664. Решите уравнение: 1) (x - 4)² = 4x - 11; 2) (x + 5)² + (x - 7)(x + 7) = 6x - 19; 3) (3x-1)(x + 4) = (2x + 3)(x + 3) – 17.

664. 1) $$(x - 4)^2 = 4x - 11$$ $$x^2 - 8x + 16 = 4x - 11$$ $$x^2 - 12x + 27 = 0$$ $$D = (-12)^2 - 4 * 1 * 27 = 144 - 108 = 36$$ $$x_1 = \frac{12 + \sqrt{36}}{2} = \frac{12 + 6}{2} = \frac{18}{2} = 9$$ $$x_2 = \frac{12 - \sqrt{36}}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3$$ 2) $$(x + 5)^2 + (x - 7)(x + 7) = 6x - 19$$ $$x^2 + 10x + 25 + x^2 - 49 = 6x - 19$$ $$2x^2 + 10x - 24 = 6x - 19$$ $$2x^2 + 4x - 5 = 0$$ $$D = 4^2 - 4 * 2 * (-5) = 16 + 40 = 56$$ $$x_1 = \frac{-4 + \sqrt{56}}{4} = \frac{-4 + 2\sqrt{14}}{4} = \frac{-2 + \sqrt{14}}{2}$$ $$x_2 = \frac{-4 - \sqrt{56}}{4} = \frac{-4 - 2\sqrt{14}}{4} = \frac{-2 - \sqrt{14}}{2}$$ 3) $$(3x - 1)(x + 4) = (2x + 3)(x + 3) - 17$$ $$3x^2 + 12x - x - 4 = 2x^2 + 6x + 3x + 9 - 17$$ $$3x^2 + 11x - 4 = 2x^2 + 9x - 8$$ $$x^2 + 2x + 4 = 0$$ $$D = 2^2 - 4 * 1 * 4 = 4 - 16 = -12$$ Т.к. дискриминант отрицательный, уравнение не имеет действительных корней.