{"id":475,"date":"2025-07-23T03:58:39","date_gmt":"2025-07-23T03:58:39","guid":{"rendered":"https:\/\/giapsgk.com\/chuyen-de\/?p=475"},"modified":"2025-07-23T05:21:47","modified_gmt":"2025-07-23T05:21:47","slug":"tong-hop-cac-cong-thuc-toan-9-hoc-ki-1-va-hoc-ki-2","status":"publish","type":"post","link":"https:\/\/giapsgk.com\/chuyen-de\/tong-hop-cac-cong-thuc-toan-9-hoc-ki-1-va-hoc-ki-2\/","title":{"rendered":"T\u1ed5ng h\u1ee3p c\u00e1c c\u00f4ng th\u1ee9c To\u00e1n 9 h\u1ecdc k\u00ec 1 v\u00e0 h\u1ecdc k\u00ec 2"},"content":{"rendered":"

To\u00e1n l\u1edbp 9 l\u00e0 giai \u0111o\u1ea1n b\u1ea3n l\u1ec1, quy\u1ebft \u0111\u1ecbnh k\u1ebft qu\u1ea3 k\u1ef3 thi chuy\u1ec3n c\u1ea5p v\u00e0o l\u1edbp 10 \u2013 m\u1ed9t k\u1ef3 thi quan tr\u1ecdng \u0111\u1ed1i v\u1edbi m\u1ecdi h\u1ecdc sinh. Trong qu\u00e1 tr\u00ecnh \u00f4n luy\u1ec7n, vi\u1ec7c n\u1eafm v\u1eefng c\u00e1c c\u00f4ng th\u1ee9c To\u00e1n 9<\/strong> l\u00e0 y\u1ebfu t\u1ed1 then ch\u1ed1t gi\u00fap h\u1ecdc sinh gi\u1ea3i b\u00e0i nhanh, ch\u00ednh x\u00e1c v\u00e0 ti\u1ebft ki\u1ec7m th\u1eddi gian.<\/p>\n

B\u00e0i vi\u1ebft n\u00e0y s\u1ebd h\u1ec7 th\u1ed1ng l\u1ea1i to\u00e0n b\u1ed9 c\u00f4ng th\u1ee9c To\u00e1n l\u1edbp 9 theo t\u1eebng ch\u01b0\u01a1ng tr\u00ecnh \u0110\u1ea1i s\u1ed1 v\u00e0 H\u00ecnh h\u1ecdc, b\u00e1m s\u00e1t s\u00e1ch gi\u00e1o khoa hi\u1ec7n h\u00e0nh. T\u1eeb c\u00f4ng th\u1ee9c nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai, \u0111\u1ecbnh l\u00fd Viet \u0111\u1ebfn c\u00e1c h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng \u2013 t\u1ea5t c\u1ea3 \u0111\u1ec1u \u0111\u01b0\u1ee3c tr\u00ecnh b\u00e0y ng\u1eafn g\u1ecdn, d\u1ec5 h\u1ecdc, d\u1ec5 \u00e1p d\u1ee5ng.<\/p>\n

Ch\u01b0\u01a1ng I: Ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh<\/strong><\/h2>\n

1. Kh\u00e1i ni\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 h\u1ec7 hai ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n<\/strong><\/h3>\n

\u0110\u1ecbnh ngh\u0129a<\/strong>:<\/p>\n

    \n
  • Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n: $ax + by = c$.<\/li>\n
  • H\u1ec7 hai ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n: $a_1x + b_1y = c_1$ v\u00e0 $a_2x + b_2y = c_2$.<\/li>\n<\/ul>\n

    Ghi ch\u00fa quan tr\u1ecdng<\/strong>: H\u1ec7 c\u00f3 nghi\u1ec7m n\u1ebfu h\u1ec7 s\u1ed1 kh\u00f4ng \u0111\u1ed3ng nh\u1ea5t v\u00e0 \u0111\u1ecbnh th\u1ee9c $a_1b_2 – a_2b_1 \\neq 0$.<\/p>\n

