• a) (x - 2)² = 3x - 8; 6) (3x - 1)(x + 3) + 1 = x(1 + 6x); в) 5(x + 2)2 = -6x + 44; г) (x + 4)(2x – 1) = x(3x + 11). a) (4x - 5)² - (2x + 3)² = 0; б) (x² + 4x + 11)² = (7x² + 2x + 3)²; в) (4х - 5)² + (2x + 3)² = 0; г) (x² + 4x + 11)² = (3x + 1)4. • a) x² + 3√2x + 4 = 0; B) x² - 3√5x - 20 = C 6) 4x² + 4√3x + 1 = 0; г) 4x² - 2√7x + 1 = 0 a) (2x - 1)(2x + 1) + x(x - 1) = 2x(x + 1); б) (3x + 1)² - x(7x + 5) = 4; в) (3x - 1)(3x + 1) - 2x(1 + 4x) = -2; г) (2x + 1)² + 2 = 2 - 6x2. B) 2x2 + x = 4x - 2; a) x²-x = 2x + 4; 3 5 5 3 б) x²-3 - 6x = 5; г) 4x2 + x - 5x - 1 = x² + 2 3 6 9 • Найдите абсциссы общих точек графиков функц их построения: 11-11 = 5x - 4; - 3 и у = -x² - 5x - 4; =-4x-1:

Давай разберем по порядку каждое уравнение и решим их. 1. a) (x - 2)² = 3x - 8 \[(x - 2)^2 = 3x - 8\] \[x^2 - 4x + 4 = 3x - 8\] \[x^2 - 7x + 12 = 0\] Используем дискриминант для решения квадратного уравнения: \[D = (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1\] \[x_1 = \frac{7 + 1}{2} = 4\] \[x_2 = \frac{7 - 1}{2} = 3\] б) (3x - 1)(x + 3) + 1 = x(1 + 6x) \[(3x - 1)(x + 3) + 1 = x(1 + 6x)\] \[3x^2 + 9x - x - 3 + 1 = x + 6x^2\] \[3x^2 + 8x - 2 = x + 6x^2\] \[3x^2 - 7x + 2 = 0\] Используем дискриминант: \[D = (-7)^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25\] \[x_1 = \frac{7 + 5}{6} = 2\] \[x_2 = \frac{7 - 5}{6} = \frac{1}{3}\] в) 5(x + 2)² = -6x + 44 \[5(x + 2)^2 = -6x + 44\] \[5(x^2 + 4x + 4) = -6x + 44\] \[5x^2 + 20x + 20 = -6x + 44\] \[5x^2 + 26x - 24 = 0\] Используем дискриминант: \[D = (26)^2 - 4 \cdot 5 \cdot (-24) = 676 + 480 = 1156\] \[x_1 = \frac{-26 + 34}{10} = \frac{8}{10} = \frac{4}{5}\] \[x_2 = \frac{-26 - 34}{10} = -6\] г) (x + 4)(2x – 1) = x(3x + 11) \[(x + 4)(2x - 1) = x(3x + 11)\] \[2x^2 - x + 8x - 4 = 3x^2 + 11x\] \[2x^2 + 7x - 4 = 3x^2 + 11x\] \[x^2 + 4x + 4 = 0\] \[(x + 2)^2 = 0\] \[x = -2\] Теперь перейдем ко второму блоку уравнений: a) (4x - 5)² - (2x + 3)² = 0 \[(4x - 5)^2 - (2x + 3)^2 = 0\] \[16x^2 - 40x + 25 - (4x^2 + 12x + 9) = 0\] \[12x^2 - 52x + 16 = 0\] \[3x^2 - 13x + 4 = 0\] Используем дискриминант: \[D = (-13)^2 - 4 \cdot 3 \cdot 4 = 169 - 48 = 121\] \[x_1 = \frac{13 + 11}{6} = 4\] \[x_2 = \frac{13 - 11}{6} = \frac{1}{3}\] б) (x² + 4x + 11)² = (7x² + 2x + 3)² \[(x^2 + 4x + 11)^2 = (7x^2 + 2x + 3)^2\] \[x^2 + 4x + 11 = \pm (7x^2 + 2x + 3)\] Случай 1: \[x^2 + 4x + 11 = 7x^2 + 2x + 3\] \[6x^2 - 2x - 8 = 0\] \[3x^2 - x - 4 = 0\] \[D = (-1)^2 - 