$(3x + 1)^2 = 3x + 1$
$(3x + 1)^2 - 3x + 1 = 0$
(3x + 1)((3x + 1) − 1) = 0
(3x + 1)3x = 0
3x = 0
x = 0
или
3x + 1 = 0
3x = −1
$x = -\frac{1}{3}$
Ответ:
$x_1 = 0$;
$x_2 = -\frac{1}{3}$.

$(3x + 1)^2 = 3(x + 1)$
$9x^2 + 6x + 1 = 3x + 3$
$9x^2 + 6x + 1 - 3x - 3 = 0$
$9x^2 - 3x - 2 = 0$
$D = 9 + 4 * 9 * 2 = 9 + 72 = 81$
$x = \frac{-3 ± \sqrt{81}}{18}$
$x_1 = \frac{-3 - 9}{18} = \frac{-12}{18} = -\frac{2}{3}$
$x_2 = \frac{-3 + 9}{18} = \frac{6}{18} = \frac{1}{3}$
Ответ:
$x_1 = -\frac{2}{3}$;
$x_2 = \frac{1}{3}$.

$(3x + 1)^2 = (2x - 5)^2$
((3x + 1) − (2x − 5))((3x + 1) + (2x − 5)) = 0
(3x + 1 − 2x + 5)(3x + 1 + 2x − 5) = 0
(x + 6)(5x − 4) = 0
x + 6 = 0
x = −6
или
5x − 4 = 0
5x = 4
x = 0,8
Ответ:
$x_1 = -6$;
$x_2 = 0,8$.

$(3x + 4)^2 = 4(x + 3)$
$9x^2 + 24x + 16 = 4x + 12$
$9x^2 + 24x - 4x + 16 - 12 = 0$
$9x^2 + 20x + 4 = 0$
$D = 10^2 - 9 * 4 = 100 - 36 = 64$
$x = \frac{-10 ± \sqrt{64}}{9}$
$x_1 = \frac{-10 - 8}{9} = \frac{-18}{9} = -2$
$x_2 = \frac{-10 + 8}{9} = -\frac{2}{9}$
Ответ:
$x_1 = -2$;
$x_2 = -\frac{2}{9}$.

$4(x + 3)^2 = (2x + 6)^2$
$4(x + 3)^2 = (2(x + 3))^2$
$4(x + 3)^2 = 4(x + 3)^2$
x ∈ R
Тождество выполняется при любом действительном x.

$(6x + 3)^2 = (x - 4)^2$
$(6x + 3)^2 - (x - 4)^2$
((6x + 3) − (x − 4))((6x + 3) + (x − 4)) = 0
(6x + 3 − x + 4)(6x + 3 + x − 4) = 0
(5x + 7)(7x − 1) = 0
5x + 7 = 0
5x = −7
$x = -\frac{7}{5} = -1\frac{2}{5}$
или
7x − 1 = 0
7x = 1
$x = \frac{1}{7}$
Ответ:
$x_1 = -1\frac{2}{5}$;
$x_2 = \frac{1}{7}$.