{"id":568784,"date":"2026-07-27T09:15:24","date_gmt":"2026-07-27T02:15:24","guid":{"rendered":"https:\/\/hoanghamobile.com\/tin-tuc\/?p=568784"},"modified":"2026-07-27T09:15:24","modified_gmt":"2026-07-27T02:15:24","slug":"phan-so-la-gi","status":"publish","type":"post","link":"https:\/\/hoanghamobile.com\/tin-tuc\/phan-so-la-gi\/","title":{"rendered":"Ph\u00e2n s\u1ed1 l\u00e0 g\u00ec? R\u00fat g\u1ecdn, quy \u0111\u1ed3ng v\u00e0 so s\u00e1nh ph\u00e2n s\u1ed1"},"content":{"rendered":"

Ph\u00e2n s\u1ed1 l\u00e0 c\u00e1ch vi\u1ebft bi\u1ec3u th\u1ecb th\u01b0\u01a1ng c\u1ee7a m\u1ed9t ph\u00e9p chia, c\u00f3 d\u1ea1ng a\/b, trong \u0111\u00f3 a l\u00e0 t\u1eed s\u1ed1, b l\u00e0 m\u1eabu s\u1ed1 v\u00e0 b kh\u00e1c 0. V\u00ed d\u1ee5 3\/4 l\u00e0 ph\u00e2n s\u1ed1 c\u00f3 t\u1eed s\u1ed1 3 v\u00e0 m\u1eabu s\u1ed1 4, bi\u1ec3u th\u1ecb 3 ph\u1ea7n b\u1eb1ng nhau trong 4 ph\u1ea7n. Ph\u00e2n s\u1ed1 l\u00e0 ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng c\u1ee7a To\u00e1n ti\u1ec3u h\u1ecdc v\u00e0 l\u1edbp 6. B\u00e0i vi\u1ebft n\u00e0y tr\u00ecnh b\u00e0y <\/span>ph\u00e2n s\u1ed1 l\u00e0 g\u00ec<\/b>, ph\u00e2n s\u1ed1 b\u1eb1ng nhau, r\u00fat g\u1ecdn, quy \u0111\u1ed3ng, so s\u00e1nh, c\u00e1c ph\u00e9p t\u00ednh v\u1edbi ph\u00e2n s\u1ed1 v\u00e0 b\u00e0i t\u1eadp c\u00f3 l\u1eddi gi\u1ea3i.<\/span><\/p>\n

Ph\u00e2n s\u1ed1 l\u00e0 g\u00ec?<\/b><\/h2>\n

Ph\u00e2n s\u1ed1 c\u00f3 d\u1ea1ng a\/b, v\u1edbi a l\u00e0 t\u1eed s\u1ed1 (s\u1ed1 \u1edf tr\u00ean), b l\u00e0 m\u1eabu s\u1ed1 (s\u1ed1 \u1edf d\u01b0\u1edbi) v\u00e0 \u0111i\u1ec1u ki\u1ec7n b kh\u00e1c 0. Ph\u00e2n s\u1ed1 bi\u1ec3u th\u1ecb vi\u1ec7c chia m\u1ed9t \u0111\u01a1n v\u1ecb th\u00e0nh b ph\u1ea7n b\u1eb1ng nhau v\u00e0 l\u1ea5y a ph\u1ea7n. V\u00ed d\u1ee5, 3\/4 ngh\u0129a l\u00e0 chia th\u00e0nh 4 ph\u1ea7n b\u1eb1ng nhau v\u00e0 l\u1ea5y 3 ph\u1ea7n. M\u1ecdi s\u1ed1 nguy\u00ean c\u0169ng c\u00f3 th\u1ec3 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng ph\u00e2n s\u1ed1 c\u00f3 m\u1eabu s\u1ed1 b\u1eb1ng 1 (v\u00ed d\u1ee5 5 = 5\/1).<\/span><\/p>\n