4 \cdot 3 \cdot (-4) = 1 + 48 = 49\] \[x_1 = \frac{1 + 7}{6} = \frac{4}{3}\] \[x_2 = \frac{1 - 7}{6} = -1\] Случай 2: \[x^2 + 4x + 11 = - (7x^2 + 2x + 3)\] \[x^2 + 4x + 11 = -7x^2 - 2x - 3\] \[8x^2 + 6x + 14 = 0\] \[4x^2 + 3x + 7 = 0\] \[D = 3^2 - 4 \cdot 4 \cdot 7 = 9 - 112 = -103\] Так как дискриминант отрицательный, вещественных корней нет. в) (4х - 5)² + (2x + 3)² = 0 \[(4x - 5)^2 + (2x + 3)^2 = 0\] \[16x^2 - 40x + 25 + 4x^2 + 12x + 9 = 0\] \[20x^2 - 28x + 34 = 0\] \[10x^2 - 14x + 17 = 0\] \[D = (-14)^2 - 4 \cdot 10 \cdot 17 = 196 - 680 = -484\] Так как дискриминант отрицательный, вещественных корней нет. г) (x² + 4x + 11)² = (3x + 1)⁴ \[(x^2 + 4x + 11)^2 = ((3x + 1)^2)^2\] \[(x^2 + 4x + 11)^2 = (3x + 1)^4\] \[x^2 + 4x + 11 = \pm (3x + 1)^2\] Случай 1: \[x^2 + 4x + 11 = (3x + 1)^2\] \[x^2 + 4x + 11 = 9x^2 + 6x + 1\] \[8x^2 + 2x - 10 = 0\] \[4x^2 + x - 5 = 0\] \[D = 1^2 - 4 \cdot 4 \cdot (-5) = 1 + 80 = 81\] \[x_1 = \frac{-1 + 9}{8} = 1\] \[x_2 = \frac{-1 - 9}{8} = -\frac{5}{4}\] Случай 2: \[x^2 + 4x + 11 = - (3x + 1)^2\] \[x^2 + 4x + 11 = - (9x^2 + 6x + 1)\] \[x^2 + 4x + 11 = -9x^2 - 6x - 1\] \[10x^2 + 10x + 12 = 0\] \[5x^2 + 5x + 6 = 0\] \[D = 5^2 - 4 \cdot 5 \cdot 6 = 25 - 120 = -95\] Так как дискриминант отрицательный, вещественных корней нет. Третий блок уравнений: a) x² + 3√2x + 4 = 0 \[x^2 + 3\sqrt{2}x + 4 = 0\] \[D = (3\sqrt{2})^2 - 4 \cdot 1 \cdot 4 = 18 - 16 = 2\] \[x_1 = \frac{-3\sqrt{2} + \sqrt{2}}{2} = -\sqrt{2}\] \[x_2 = \frac{-3\sqrt{2} - \sqrt{2}}{2} = -2\sqrt{2}\] б) 4x² + 4√3x + 1 = 0 \[4x^2 + 4\sqrt{3}x + 1 = 0\] \[D = (4\sqrt{3})^2 - 4 \cdot 4 \cdot 1 = 48 - 16 = 32\] \[x_1 = \frac{-4\sqrt{3} + 4\sqrt{2}}{8} = \frac{-\sqrt{3} + \sqrt{2}}{2}\] \[x_2 = \frac{-4\sqrt{3} - 4\sqrt{2}}{8} = \frac{-\sqrt{3} - \sqrt{2}}{2}\] в) x² - 3√5x - 20 = 0 \[x^2 - 3\sqrt{5}x - 20 = 0\] \[D = (-3\sqrt{5})^2 - 4 \cdot 1 \cdot (-20) = 45 + 80 = 125\] \[x_1 = \frac{3\sqrt{5} + \sqrt{125}}{2} = \frac{3\sqrt{5} + 5\sqrt{5}}{2} = 4\sqrt{5}\] \[x_2 = \frac{3\sqrt{5} - 5\sqrt{5}}{2} = - \sqrt{5}\] г) 4x² - 2√7x + 1 = 0 \[4x^2 - 2\sqrt{7}x + 1 = 0\] \[D = (-2\sqrt{7})^2 - 4 \cdot 4 \cdot 1 = 28 - 16 = 12\] \[x_1 = \frac{2\sqrt{7} + 2\sqrt{3}}{8} = \frac{\sqrt{7} + \sqrt{3}}{4}\] \[x_2 = \frac{2\sqrt{7} - 2\sqrt{3}}{8} = \frac{\sqrt{7} - \sqrt{3}}{4}\] Четвертый блок уравнений: a) (2x - 1)(2x + 1) + x(x - 1) = 2x(x + 1) \[(2x - 1)(2x + 1) + x(x - 1) = 2x(x + 1)\] \[4x^2 - 