    C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

      \n
    • Nghi\u1ec7m c\u1ee7a h\u1ec7: T\u00ecm $x, y$ th\u1ecfa m\u00e3n c\u1ea3 hai ph\u01b0\u01a1ng tr\u00ecnh.<\/li>\n<\/ul>\n

      2. Gi\u1ea3i h\u1ec7 hai ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n<\/strong><\/h3>\n

      \u0110\u1ecbnh ngh\u0129a<\/strong>: Gi\u1ea3i h\u1ec7 b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p th\u1ebf, ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng \u0111\u1ea1i s\u1ed1.<\/p>\n

      Ghi ch\u00fa quan tr\u1ecdng<\/strong>:<\/p>\n

        \n
      • Ph\u01b0\u01a1ng ph\u00e1p th\u1ebf: T\u1eeb m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh, bi\u1ec3u di\u1ec5n $x$ ho\u1eb7c $y$, thay v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh c\u00f2n l\u1ea1i.<\/li>\n
      • Ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng: Nh\u00e2n c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 lo\u1ea1i bi\u1ebfn.<\/li>\n<\/ul>\n

        C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

          \n
        • Nghi\u1ec7m: $x = \\frac{c_1b_2 – c_2b_1}{a_1b_2 – a_2b_1}$, $y = \\frac{a_1c_2 – a_2c_1}{a_1b_2 – a_2b_1}$ (n\u1ebfu $a_1b_2 – a_2b_1 \\neq 0$).<\/li>\n<\/ul>\n

          3. Gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh<\/strong><\/h3>\n

          \u0110\u1ecbnh ngh\u0129a<\/strong>: L\u1eadp h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh t\u1eeb b\u00e0i to\u00e1n th\u1ef1c t\u1ebf.<\/p>\n

          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: X\u00e1c \u0111\u1ecbnh hai \u1ea9n s\u1ed1 v\u00e0 m\u1ed1i quan h\u1ec7 gi\u1eefa ch\u00fang.<\/p>\n

          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>: X\u00e2y d\u1ef1ng h\u1ec7 $ax + by = m$, $cx + dy = n$ d\u1ef1a tr\u00ean d\u1eef ki\u1ec7n.<\/p>\n

          Ch\u01b0\u01a1ng II: Ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n<\/strong><\/h2>\n

          4. Ph\u01b0\u01a1ng tr\u00ecnh quy v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n<\/strong><\/h3>\n

          \u0110\u1ecbnh ngh\u0129a<\/strong>: Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 th\u1ec3 bi\u1ebfn \u0111\u1ed5i v\u1ec1 d\u1ea1ng $ax + b = 0$.<\/p>\n

          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: Lo\u1ea1i ph\u00e2n s\u1ed1, c\u0103n, ho\u1eb7c l\u0169y th\u1eeba \u0111\u1ec3 quy v\u1ec1 d\u1ea1ng \u0111\u01a1n gi\u1ea3n.<\/p>\n

          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>: Nghi\u1ec7m: $x = -\\frac{b}{a}$ (n\u1ebfu $a \\neq 0$).<\/p>\n

          5. B\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u00e0 t\u00ednh ch\u1ea5t<\/strong><\/h3>\n

          \u0110\u1ecbnh ngh\u0129a<\/strong>: B\u1ea5t \u0111\u1eb3ng th\u1ee9c l\u00e0 m\u1ed1i quan h\u1ec7 $a > b$ (ho\u1eb7c $\\geq, <, \\leq$).<\/p>\n

          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: T\u00ednh ch\u1ea5t: N\u1ebfu $a > b$ th\u00ec $a + c > b + c$, $a \\cdot c > b \\cdot c$ (n\u1ebfu $c > 0$).<\/p>\n

          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

            \n
          • $a > b \\implies a + c > b + c$.<\/li>\n
          • $a > b$ v\u00e0 $c > 0 \\implies ac > bc$.<\/li>\n<\/ul>\n

            6. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n<\/strong><\/h3>\n

            \u0110\u1ecbnh ngh\u0129a<\/strong>: B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $ax + b > c$ (ho\u1eb7c $\\geq, <, \\leq$).<\/p>\n

