Решите уравнение:
1) $(3x - 5)^2 - 49 = 0$;
2) $(4x + 7)^2 - 9x^2 = 0$;
3) $(a - 1)^2 - (2a + 9)^2 = 0$;
4) $25(3b + 1)^2 - 16(2b - 1)^2 = 0$.
$(3x - 5)^2 - 49 = 0$
$((3x - 5) - 7)((3x + 5) - 7) = 0$
$(3x_1 - 5) - 7 = 0$
$3x_1 - 12 = 0$
$3x_1 = 12$
$x_1 = 12 : 3$
$x_1 = 4$;
$(3x_2 - 5) + 7 = 0$
$3x_2 + 12 = 0$
$3x_2 = -12$
$x_2 = -12 : 3$
$x_2 = -4$.
$(4x + 7)^2 - 9x^2 = 0$
$((4x + 7) - 3x)((4x + 7) + 3x) = 0$
$((4x_1 + 7) - 3x_1) = 0$
$4x_1 + 7 - 3x_1 = 0$
$x_1 = -7$;
$((4x_2 + 7) + 3x_2) = 0$
$7x_2 = -7$
$x_2 = -7 : 7$
$x_2 = -1$.
$(a - 1)^2 - (2a + 9)^2 = 0$
$((a - 1) - (2a + 9))((a - 1) + (2a + 9)) = 0$
$a_1 - 1 - 2a_1 - 9 = 0$
$-a_1 - 10 = 0$
$-a_1 = 10$
$a_1 = -10$;
$a_2 - 1 + 2a_2 + 9 = 0$
$3a_2 + 8 = 0$
$3a_2 = -8$
$a_2 = -\frac{8}{3} = -2\frac{2}{3}$.
$25(3b + 1)^2 - 16(2b - 1)^2 = 0$
$5^2(3b + 1)^2 - 4^2(2b - 1)^2 = 0$
$(15b + 5)^2 - (8b - 4)^2 = 0$
$((15b + 5) - (8b - 4))((15b + 5) + (8b - 4)) = 0$
$15b_1 + 5 - 8b_1 + 4 = 0$
$7b_1 = -9$
$b_1 = -\frac{9}{7} = -1\frac{2}{7}$;
$15b_2 + 5 + 8b_2 - 4 = 0$
$23b_2 = -1$
$b_2 = -\frac{1}{23}$.
Пожауйста, оцените решение