Решите уравнение, используя разложение на множители:
1) (2x − 9)(x + 6) − x(x + 6) = 0;
2) (3x + 4)(x − 10) + (10 − x)(x − 8) = 0;
3) $3(3x + 1)^2 - 4(3x + 1) = 0$;
4) (9x − 12) − x(9x − 12) = 0.
(2x − 9)(x + 6) − x(x + 6) = 0
(x + 6)(2x − 9 − x) = 0
(x + 6)(x − 9) = 0
$x_1 + 6 = 0 : (x - 9)$
$x_1 + 6 = 0$
$x_1 = -6$;
$x_2 - 9 = 0 : (x + 6)$
$x_2 - 9 = 0$
$x_2 = 9$.
(3x + 4)(x − 10) + (10 − x)(x − 8) = 0
(3x + 4)(x − 10) − (x − 10)(x − 8) = 0
(x − 10)(3x + 4 − x + 8) = 0
(x − 10)(2x + 12) = 0
$x_1 - 10 = 0 : (2x + 12)$
$x_1 - 10 = 0$
$x_1 = 10$;
$2x_2 + 12 = 0 : (x - 10)$
$2x_2 + 12 = 0$
$2x_2 = -12$
$x_2 = -12 : 2$
$x_2 = -6$.
$3(3x + 1)^2 - 4(3x + 1) = 0$
(3x + 1)(3(3x + 1) − 4) = 0
(3x + 1)(9x + 3 − 4) = 0
(3x + 1)(9x − 1) = 0
$3x_1 + 1 = 0 : (9x - 1)$
$3x_1 + 1 = 0$
$3x_1 = -1$
$x_1 = -\frac{1}{3}$;
$9x_2 - 1 = 0 : (3x + 1)$
$9x_2 - 1 = 0$
$9x_2 = 1$
$x_2 = \frac{1}{9}$.
(9x − 12) − x(9x − 12) = 0
(9x − 12)(1 − x) = 0
$9x_1 - 12 = 0 : (1 - x)$
$9x_1 - 12 = 0$
$9x_1 = 12$
$x_1 = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3}$;
$1 - x_2 = 0 : (9x - 12)$
$1 - x_2 = 0$
$-x_2 = -1$
$x_2 = 1$.
Пожауйста, оцените решение