\"phan-so-la-gi-3\"<\/p>\n

Ph\u00e2n s\u1ed1 b\u1eb1ng nhau<\/b><\/h2>\n

Hai ph\u00e2n s\u1ed1 a\/b v\u00e0 c\/d \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 b\u1eb1ng nhau n\u1ebfu a \u00d7 d = b \u00d7 c. T\u1eeb \u0111\u00f3 ta c\u00f3 t\u00ednh ch\u1ea5t c\u01a1 b\u1ea3n: n\u1ebfu nh\u00e2n (ho\u1eb7c chia) c\u1ea3 t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 c\u1ee7a m\u1ed9t ph\u00e2n s\u1ed1 v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 kh\u00e1c 0 th\u00ec \u0111\u01b0\u1ee3c m\u1ed9t ph\u00e2n s\u1ed1 b\u1eb1ng ph\u00e2n s\u1ed1 \u0111\u00e3 cho. V\u00ed d\u1ee5: 1\/2 = 2\/4 = 3\/6 v\u00ec t\u1ea5t c\u1ea3 \u0111\u1ec1u bi\u1ec3u th\u1ecb c\u00f9ng m\u1ed9t gi\u00e1 tr\u1ecb.<\/span><\/p>\n

R\u00fat g\u1ecdn ph\u00e2n s\u1ed1 l\u00e0 g\u00ec<\/b><\/h2>\n

R\u00fat g\u1ecdn ph\u00e2n s\u1ed1 l\u00e0 chia c\u1ea3 t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 cho m\u1ed9t \u01b0\u1edbc chung c\u1ee7a ch\u00fang \u0111\u1ec3 \u0111\u01b0\u1ee3c ph\u00e2n s\u1ed1 \u0111\u01a1n gi\u1ea3n h\u01a1n nh\u01b0ng v\u1eabn b\u1eb1ng ph\u00e2n s\u1ed1 ban \u0111\u1ea7u. Ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n l\u00e0 ph\u00e2n s\u1ed1 m\u00e0 t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 nguy\u00ean t\u1ed1 c\u00f9ng nhau (kh\u00f4ng r\u00fat g\u1ecdn \u0111\u01b0\u1ee3c n\u1eefa). V\u00ed d\u1ee5: 12\/18 = (12:6)\/(18:6) = 2\/3, v\u00e0 2\/3 l\u00e0 ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n.<\/span><\/p>\n

C\u00e1ch l\u00e0m t\u1eebng b\u01b0\u1edbc: (1) T\u00ecm \u01b0\u1edbc chung l\u1edbn nh\u1ea5t (\u01afCLN) c\u1ee7a t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1; (2) Chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu cho \u01afCLN \u0111\u00f3. V\u00ed d\u1ee5 r\u00fat g\u1ecdn 24\/36: \u01afCLN(24, 36) = 12, n\u00ean 24\/36 = (24:12)\/(36:12) = 2\/3.<\/span><\/p>\n

\"phan-so-la-gi-1\"<\/p>\n

Quy \u0111\u1ed3ng m\u1eabu s\u1ed1 c\u00e1c ph\u00e2n s\u1ed1<\/b><\/h2>\n

Quy \u0111\u1ed3ng m\u1eabu s\u1ed1 l\u00e0 bi\u1ebfn \u0111\u1ed5i c\u00e1c ph\u00e2n s\u1ed1 v\u1ec1 c\u00f9ng m\u1ed9t m\u1eabu s\u1ed1 chung \u0111\u1ec3 d\u1ec5 so s\u00e1nh v\u00e0 t\u00ednh to\u00e1n. M\u1eabu s\u1ed1 chung th\u01b0\u1eddng l\u00e0 b\u1ed9i chung nh\u1ecf nh\u1ea5t (BCNN) c\u1ee7a c\u00e1c m\u1eabu s\u1ed1. V\u00ed d\u1ee5, quy \u0111\u1ed3ng 1\/4 v\u00e0 1\/6: BCNN(4, 6) = 12, n\u00ean 1\/4 = 3\/12 v\u00e0 1\/6 = 2\/12.<\/span><\/p>\n

C\u00e1c b\u01b0\u1edbc quy \u0111\u1ed3ng: (1) T\u00ecm BCNN c\u1ee7a c\u00e1c m\u1eabu s\u1ed1 l\u00e0m m\u1eabu s\u1ed1 chung; (2) V\u1edbi m\u1ed7i ph\u00e2n s\u1ed1, l\u1ea5y m\u1eabu s\u1ed1 chung chia cho m\u1eabu s\u1ed1 c\u1ee7a ph\u00e2n s\u1ed1 \u0111\u00f3 \u0111\u1ec3 t\u00ecm th\u1eeba s\u1ed1 ph\u1ee5; (3) Nh\u00e2n c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a ph\u00e2n s\u1ed1 v\u1edbi th\u1eeba s\u1ed1 ph\u1ee5 t\u01b0\u01a1ng \u1ee9ng.<\/span><\/p>\n

