111\u1ec1.<\/p>\n
V\u1edbi ph\u01b0\u01a1ng ph\u00e1p n\u00e0y, anh em ch\u00fang ta t\u00ednh theo t\u1ed5ng 2 ch\u1eef s\u1ed1 con \u0111\u1ec1 g\u1ea7n nh\u1ea5t v\u00e0 so s\u00e1nh v\u1edbi 10 \u0111\u1ec3 l\u1ea5y d\u00e0n. T\u1ee9c l\u00e0:<\/span><\/p>\n
Theo c\u00f4ng th\u1ee9c tr\u00ean, anh em c\u00f3 th\u1ec3 thay s\u1ed1 v\u00e0o x \u0111\u1ec3 t\u1ea1o ra d\u00e0n m\u1edbi.V\u00ed d\u1ee5 v\u1edbi x l\u00e0 0, ta c\u00f3 d\u00e0n 0 \u2013 (0+5). V\u1eady ta c\u00f3 d\u00e0n \u0111\u1ec1 0-5.<\/span><\/p>\n
T<\/span>\u01b0\u01a1ng t\u1ef1 ch\u00fang ta s\u1ebd c\u00f3 c\u00e1c b\u1ed9 d\u00e0n l\u00f4 \u0111\u1ec1 theo c\u00f4ng th\u1ee9c tr\u00ean nh\u01b0 sau:<\/p>\n
T\u1ed5ng c\u1ea3 th\u1ea3y c\u00f3 55 con s\u1ed1. Anh \u1eb9m c\u00f3 th\u1ec3 ch\u1ecdn nu\u00f4i min 1 tu\u1ea7n ho\u1eb7c max 10 ng\u00e0y.<\/p>\n