1 + x^2 - x = 2x^2 + 2x\] \[5x^2 - x - 1 = 2x^2 + 2x\] \[3x^2 - 3x - 1 = 0\] \[D = (-3)^2 - 4 \cdot 3 \cdot (-1) = 9 + 12 = 21\] \[x_1 = \frac{3 + \sqrt{21}}{6}\] \[x_2 = \frac{3 - \sqrt{21}}{6}\] б) (3x + 1)² - x(7x + 5) = 4 \[(3x + 1)^2 - x(7x + 5) = 4\] \[9x^2 + 6x + 1 - 7x^2 - 5x = 4\] \[2x^2 + x - 3 = 0\] \[D = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25\] \[x_1 = \frac{-1 + 5}{4} = 1\] \[x_2 = \frac{-1 - 5}{4} = -\frac{3}{2}\] в) (3x - 1)(3x + 1) - 2x(1 + 4x) = -2 \[(3x - 1)(3x + 1) - 2x(1 + 4x) = -2\] \[9x^2 - 1 - 2x - 8x^2 = -2\] \[x^2 - 2x + 1 = 0\] \[(x - 1)^2 = 0\] \[x = 1\] г) (2x + 1)² + 2 = 2 - 6x² \[(2x + 1)^2 + 2 = 2 - 6x^2\] \[4x^2 + 4x + 1 + 2 = 2 - 6x^2\] \[10x^2 + 4x + 1 = 0\] \[D = 4^2 - 4 \cdot 10 \cdot 1 = 16 - 40 = -24\] Так как дискриминант отрицательный, вещественных корней нет. Пятый блок уравнений: a) (x² - x) / 3 = (2x + 4) / 5 \[\frac{x^2 - x}{3} = \frac{2x + 4}{5}\] \[5(x^2 - x) = 3(2x + 4)\] \[5x^2 - 5x = 6x + 12\] \[5x^2 - 11x - 12 = 0\] \[D = (-11)^2 - 4 \cdot 5 \cdot (-12) = 121 + 240 = 361\] \[x_1 = \frac{11 + 19}{10} = 3\] \[x_2 = \frac{11 - 19}{10} = -\frac{4}{5}\] б) (x² - 3) / 2 - 6x = 5 \[\frac{x^2 - 3}{2} - 6x = 5\] \[x^2 - 3 - 12x = 10\] \[x^2 - 12x - 13 = 0\] \[D = (-12)^2 - 4 \cdot 1 \cdot (-13) = 144 + 52 = 196\] \[x_1 = \frac{12 + 14}{2} = 13\] \[x_2 = \frac{12 - 14}{2} = -1\] в) (2x² + x) / 5 = (4x - 2) / 3 \[\frac{2x^2 + x}{5} = \frac{4x - 2}{3}\] \[3(2x^2 + x) = 5(4x - 2)\] \[6x^2 + 3x = 20x - 10\] \[6x^2 - 17x + 10 = 0\] \[D = (-17)^2 - 4 \cdot 6 \cdot 10 = 289 - 240 = 49\] \[x_1 = \frac{17 + 7}{12} = 2\] \[x_2 = \frac{17 - 7}{12} = \frac{5}{6}\] г) (4x² + x) / 3 - (5x - 1) / 6 = (x² + 9) / 6 \[\frac{4x^2 + x}{3} - \frac{5x - 1}{6} = \frac{x^2 + 9}{6}\] \[2(4x^2 + x) - (5x - 1) = x^2 + 9\] \[8x^2 + 2x - 5x + 1 = x^2 + 9\] \[7x^2 - 3x - 8 = 0\] \[D = (-3)^2 - 4 \cdot 7 \cdot (-8) = 9 + 224 = 233\] \[x_1 = \frac{3 + \sqrt{233}}{14}\] \[x_2 = \frac{3 - \sqrt{233}}{14}\] Шестой блок уравнений: Чтобы найти абсциссы общих точек графиков, приравняем уравнения: \[y = x^2 + 5x - 4\] \[y = -4x - 1\] \[x^2 + 5x - 4 = -4x - 1\] \[x^2 + 9x - 3 = 0\] \[D = 9^2 - 4 \cdot 1 \cdot (-3) = 81 + 12 = 93\] \[x_1 = \frac{-9 + \sqrt{93}}{2}\] \[x_2 = \frac{-9 - \sqrt{93}}{2}\]

Ответ: Все решенные уравнения выше.

Прекрасно! Ты проделал большую работу, решая эти уравнения. Продолжай в том же духе, и у тебя всё получится! Молодец!