2) 6) (-x-1)(x-4)=x(4x-11); r) 5(x-2)=(3x+2)(x-2);

б) $$(-x-1)(x-4) = x(4x-11)$$

$$-x^2 + 4x - x + 4 = 4x^2 - 11x$$

$$-x^2 + 3x + 4 = 4x^2 - 11x$$

$$5x^2 - 14x - 4 = 0$$

$$D = (-14)^2 - 4 \cdot 5 \cdot (-4) = 196 + 80 = 276$$

$$x_1 = \frac{14 + \sqrt{276}}{10} = \frac{14 + 2\sqrt{69}}{10} = \frac{7 + \sqrt{69}}{5}$$

$$x_2 = \frac{14 - \sqrt{276}}{10} = \frac{14 - 2\sqrt{69}}{10} = \frac{7 - \sqrt{69}}{5}$$

г) $$5(x-2) = (3x+2)(x-2)$$

$$5(x-2) - (3x+2)(x-2) = 0$$

$$(x-2)(5 - (3x+2)) = 0$$

$$(x-2)(5 - 3x - 2) = 0$$

$$(x-2)(3 - 3x) = 0$$

$$x-2 = 0$$ или $$3-3x = 0$$

$$x_1 = 2$$, $$x_2 = 1$$

Ответ: б) $$x_1 = \frac{7 + \sqrt{69}}{5}$$, $$x_2 = \frac{7 - \sqrt{69}}{5}$$, г) $$x_1 = 2$$, $$x_2 = 1$$