            Ghi ch\u00fa quan tr\u1ecdng<\/strong>: Khi nh\u00e2n\/chia c\u1ea3 v\u1ebf b\u1eb1ng s\u1ed1 \u00e2m, \u0111\u1ed5i d\u1ea5u b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh.<\/p>\n

            C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>: Nghi\u1ec7m: $x > \\frac{c – b}{a}$ (n\u1ebfu $a > 0$); $x < \\frac{c – b}{a}$ (n\u1ebfu $a < 0$).<\/p>\n

            Ch\u01b0\u01a1ng III: C\u0103n b\u1eadc hai v\u00e0 c\u0103n b\u1eadc ba<\/strong><\/h2>\n

            7. C\u0103n b\u1eadc hai v\u00e0 c\u0103n th\u1ee9c b\u1eadc hai<\/strong><\/h3>\n

            \u0110\u1ecbnh ngh\u0129a<\/strong>: C\u0103n b\u1eadc hai c\u1ee7a $a$ l\u00e0 $x$ sao cho $x^2 = a$ ($a \\geq 0$).<\/p>\n

            Ghi ch\u00fa quan tr\u1ecdng<\/strong>: $\\sqrt{a} \\geq 0$, $\\sqrt{a} \\cdot \\sqrt{b} = \\sqrt{ab}$ (n\u1ebfu $a, b \\geq 0$).<\/p>\n

            C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

              \n
            • $\\sqrt{a^2} = |a|$.<\/li>\n
            • $(\\sqrt{a})^2 = a$ (n\u1ebfu $a \\geq 0$).<\/li>\n<\/ul>\n

              8. Khai c\u0103n b\u1eadc hai v\u1edbi ph\u00e9p nh\u00e2n v\u00e0 ph\u00e9p chia<\/strong><\/h3>\n

              \u0110\u1ecbnh ngh\u0129a<\/strong>: $\\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}}$ (n\u1ebfu $a, b > 0$).<\/p>\n

              Ghi ch\u00fa quan tr\u1ecdng<\/strong>: R\u00fat g\u1ecdn ph\u00e2n th\u1ee9c ch\u1ee9a c\u0103n.<\/p>\n

              C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

                \n
              • $\\sqrt{ab} = \\sqrt{a} \\cdot \\sqrt{b}$ (n\u1ebfu $a, b \\geq 0$).<\/li>\n
              • $\\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}}$ (n\u1ebfu $b \\neq 0$).<\/li>\n<\/ul>\n

                9. Bi\u1ebfn \u0111\u1ed5i \u0111\u01a1n gi\u1ea3n v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c ch\u1ee9a c\u0103n th\u1ee9c b\u1eadc hai<\/strong><\/h3>\n

                \u0110\u1ecbnh ngh\u0129a<\/strong>: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c nh\u01b0 $\\sqrt{a + b} + \\sqrt{a – b}$.<\/p>\n

                Ghi ch\u00fa quan tr\u1ecdng<\/strong>: Lo\u1ea1i m\u1eabu c\u00f3 ch\u1ee9a c\u0103n.<\/p>\n

                C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\u00a0$\\frac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\frac{a}{b}}$ (n\u1ebfu $a, b > 0$).<\/p>\n

                10. C\u0103n b\u1eadc ba v\u00e0 c\u0103n th\u1ee9c b\u1eadc ba<\/strong><\/h3>\n

                \u0110\u1ecbnh ngh\u0129a<\/strong>: C\u0103n b\u1eadc ba c\u1ee7a $a$ l\u00e0 $x$ sao cho $x^3 = a$.<\/p>\n

                Ghi ch\u00fa quan tr\u1ecdng<\/strong>: $\\sqrt[3]{a}$ t\u1ed3n t\u1ea1i v\u1edbi m\u1ecdi $a$ th\u1ef1c.<\/p>\n

                C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

                  \n
                • $\\sqrt[3]{ab} = \\sqrt[3]{a} \\cdot \\sqrt[3]{b}$.<\/li>\n
                • $\\sqrt[3]{\\frac{a}{b}} = \\frac{\\sqrt[3]{a}}{\\sqrt[3]{b}}$ (n\u1ebfu $b \\neq 0$).<\/li>\n<\/ul>\n

