629. Решите уравнение: 1) (3x – 2)(3x + 2) + (4x – 5)² = 10x + 21; 2) (2x – 1)(x + 8) – (x – 1)(x + 1) = 15x.

1. $$\begin{aligned} (3x - 2)(3x + 2) + (4x - 5)^2 &= 10x + 21 \\ 9x^2 - 4 + 16x^2 - 40x + 25 &= 10x + 21 \\ 25x^2 - 40x + 21 &= 10x + 21 \\ 25x^2 - 50x &= 0 \\ 25x(x - 2) &= 0 \\ x = 0 \quad \text{или} \quad x - 2 &= 0 \\ x = 0 \quad \text{или} \quad x &= 2 \end{aligned}$$

2. $$\begin{aligned} (2x - 1)(x + 8) - (x - 1)(x + 1) &= 15x \\ 2x^2 + 16x - x - 8 - (x^2 - 1) &= 15x \\ 2x^2 + 15x - 8 - x^2 + 1 &= 15x \\ x^2 + 15x - 7 &= 15x \\ x^2 - 7 &= 0 \\ x^2 &= 7 \\ x &= \pm \sqrt{7} \end{aligned}$$

Ответ:

1. $$x = 0, x = 2$$
2. $$x = \pm \sqrt{7}$$