\u00ab<\/mark><\/strong><\/p>\n\n\n\nX\u00e1c su\u1ea5t tr\u00fang ba c\u00e0ng c\u1ef1c th\u1ea5p n\u00ean d\u1ef1 \u0111o\u00e1n ba c\u00e0ng h\u00f4m nay<\/strong> lu\u00f4n l\u00e0 b\u00e0i to\u00e1n kh\u00f3 m\u00e0 kh\u00f4ng ph\u1ea3i ai c\u0169ng gi\u1ea3i \u0111\u01b0\u1ee3c. M\u1ed9t s\u1ed1 ph\u01b0\u01a1ng ph\u00e1p ch\u00fang t\u00f4i chia s\u1ebb d\u01b0\u1edbi \u0111\u00e2y hy v\u1ecdng s\u1ebd gi\u00fap anh em b\u1eaft c\u1ea7u 3 c\u00e0ng B\u1eafc – Trung – Nam d\u1ec5 d\u00e0ng h\u01a1n.<\/p>\n\n\n\nC\u00e1ch 1: D\u1ef1 \u0111o\u00e1n 3 c\u00e0ng b\u1eb1ng c\u00e1ch gh\u00e9p s\u1ed1 theo th\u1ee9 ng\u00e0y ng\u00e0y 06\/01\/2025<\/strong><\/h3>\n\n\n\n\u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng ph\u00e1p \u0111\u00e1nh ba c\u00e0ng d\u1ef1a v\u00e0o th\u1ee9 ng\u00e0y. V\u00ed d\u1ee5 nh\u01b0: th\u1ee9 2 ng\u00e0y 14 th\u00ec ch\u00fang ta s\u1ebd \u0111\u00e1nh 3 c\u00e0ng 214. Th\u1ee9 3 ng\u00e0y 15 th\u00ec \u0111\u00e1nh ba c\u00e0ng 315\u2026. Tuy nhi\u00ean, d\u1ef1 \u0111o\u00e1n 3 c\u00e0ng h\u00f4m nay theo c\u00e1ch gh\u00e9p th\u1ee9 ng\u00e0y t\u1ef7 l\u1ec7 tr\u00fang kh\u00f4ng cao. M\u1eb7t kh\u00e1c, n\u1ebfu ch\u01a1i theo h\u00ecnh th\u1ee9c n\u00e0y nh\u1ea5t \u0111\u1ecbnh c\u1ea7n nhi\u1ec1u v\u1ed1n \u0111\u1ec3 nu\u00f4i khung cho nh\u1eefng ng\u00e0y ti\u1ebfp theo.<\/p>\n\n\n\n
C\u00e1ch 2: D\u1ef1 \u0111o\u00e1n 3 c\u00e0ng theo t\u1ed5ng t\u00edch c\u1ee7a k\u1ef3 quay tr\u01b0\u1edbc<\/em> ng\u00e0y 06\/01\/2025<\/strong><\/h3>\n\n\n\nC\u00e1ch d\u1ef1 \u0111o\u00e1n 3 c\u00e0ng h\u00f4m nay<\/strong> d\u1ef1a theo t\u1ed5ng t\u00edch c\u0169ng ph\u1ea3i d\u1ef1a v\u00e0o ng\u00e0y th\u00e1ng \u0111\u1ec3 t\u00ecm ra c\u1eb7p s\u1ed1 chu\u1ea9n. V\u00ed d\u1ee5: H\u00f4m nay l\u00e0 ng\u00e0y 12 th\u00ec t\u00edch c\u1ee7a ch\u00fang l\u00e0 1×2=2. 2 s\u1ebd l\u00e0 con s\u1ed1 th\u1ee9 nh\u1ea5t c\u1ee7a 3 c\u00e0ng. Ti\u1ebfp \u0111\u00f3, c\u1ed9ng t\u1ed5ng: 1+2=3, \u0111\u00e2y l\u00e0 con s\u1ed1 th\u1ee9 hai. C\u00f2n s\u1ed1 th\u1ee9 3 l\u00e0 t\u1ed5ng c\u1ee7a hai s\u1ed1 v\u1eeba t\u00ecm ra: 2 +3 = 5.<\/p>\n\n\n\nV\u1eady, s\u1ed1 3 c\u00e0ng b\u1ea1n n\u00ean \u0111\u00e1nh v\u00e0o ng\u00e0y h\u00f4m sau l\u00e0 235. Trong tr\u01b0\u1eddng h\u1ee3p t\u1ed5ng, t\u00edch c\u1ee7a c\u00e1c ph\u00e9p t\u00ednh tr\u00ean l\u1edbn h\u01a1n 10 th\u00ec ch\u00fang ta v\u1eabn \u00e1p d\u1ee5ng c\u00e1ch t\u00ednh tr\u00ean sau \u0111\u00f3 l\u1ea5y 3 s\u1ed1 cu\u1ed1i c\u1ee7a ph\u00e9p t\u00ednh \u0111\u00f3.<\/p>\n\n\n\n
C\u00e1ch 3: \u0110\u00e1nh 3 c\u00e0ng d\u1ef1a theo s\u1ed1 \u0111\u1ec1<\/em><\/strong><\/h3>\n\n\n\n