548. Решите уравнение: 1) (3x-5)² - 49 = 0; 2) (4x + 7)2-9x2 = 0; 3) (a-1)² - (2a + 9)² = 0; 4) 25(3b + 1)² - 16(2b-1)² = 0.

Решим каждое уравнение:

1) $$(3x - 5)^2 - 49 = 0$$

$$ (3x - 5)^2 - 7^2 = 0 $$

Применим формулу разности квадратов: $$a^2 - b^2 = (a - b)(a + b)$$.

$$ (3x - 5 - 7)(3x - 5 + 7) = 0 $$

$$ (3x - 12)(3x + 2) = 0 $$

$$ 3x - 12 = 0 $$ или $$ 3x + 2 = 0 $$

$$ 3x = 12 $$ или $$ 3x = -2 $$

$$ x = 4 $$ или $$ x = -\frac{2}{3} $$

2) $$(4x + 7)^2 - 9x^2 = 0$$

$$ (4x + 7)^2 - (3x)^2 = 0 $$

Применим формулу разности квадратов.

$$ (4x + 7 - 3x)(4x + 7 + 3x) = 0 $$

$$ (x + 7)(7x + 7) = 0 $$

$$ x + 7 = 0 $$ или $$ 7x + 7 = 0 $$

$$ x = -7 $$ или $$ 7x = -7 $$

$$ x = -7 $$ или $$ x = -1 $$

3) $$(a - 1)^2 - (2a + 9)^2 = 0$$

$$ (a - 1 - (2a + 9))(a - 1 + 2a + 9) = 0 $$

$$ (a - 1 - 2a - 9)(3a + 8) = 0 $$

$$ (-a - 10)(3a + 8) = 0 $$

$$ -a - 10 = 0 $$ или $$ 3a + 8 = 0 $$

$$ a = -10 $$ или $$ 3a = -8 $$

$$ a = -10 $$ или $$ a = -\frac{8}{3} $$

4) $$25(3b + 1)^2 - 16(2b - 1)^2 = 0$$

$$ (5(3b + 1))^2 - (4(2b - 1))^2 = 0 $$

$$ (15b + 5)^2 - (8b - 4)^2 = 0 $$

$$ (15b + 5 - (8b - 4))(15b + 5 + 8b - 4) = 0 $$

$$ (15b + 5 - 8b + 4)(23b + 1) = 0 $$

$$ (7b + 9)(23b + 1) = 0 $$

$$ 7b + 9 = 0 $$ или $$ 23b + 1 = 0 $$

$$ 7b = -9 $$ или $$ 23b = -1 $$

$$ b = -\frac{9}{7} $$ или $$ b = -\frac{1}{23} $$

Ответ: 1) $$x = 4$$, $$x = -\frac{2}{3}$$; 2) $$x = -7$$, $$x = -1$$; 3) $$a = -10$$, $$a = -\frac{8}{3}$$; 4) $$b = -\frac{9}{7}$$, $$b = -\frac{1}{23}$$