521. Решите уравнение: a) x² - 5 = (x + 5)(2x - 1); б) 2x - (x + 1)² = 3x² - 6;

a) $$x^2 - 5 = (x + 5)(2x - 1)$$

$$x^2 - 5 = 2x^2 - x + 10x - 5$$

$$x^2 - 5 = 2x^2 + 9x - 5$$

$$x^2 - 2x^2 - 9x - 5 + 5 = 0$$

$$-x^2 - 9x = 0$$

$$-x(x + 9) = 0$$

$$x_1 = 0, x_2 = -9$$

Ответ: x₁ = 0, x₂ = -9

б) $$2x - (x + 1)^2 = 3x^2 - 6$$

$$2x - (x^2 + 2x + 1) = 3x^2 - 6$$

$$2x - x^2 - 2x - 1 = 3x^2 - 6$$

$$-x^2 - 1 = 3x^2 - 6$$

$$-x^2 - 3x^2 - 1 + 6 = 0$$

$$-4x^2 + 5 = 0$$

$$-4x^2 = -5$$

$$x^2 = \frac{5}{4}$$

$$x_1 = \frac{\sqrt{5}}{2}, x_2 = -\frac{\sqrt{5}}{2}$$

Ответ: $$x_1 = \frac{\sqrt{5}}{2}, x_2 = -\frac{\sqrt{5}}{2}$$