4. Решите уравнение: 1) $$4(3y + 1)^2 - 27 = (4y + 9)(4y - 9) + 2(5y + 2)(2y - 7)$$; 2) $$(4x - 3)^2 - 25x^2 = 0$$; 3) $$(x - 2)^2 + (x + 7)^2 = 2(2-x)(x + 7)$$.

1) $$4(3y + 1)^2 - 27 = (4y + 9)(4y - 9) + 2(5y + 2)(2y - 7)$$
$$4(9y^2 + 6y + 1) - 27 = (16y^2 - 81) + 2(10y^2 - 35y + 4y - 14)$$
$$36y^2 + 24y + 4 - 27 = 16y^2 - 81 + 2(10y^2 - 31y - 14)$$
$$36y^2 + 24y - 23 = 16y^2 - 81 + 20y^2 - 62y - 28$$
$$36y^2 + 24y - 23 = 36y^2 - 62y - 109$$
$$24y + 62y = -109 + 23$$
$$86y = -86$$
$$y = -1$$

2) $$(4x - 3)^2 - 25x^2 = 0$$
$$(4x - 3)^2 - (5x)^2 = 0$$
$$(4x - 3 - 5x)(4x - 3 + 5x) = 0$$
$$(-x - 3)(9x - 3) = 0$$
$$-x - 3 = 0$$ или $$9x - 3 = 0$$
$$x = -3$$ или $$9x = 3 Rightarrow x = \frac{1}{3}$$

3) $$(x - 2)^2 + (x + 7)^2 = 2(2-x)(x + 7)$$
$$x^2 - 4x + 4 + x^2 + 14x + 49 = 2(2x + 14 - x^2 - 7x)$$
$$2x^2 + 10x + 53 = 2(-x^2 - 5x + 14)$$
$$2x^2 + 10x + 53 = -2x^2 - 10x + 28$$
$$4x^2 + 20x + 25 = 0$$
$$(2x + 5)^2 = 0$$
$$2x + 5 = 0$$
$$2x = -5$$
$$x = -\frac{5}{2}$$