Найдите корень уравнения:
а) $5 + x^2 = (x + 1)(x + 6)$;
б) 2x(x − 8) = (x + 1)(2x − 3);
в) (3x − 2)(x + 4) − 3(x + 5)(x − 1) = 0;
г) $x^2 + x(6x - 2x) = (x - 1)(2 - x) - 2$.
$5 + x^2 = (x + 1)(x + 6)$
$5 + x^2 = x^2 + 6x + x + 6$
$x^2 - x^2 - 6x - x = 6 - 5$
−7x = 1
$x = -\frac{1}{7}$
$2x(x - 8) = (x + 1)(2x - 3)$
$2x^2 - 16x = 2x^2 - 3x + 2x - 3$
$2x^2 - 16x - 2x^2 + 3x - 2x = -3$
−15x = −3
$x = \frac{3}{15} = \frac{1}{5}$
$(3x - 2)(x + 4) - 3(x + 5)(x - 1) = 0$
$3x^2 + 12x - 2x - 8 - 3(x^2 - x + 5x - 5) = 0$
$3x^2 + 10x - 8 - 3(x^2 + 4x - 5) = 0$
$3x^2 + 10x - 8 - 3x^2 - 12x + 15 = 0$
−2x = − 7
$x = \frac{7}{2} = 3\frac{1}{2}$
$x^2 + x(6x - 2x) = (x - 1)(2 - x) - 2$
$x^2 + 6x - 2x^2 = (2x - x^2 - 2 + x) - 2$
$x^2 + 6x - 2x^2 = 3x - x^2 - 4$
$x^2 + 6x - 2x^2 - x + x^2 = -4$
3x = −4
$x = -\frac{4}{3} = -1\frac{1}{3}$
Пожауйста, оцените решение