So s\u00e1nh hai ph\u00e2n s\u1ed1 l\u00e0 g\u00ec<\/b><\/h2>\n

\u0110\u1ec3 so s\u00e1nh hai ph\u00e2n s\u1ed1, ta th\u1ef1c hi\u1ec7n nh\u01b0 sau:<\/span><\/p>\n

    \n
  • N\u1ebfu hai ph\u00e2n s\u1ed1 c\u00f9ng m\u1eabu s\u1ed1: Ph\u00e2n s\u1ed1 n\u00e0o c\u00f3 t\u1eed s\u1ed1 l\u1edbn h\u01a1n th\u00ec ph\u00e2n s\u1ed1 \u0111\u00f3 l\u1edbn h\u01a1n.<\/span><\/li>\n
  • N\u1ebfu hai ph\u00e2n s\u1ed1 kh\u00e1c m\u1eabu s\u1ed1: C\u00f3 th\u1ec3 \u00e1p d\u1ee5ng m\u1ed9t trong hai c\u00e1ch:<\/span>\n
      \n
    • Quy \u0111\u1ed3ng m\u1eabu s\u1ed1, r\u1ed3i so s\u00e1nh c\u00e1c t\u1eed s\u1ed1.<\/span><\/li>\n
    • Nh\u00e2n ch\u00e9o:<\/span>\n
        \n
      • N\u1ebfu a \u00d7 d > b \u00d7 c th\u00ec a\/b > c\/d.<\/span><\/li>\n
      • N\u1ebfu a \u00d7 d < b \u00d7 c th\u00ec a\/b < c\/d.<\/span><\/li>\n
      • N\u1ebfu a \u00d7 d = b \u00d7 c th\u00ec a\/b = c\/d.<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n

        V\u00ed d\u1ee5: So s\u00e1nh 2\/3 v\u00e0 3\/4. Ta c\u00f3:<\/span><\/p>\n

          \n
        • 2 \u00d7 4 = 8<\/span><\/li>\n
        • 3 \u00d7 3 = 9<\/span><\/li>\n
        • V\u00ec 8 < 9, suy ra 2\/3 < 3\/4.<\/span><\/li>\n<\/ul>\n

          \"phan-so-la-gi-2\"<\/p>\n

          C\u00e1c ph\u00e9p t\u00ednh v\u1edbi ph\u00e2n s\u1ed1<\/b><\/h2>\n

          Sau khi hi\u1ec3u b\u1ea3n ch\u1ea5t ph\u00e2n s\u1ed1, b\u1ea1n c\u1ea7n n\u1eafm b\u1ed1n ph\u00e9p t\u00ednh c\u01a1 b\u1ea3n. B\u1ea3ng d\u01b0\u1edbi \u0111\u00e2y t\u00f3m t\u1eaft quy t\u1eafc v\u00e0 v\u00ed d\u1ee5 ng\u1eafn:<\/span><\/p>\n\n\n\n\n\n\n\n\n\n
          Ph\u00e9p t\u00ednh<\/th>\nQuy t\u1eafc<\/th>\nV\u00ed d\u1ee5<\/th>\n<\/tr>\n<\/thead>\n
          C\u1ed9ng (c\u00f9ng m\u1eabu)<\/td>\nC\u1ed9ng t\u1eed, gi\u1eef nguy\u00ean m\u1eabu<\/td>\n2\/7 + 3\/7 = 5\/7<\/td>\n<\/tr>\n
          C\u1ed9ng (kh\u00e1c m\u1eabu)<\/td>\nQuy \u0111\u1ed3ng r\u1ed3i c\u1ed9ng t\u1eed<\/td>\n1\/2 + 1\/3 = 5\/6<\/td>\n<\/tr>\n
          Tr\u1eeb<\/td>\nQuy \u0111\u1ed3ng r\u1ed3i tr\u1eeb t\u1eed<\/td>\n3\/4 \u2212 1\/6 = 7\/12<\/td>\n<\/tr>\n
          Nh\u00e2n<\/td>\nNh\u00e2n t\u1eed v\u1edbi t\u1eed, m\u1eabu v\u1edbi m\u1eabu<\/td>\n2\/3 \u00d7 3\/5 = 6\/15 = 2\/5<\/td>\n<\/tr>\n
          Chia<\/td>\nNh\u00e2n v\u1edbi ph\u00e2n s\u1ed1 ngh\u1ecbch \u0111\u1ea3o<\/td>\n2\/3 : 4\/5 = 2\/3 \u00d7 5\/4 = 10\/12 = 5\/6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          V\u00ed d\u1ee5 c\u00f3 l\u1eddi gi\u1ea3i t\u1eebng b\u01b0\u1edbc \u2013 T\u00ednh 1\/2 + 1\/3:<\/span><\/p>\n