                  Ch\u01b0\u01a1ng IV: H\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng<\/strong><\/h2>\n

                  11. T\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn<\/strong><\/h3>\n

                  \u0110\u1ecbnh ngh\u0129a<\/strong>: $\\sin A = \\frac{a}{c}$, $\\cos A = \\frac{b}{c}$, $\\tan A = \\frac{a}{b}$ (v\u1edbi $a, b, c$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 c\u1ea1nh \u0111\u1ed1i, c\u1ea1nh k\u1ec1, c\u1ea1nh huy\u1ec1n).<\/p>\n

                  Ghi ch\u00fa quan tr\u1ecdng<\/strong>: $\\tan A = \\frac{\\sin A}{\\cos A}$.<\/p>\n

                  C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

                    \n
                  • $\\sin^2 A + \\cos^2 A = 1$.<\/li>\n
                  • $\\tan A = \\frac{\\sin A}{\\cos A}$.<\/li>\n<\/ul>\n

                    12. M\u1ed9t s\u1ed1 h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh, g\u00f3c trong tam gi\u00e1c vu\u00f4ng v\u00e0 \u1ee9ng d\u1ee5ng<\/strong><\/h3>\n

                    \u0110\u1ecbnh ngh\u0129a<\/strong>: \u0110\u1ecbnh l\u00fd Pytago: $a^2 + b^2 = c^2$.<\/p>\n

                    Ghi ch\u00fa quan tr\u1ecdng<\/strong>: \u1ee8ng d\u1ee5ng trong t\u00ednh to\u00e1n c\u1ea1nh, g\u00f3c.<\/p>\n

                    C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

                      \n
                    • $a^2 + b^2 = c^2$.<\/li>\n
                    • Di\u1ec7n t\u00edch: $S = \\frac{1}{2}ab$.<\/li>\n<\/ul>\n

                      Ch\u01b0\u01a1ng V: \u0110\u01b0\u1eddng tr\u00f2n<\/strong><\/h2>\n

                      13. M\u1edf \u0111\u1ea7u v\u1ec1 \u0111\u01b0\u1eddng tr\u00f2n<\/strong><\/h3>\n

                      \u0110\u1ecbnh ngh\u0129a<\/strong>: \u0110\u01b0\u1eddng tr\u00f2n l\u00e0 t\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m c\u00e1ch t\u00e2m m\u1ed9t kho\u1ea3ng b\u1eb1ng b\u00e1n k\u00ednh $r$.<\/p>\n

                      Ghi ch\u00fa quan tr\u1ecdng<\/strong>: Chu vi $C = 2\\pi r$, di\u1ec7n t\u00edch $S = \\pi r^2$.<\/p>\n

                      C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

                        \n
                      • $C = 2\\pi r$.<\/li>\n
                      • $S = \\pi r^2$.<\/li>\n<\/ul>\n

                        14. Cung v\u00e0 d\u00e2y c\u1ee7a m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n<\/strong><\/h3>\n

                        \u0110\u1ecbnh ngh\u0129a<\/strong>: Cung l\u00e0 ph\u1ea7n cung tr\u00f2n, d\u00e2y l\u00e0 \u0111o\u1ea1n th\u1eb3ng n\u1ed1i hai \u0111i\u1ec3m tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n.<\/p>\n

                        Ghi ch\u00fa quan tr\u1ecdng<\/strong>: G\u00f3c \u1edf t\u00e2m li\u00ean quan \u0111\u1ebfn cung.<\/p>\n

                        C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>: \u0110\u1ed9 d\u00e0i cung: $l = \\frac{\\theta}{360} \\cdot 2\\pi r$ (g\u00f3c $\\theta$ t\u00ednh b\u1eb1ng \u0111\u1ed9).<\/p>\n

                        15. \u0110\u1ed9 d\u00e0i c\u1ee7a cung tr\u00f2n. Di\u1ec7n t\u00edch h\u00ecnh qu\u1ea1t tr\u00f2n v\u00e0 h\u00ecnh v\u00e0nh khuy\u00ean<\/strong><\/h3>\n

