126. Решите уравнение: 1) (2x + 5) (x + 2) = 21; 2) (x + 3)(x-1) – (3x + 1)(x – 7) = x(x + 18); 3) (4x – 3)² + (2x – 1)(2x + 1) = 24.

1) $$(2x + 5)(x + 2) = 21$$

$$2x^2 + 4x + 5x + 10 - 21 = 0$$

$$2x^2 + 9x - 11 = 0$$

$$D = 9^2 - 4 \cdot 2 \cdot (-11) = 81 + 88 = 169$$

$$x_1 = \frac{-9 + \sqrt{169}}{2 \cdot 2} = \frac{-9 + 13}{4} = \frac{4}{4} = 1$$

$$x_2 = \frac{-9 - \sqrt{169}}{2 \cdot 2} = \frac{-9 - 13}{4} = \frac{-22}{4} = -\frac{11}{2}$$

Ответ: $$x_1 = 1, x_2 = -\frac{11}{2}$$

2) $$(x + 3)(x - 1) - (3x + 1)(x - 7) = x(x + 18)$$

$$x^2 - x + 3x - 3 - (3x^2 - 21x + x - 7) = x^2 + 18x$$

$$x^2 + 2x - 3 - 3x^2 + 20x + 7 = x^2 + 18x$$

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

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

$$D = (-4)^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64$$

$$x_1 = \frac{4 + \sqrt{64}}{2 \cdot 3} = \frac{4 + 8}{6} = \frac{12}{6} = 2$$

$$x_2 = \frac{4 - \sqrt{64}}{2 \cdot 3} = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}$$

Ответ: $$x_1 = 2, x_2 = -\frac{2}{3}$$

3) $$(4x - 3)^2 + (2x - 1)(2x + 1) = 24$$

$$16x^2 - 24x + 9 + 4x^2 - 1 = 24$$

$$20x^2 - 24x + 8 - 24 = 0$$

$$20x^2 - 24x - 16 = 0$$

$$5x^2 - 6x - 4 = 0$$

$$D = (-6)^2 - 4 \cdot 5 \cdot (-4) = 36 + 80 = 116$$

$$x_1 = \frac{6 + \sqrt{116}}{2 \cdot 5} = \frac{6 + 2\sqrt{29}}{10} = \frac{3 + \sqrt{29}}{5}$$

$$x_2 = \frac{6 - \sqrt{116}}{2 \cdot 5} = \frac{6 - 2\sqrt{29}}{10} = \frac{3 - \sqrt{29}}{5}$$

Ответ: $$x_1 = \frac{3 + \sqrt{29}}{5}, x_2 = \frac{3 - \sqrt{29}}{5}$$