            \n
          • B\u01b0\u1edbc 1: T\u00ecm m\u1eabu s\u1ed1 chung. BCNN(2, 3) = 6.<\/span><\/li>\n
          • B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng: 1\/2 = 3\/6 v\u00e0 1\/3 = 2\/6.<\/span><\/li>\n
          • B\u01b0\u1edbc 3: C\u1ed9ng t\u1eed s\u1ed1: 3\/6 + 2\/6 = 5\/6. K\u1ebft qu\u1ea3 5\/6 \u0111\u00e3 t\u1ed1i gi\u1ea3n.<\/span><\/li>\n<\/ul>\n

            B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng c\u00f3 l\u1eddi gi\u1ea3i<\/b><\/h2>\n

            B\u00e0i 1: R\u00fat g\u1ecdn ph\u00e2n s\u1ed1 15\/20. L\u1eddi gi\u1ea3i: \u01afCLN(15, 20) = 5, n\u00ean 15\/20 = 3\/4.<\/span><\/p>\n

            B\u00e0i 2: Hai ph\u00e2n s\u1ed1 2\/3 v\u00e0 6\/9 c\u00f3 b\u1eb1ng nhau kh\u00f4ng? L\u1eddi gi\u1ea3i: 2 \u00d7 9 = 18 v\u00e0 3 \u00d7 6 = 18, b\u1eb1ng nhau n\u00ean hai ph\u00e2n s\u1ed1 b\u1eb1ng nhau.<\/span><\/p>\n

            B\u00e0i 3: Quy \u0111\u1ed3ng 2\/3 v\u00e0 1\/5. L\u1eddi gi\u1ea3i: BCNN(3, 5) = 15, n\u00ean 2\/3 = 10\/15 v\u00e0 1\/5 = 3\/15.<\/span><\/p>\n

            B\u00e0i 4: So s\u00e1nh 3\/5 v\u00e0 4\/7. L\u1eddi gi\u1ea3i: 3 \u00d7 7 = 21 v\u00e0 5 \u00d7 4 = 20; v\u00ec 21 > 20 n\u00ean 3\/5 > 4\/7.<\/span><\/p>\n

            B\u00e0i 5: T\u00ednh 3\/4 \u2212 1\/6. L\u1eddi gi\u1ea3i: BCNN(4, 6) = 12; 3\/4 = 9\/12 v\u00e0 1\/6 = 2\/12; 9\/12 \u2212 2\/12 = 7\/12.<\/span><\/p>\n

            B\u00e0i 6 (tr\u1eafc nghi\u1ec7m): Ph\u00e2n s\u1ed1 8\/12 r\u00fat g\u1ecdn th\u00e0nh ph\u00e2n s\u1ed1 n\u00e0o? A. 2\/3 B. 4\/6 C. 3\/4 D. 1\/2. \u0110\u00e1p \u00e1n: A (v\u00ec \u01afCLN(8, 12) = 4, 8\/12 = 2\/3).<\/span><\/p>\n

            C\u00e1c lo\u1ea1i ph\u00e2n s\u1ed1 v\u00e0 li\u00ean h\u1ec7 v\u1edbi h\u1ed7n s\u1ed1<\/b><\/h2>\n

            D\u1ef1a v\u00e0o gi\u00e1 tr\u1ecb c\u1ee7a t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1, ph\u00e2n s\u1ed1 \u0111\u01b0\u1ee3c chia th\u00e0nh c\u00e1c lo\u1ea1i sau:<\/span><\/p>\n

