520. Найдите корни уравнения: a) (x + 3)(x – 4) = −12; б) 1⅔t + (2t + 1)(⅓t – 1) = 0; в) 3x(2x + 3) = 2x(x + 4,5) + 2; г) (x – 1)(x + 1) = 2(x²– 3).

а) $$(x + 3)(x - 4) = -12$$

$$x^2 - 4x + 3x - 12 + 12 = 0$$

$$x^2 - x = 0$$

$$x(x - 1) = 0$$

$$x_1 = 0, x_2 = 1$$

б) $$1\frac{2}{3}t + (2t + 1)(\frac{1}{3}t - 1) = 0$$

$$\frac{5}{3}t + \frac{2}{3}t^2 - 2t + \frac{1}{3}t - 1 = 0$$

$$\frac{2}{3}t^2 + (\frac{5}{3} - 2 + \frac{1}{3})t - 1 = 0$$

$$\frac{2}{3}t^2 - \frac{2}{3}t - 1 = 0$$

$$2t^2 - 2t - 3 = 0$$

$$D = 4 + 4 \cdot 2 \cdot 3 = 4 + 24 = 28$$

$$t_1 = \frac{2 + \sqrt{28}}{4} = \frac{2 + 2\sqrt{7}}{4} = \frac{1 + \sqrt{7}}{2}$$

$$t_2 = \frac{1 - \sqrt{7}}{2}$$

в) $$3x(2x + 3) = 2x(x + 4,5) + 2$$

$$6x^2 + 9x = 2x^2 + 9x + 2$$

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

$$x^2 = \frac{1}{2}$$

$$x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$$

г) $$(x - 1)(x + 1) = 2(x^2 - 3)$$

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

$$x^2 = 5$$

$$x = \pm \sqrt{5}$$

Ответ: а) $$x_1 = 0, x_2 = 1$$; б) $$t_1 = \frac{1 + \sqrt{7}}{2}$$, $$t_2 = \frac{1 - \sqrt{7}}{2}$$; в) $$x = \pm \frac{\sqrt{2}}{2}$$; г) $$x = \pm \sqrt{5}$$