Li\u1ec7u c\u00f3 d\u00e0n \u0111\u1ec1 38 s\u1ed1 hay kh\u00f4ng ? N\u1ebfu nh\u01b0 ch\u00fang ta t\u1ef1 ngh\u0129 ra th\u00ec ch\u1eafc ch\u1eafn c\u00f3. Nh\u01b0ng n\u1ebfu ta theo c\u00e1c nguy\u00ean t\u1eafc nh\u01b0 l\u00e0 l\u1ea5y d\u00e0n \u0111\u1ec1 (x \u2013 x+5) nh\u01b0: 0 \u2013 5, 1 \u2013 6,\u2026hay theo nh\u01b0 c\u00e1ch soi d\u00e0n \u0111\u1ec1 20 s\u1ed1 khung 3 ng\u00e0y th\u00ec ch\u00fang ta c\u0169ng kh\u00f3 m\u00e0 l\u1ea5y ra \u0111\u01b0\u1ee3c t\u1edbi 36 s\u1ed1.<\/p>\n
T\u1ea5t nhi\u00ean l\u00e0 c\u00f3 r\u1ed3i, ngo\u00e0i vi\u1ec7c t\u1ef1 ngh\u0129 ra. Ch\u00fang ta c\u00f3 th\u1ec3 gh\u00e9p l\u1ea1i b\u1eb1ng nhi\u1ec1u ph\u01b0\u01a1ng ph\u00e1p kh\u00e1c nhau. V\u1edbi c\u00e1ch ph\u01b0\u01a1ng ph\u00e1p D\u00e0n \u0110\u1ec1 20 S\u1ed1 Khung 3 Ng\u00e0y nh\u01b0 tr\u00ean, thay v\u00ec l\u1ea5y 5 s\u1ed1 \u0111\u1ea7u v\u00e0 3 s\u1ed1 \u0111u\u00f4i, ta s\u1ebd l\u1ea5y t\u1edbi 7 s\u1ed1 \u0111\u1ea7u v\u00e0 6 s\u1ed1 \u0111u\u00f4i.<\/p>\n
Ho\u1eb7c ch\u00fang ta c\u0169ng s\u1ebd c\u00f3 nh\u1eefng c\u00e1ch kh\u00e1c nh\u01b0 l\u1ea5y d\u00e0n \u0111\u1ec1 t\u1ed5ng tr\u00ean 10 ho\u1eb7c d\u01b0\u1edbi 10. D\u01b0\u1edbi \u0111\u00e2y ch\u00fang t\u00f4i s\u1ebd \u0111\u01b0a ra m\u1ed9t s\u1ed1 v\u00ed d\u1ee5 \u0111\u1ec3 anh ch\u1ecb em bi\u1ebft \u0111\u01b0\u1ee3c d\u00e0n \u0111\u1ec1 38 s\u1ed1 c\u00f3 bao nhi\u00eau con b\u1eb1ng c\u00e1ch k\u1ebft h\u1ee3p ph\u01b0\u01a1ng ph\u00e1p nh\u00e9.<\/p>\n
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T\u1ea1o D\u00e0n \u0110\u1ec1 38 S\u1ed1 D\u1ef1a v\u00e0o ph\u01b0\u01a1ng ph\u00e1p t\u1ea1o D\u00e0n \u0110\u1ec1 20 S\u1ed1 Khung 3 Ng\u00e0y<\/h3>\n
\u0110\u1ea7u ti\u00ean, \u0111\u1ec3 anh ch\u1ecb em n\u00e0o ch\u01b0a bi\u1ebft, h\u00e3y truy c\u1eadp \u0111\u1ec3 xem l\u1ea1i d\u00e0n \u0111\u1ec1 20 s\u1ed1 khung 3 ng\u00e0y tr\u01b0\u1edbc khi \u0111\u1ecdc ti\u1ebfp.<\/p>\n
B\u01b0\u1edbc 1: ta d\u1ef1a v\u00e0o b\u1ea3ng t\u1ea7n su\u1ea5t c\u1ee7a x\u1ed5 s\u1ed1 mi\u1ec1n B\u1eafc ng\u00e0y 10\/04\/2024 \u0111\u1ec3 l\u1ea5y th\u00f4ng tin<\/p>\n
B\u01b0\u1edbc 2: T\u1eeb \u0111\u00e2y ta l\u1ea5y ra 6 s\u1ed1 \u0111\u1ea7u v\u00e0 6 s\u1ed1 \u0111u\u00f4i v\u00e0 li\u1ec7t k\u00ea trong b\u1ea3ng sau:<\/p>\n