                        \u0110\u1ecbnh ngh\u0129a<\/strong>: H\u00ecnh qu\u1ea1t tr\u00f2n l\u00e0 ph\u1ea7n \u0111\u01b0\u1eddng tr\u00f2n gi\u1edbi h\u1ea1n b\u1edfi hai b\u00e1n k\u00ednh v\u00e0 cung.<\/p>\n

                        Ghi ch\u00fa quan tr\u1ecdng<\/strong>: H\u00ecnh v\u00e0nh khuy\u00ean l\u00e0 ph\u1ea7n gi\u1eefa hai \u0111\u01b0\u1eddng tr\u00f2n \u0111\u1ed3ng t\u00e2m.<\/p>\n

                        C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n

                          \n
                        • \u0110\u1ed9 d\u00e0i cung: $l = r \\theta$ (n\u1ebfu $\\theta$ t\u00ednh b\u1eb1ng radian).<\/li>\n
                        • Di\u1ec7n t\u00edch h\u00ecnh qu\u1ea1t: $S = \\frac{\\theta}{360} \\cdot \\pi r^2$.<\/li>\n
                        • Di\u1ec7n t\u00edch v\u00e0nh khuy\u00ean: $S = \\pi (R^2 – r^2)$.<\/li>\n<\/ul>\n

                          16. V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n<\/strong><\/h3>\n

                          \u0110\u1ecbnh ngh\u0129a<\/strong>: \u0110\u01b0\u1eddng th\u1eb3ng c\u1eaft, ti\u1ebfp x\u00fac, ho\u1eb7c n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n.<\/p>\n

                          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: D\u1ef1a v\u00e0o kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng.<\/p>\n

                          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>: Kho\u1ea3ng c\u00e1ch: $d = \\frac{|ax_0 + by_0 + c|}{\\sqrt{a^2 + b^2}}$, v\u1edbi $(x_0, y_0)$ l\u00e0 t\u00e2m.<\/p>\n

                          17. V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n<\/strong><\/h3>\n

                          \u0110\u1ecbnh ngh\u0129a<\/strong>: Hai \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau, ti\u1ebfp x\u00fac, ho\u1eb7c n\u1eb1m ngo\u00e0i nhau.<\/p>\n

                          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: D\u1ef1a v\u00e0o kho\u1ea3ng c\u00e1ch gi\u1eefa hai t\u00e2m v\u00e0 t\u1ed5ng\/hi\u1ec7u b\u00e1n k\u00ednh.<\/p>\n

                          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>: $|d – (r_1 + r_2)|$ x\u00e1c \u0111\u1ecbnh v\u1ecb tr\u00ed.<\/p>\n

                          Ch\u01b0\u01a1ng VI: H\u00e0m s\u1ed1 $y = ax^2$ ($a \\neq 0$) v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u1ed9t \u1ea9n<\/strong><\/h2>\n

                          18. H\u00e0m s\u1ed1 $y = ax^2$ ($a \\neq 0$)<\/strong><\/h3>\n

                          \u0110\u1ecbnh ngh\u0129a<\/strong>: H\u00e0m s\u1ed1 $y = ax^2$ v\u1edbi $a \\neq 0$.<\/p>\n

                          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: \u0110\u1ed3 th\u1ecb l\u00e0 parabol, \u0111\u1ec9nh t\u1ea1i $(0, 0)$ n\u1ebfu $a > 0$ m\u1edf l\u00ean, $a < 0$ m\u1edf xu\u1ed1ng.<\/p>\n

                          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:\u00a0$y = ax^2$.<\/p>\n

                          19. Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u1ed9t \u1ea9n<\/strong><\/h3>\n

                          \u0110\u1ecbnh ngh\u0129a<\/strong>: Ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $ax^2 + bx + c = 0$ ($a \\neq 0$).<\/p>\n

                          Ghi ch\u00fa quan tr\u1ecdng<\/strong>: S\u1ed1 nghi\u1ec7m ph\u1ee5 thu\u1ed9c v\u00e0o $\\Delta = b^2 – 4ac$.<\/p>\n

                          C\u00f4ng th\u1ee9c c\u1ea7n nh\u1edb<\/strong>:<\/p>\n