1) x²+3x-18-05) x²+5x-6 24+2(513-4ابر 2) x246x-27-06) x²-9x-1810-8-614) 3x+18 3) x²-74-18-07-13x-22-11-7-8 15)x+20 4)-9-10-08 2-9-2012) •6-516-12

Привет! Смотри, тут у нас уравнения. Сейчас разберёмся, как их решать: 1) \(x^2 + 3x - 18 = 0\) \(D = 3^2 - 4 \cdot 1 \cdot (-18) = 9 + 72 = 81\) \(x_1 = \frac{-3 + 9}{2} = 3\) \(x_2 = \frac{-3 - 9}{2} = -6\) 2) \(x^2 + 6x - 27 = 0\) \(D = 6^2 - 4 \cdot 1 \cdot (-27) = 36 + 108 = 144\) \(x_1 = \frac{-6 + 12}{2} = 3\) \(x_2 = \frac{-6 - 12}{2} = -9\) 3) \(x^2 - 7x - 18 = 0\) \(D = (-7)^2 - 4 \cdot 1 \cdot (-18) = 49 + 72 = 121\) \(x_1 = \frac{7 + 11}{2} = 9\) \(x_2 = \frac{7 - 11}{2} = -2\) 4) \(x^2 - 9x - 10 = 0\) \(D = (-9)^2 - 4 \cdot 1 \cdot (-10) = 81 + 40 = 121\) \(x_1 = \frac{9 + 11}{2} = 10\) \(x_2 = \frac{9 - 11}{2} = -1\) 5) \(x^2 + 5x = -6\) \(\Rightarrow x^2 + 5x + 6 = 0\) \(D = 5^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1\) \(x_1 = \frac{-5 + 1}{2} = -2\) \(x_2 = \frac{-5 - 1}{2} = -3\) 6) \(x^2 - 9x = -18\) \(\Rightarrow x^2 - 9x + 18 = 0\) \(D = (-9)^2 - 4 \cdot 1 \cdot 18 = 81 - 72 = 9\) \(x_1 = \frac{9 + 3}{2} = 6\) \(x_2 = \frac{9 - 3}{2} = 3\) 9) \(x^2 + 4 = 5x\) \(\Rightarrow x^2 - 5x + 4 = 0\) \(D = (-5)^2 - 4 \cdot 1 \cdot 4 = 25 - 16 = 9\) \(x_1 = \frac{5 + 3}{2} = 4\) \(x_2 = \frac{5 - 3}{2} = 1\) 10) \(x^2 + 8 = 6x\) \(\Rightarrow x^2 - 6x + 8 = 0\) \(D = (-6)^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4\) \(x_1 = \frac{6 + 2}{2} = 4\) \(x_2 = \frac{6 - 2}{2} = 2\) 11) \(x^2 + 7 = 8x\) \(\Rightarrow x^2 - 8x + 7 = 0\) \(D = (-8)^2 - 4 \cdot 1 \cdot 7 = 64 - 28 = 36\) \(x_1 = \frac{8 + 6}{2} = 7\) \(x_2 = \frac{8 - 6}{2} = 1\) 12) \(x^2 + 6 = 5x\) \(\Rightarrow x^2 - 5x + 6 = 0\) \(D = (-5)^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1\) \(x_1 = \frac{5 + 1}{2} = 3\) \(x_2 = \frac{5 - 1}{2} = 2\) 7) \(x^2 - 13x = -22\) \(\Rightarrow x^2 - 13x + 22 = 0\) \(D = (-13)^2 - 4 \cdot 1 \cdot 22 = 169 - 88 = 81\) \(x_1 = \frac{13 + 9}{2} = 11\) \(x_2 = \frac{13 - 9}{2} = 2\) 8) \(x^2 - 9x = -20\) \(\Rightarrow x^2 - 9x + 20 = 0\) \(D = (-9)^2 - 4 \cdot 1 \cdot 20 = 81 - 80 = 1\) \(x_1 = \frac{9 + 1}{2} = 5\) \(x_2 = \frac{9 - 1}{2} = 4\) 13) \(x^2 = -2x + 24\) \(\Rightarrow x^2 + 2x - 24 = 0\) \(D = 2^2 - 4 \cdot 1 \cdot (-24) = 4 + 96 = 100\) \(x_1 = \frac{-2 + 10}{2} = 4\) \(x_2 = \frac{-2 - 10}{2} = -6\) 14) \(x^2 = 3x + 18\) \(\Rightarrow x^2 - 3x - 18 = 0\) \(D = (-3)^2 - 4 \cdot 1 \cdot (-18) = 9 + 72 = 81\) \(x_1 = \frac{3 + 9}{2} = 6\) \(x_2 = \frac{3 - 9}{2} = -3\) 15) \(x^2 = -x + 20\) \(\Rightarrow x^2 + x - 20 = 0\) \(D = 1^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81\) \(x_1 = \frac{-1 + 9}{2} = 4\) \(x_2 = \frac{-1 - 9}{2} = -5\) 16) \(x^2 = x + 12\) \(\Rightarrow x^2 - x - 12 = 0\) \(D = (-1)^2 - 4 \cdot 1 \cdot (-12) = 1 + 48 = 49\) \(x_1 = \frac{1 + 7}{2} = 4\) \(x_2 = \frac{1 - 7}{2} = -3\) В уравнениях я использовала дискриминант (\(D\)) и формулу для нахождения корней квадратного уравнения: \(x = \frac{-b \pm \sqrt{D}}{2a}\) Надеюсь, теперь тебе всё понятно!