              \n
            • Ph\u00e2n s\u1ed1 nh\u1ecf h\u01a1n 1: T\u1eed s\u1ed1 nh\u1ecf h\u01a1n m\u1eabu s\u1ed1. V\u00ed d\u1ee5: 3\/4.<\/span><\/li>\n
            • Ph\u00e2n s\u1ed1 b\u1eb1ng 1: T\u1eed s\u1ed1 b\u1eb1ng m\u1eabu s\u1ed1. V\u00ed d\u1ee5: 5\/5 = 1.<\/span><\/li>\n
            • Ph\u00e2n s\u1ed1 l\u1edbn h\u01a1n 1: T\u1eed s\u1ed1 l\u1edbn h\u01a1n m\u1eabu s\u1ed1. Lo\u1ea1i ph\u00e2n s\u1ed1 n\u00e0y c\u00f3 th\u1ec3 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng h\u1ed7n s\u1ed1. V\u00ed d\u1ee5: 7\/2 = 3 1\/2.<\/span><\/li>\n<\/ul>\n

              L\u01b0u \u00fd: Vi\u1ec7c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa ph\u00e2n s\u1ed1 v\u00e0 h\u1ed7n s\u1ed1 gi\u00fap bi\u1ec3u di\u1ec5n s\u1ed1 m\u1ed9t c\u00e1ch tr\u1ef1c quan h\u01a1n v\u00e0 thu\u1eadn ti\u1ec7n khi th\u1ef1c hi\u1ec7n c\u00e1c ph\u00e9p t\u00ednh. Ch\u1eb3ng h\u1ea1n:<\/span><\/p>\n

                \n
              • 7\/2 = 3 1\/2<\/span><\/li>\n
              • 2 3\/5 = 13\/5<\/span><\/li>\n<\/ul>\n

                \"phan-so-la-gi-4\"<\/p>\n

                C\u00e2u h\u1ecfi th\u01b0\u1eddng g\u1eb7p v\u1ec1 ph\u00e2n s\u1ed1<\/b><\/h2>\n

                Ph\u00e2n s\u1ed1 l\u00e0 g\u00ec?<\/b><\/h3>\n

                L\u00e0 c\u00e1ch vi\u1ebft d\u1ea1ng a\/b v\u1edbi a l\u00e0 t\u1eed s\u1ed1, b l\u00e0 m\u1eabu s\u1ed1 v\u00e0 b kh\u00e1c 0.<\/span><\/p>\n

                Khi n\u00e0o hai ph\u00e2n s\u1ed1 b\u1eb1ng nhau?<\/b><\/h3>\n

                Khi a \u00d7 d = b \u00d7 c, t\u1ee9c t\u00edch ch\u00e9o b\u1eb1ng nhau.<\/span><\/p>\n

                Ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n l\u00e0 g\u00ec?<\/b><\/h3>\n

                L\u00e0 ph\u00e2n s\u1ed1 c\u00f3 t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 nguy\u00ean t\u1ed1 c\u00f9ng nhau, kh\u00f4ng r\u00fat g\u1ecdn \u0111\u01b0\u1ee3c n\u1eefa.<\/span><\/p>\n

                Quy \u0111\u1ed3ng m\u1eabu s\u1ed1 \u0111\u1ec3 l\u00e0m g\u00ec?<\/b><\/h3>\n

                \u0110\u1ec3 \u0111\u01b0a c\u00e1c ph\u00e2n s\u1ed1 v\u1ec1 c\u00f9ng m\u1eabu, thu\u1eadn ti\u1ec7n cho so s\u00e1nh v\u00e0 t\u00ednh c\u1ed9ng, tr\u1eeb.<\/span><\/p>\n

                C\u00e1ch so s\u00e1nh hai ph\u00e2n s\u1ed1 kh\u00e1c m\u1eabu?<\/b><\/h3>\n

                Quy \u0111\u1ed3ng m\u1eabu s\u1ed1 r\u1ed3i so s\u00e1nh t\u1eed, ho\u1eb7c d\u00f9ng c\u00e1ch nh\u00e2n ch\u00e9o.<\/span><\/p>\n

                Mu\u1ed1n chia hai ph\u00e2n s\u1ed1 th\u00ec l\u00e0m th\u1ebf n\u00e0o?<\/b><\/h3>\n

                L\u1ea5y ph\u00e2n s\u1ed1 th\u1ee9 nh\u1ea5t nh\u00e2n v\u1edbi ph\u00e2n s\u1ed1 ngh\u1ecbch \u0111\u1ea3o c\u1ee7a ph\u00e2n s\u1ed1 th\u1ee9 hai. V\u00ed d\u1ee5 2\/3 : 4\/5 = 2\/3 \u00d7 5\/4 = 5\/6.<\/span><\/p>\n