Решите пожалуйста алгебру

1.
$\frac{6x}{ x^{2} -y^2} - \frac{3}{x-y} - \frac{4}{x+y} = \frac{6x-3*(x+y)-4*(x-y)}{ x^{2} -y^2} = \frac{6x-3x-3y-4x+4y}{ x^{2} -y^2}= \frac{-x+y}{ x^{2} -y^2}=$ $\frac{-(x-y)}{(x-y)(x+y)} = - \frac{1}{x+y}$

2.
2х(х-2) = (х+1)² - 9
2х² - 4х = х² + 2х + 1 - 9
2х² - 4х - х² - 2х - 1 + 9 = 0
х² - 6х + 8 = 0
D = b² - 4ac
D = 36 - 4·1·8 = 36 - 32 = 4
√D = √4 = 2
x₁ = (6 + 2)/2 = 8/2 = 4
x₁ = 4
x₂ = (6-2)/2 = 4/2= 2
x₂ = 2
Ответ: { 2; 4)

3.
$\frac{-6x-8}{3} \leq \frac{3+5x}{4}$
$\frac{(-6x-8)*12}{3} \leq \frac{(3+5x)*12}{4}$
(-6x-8)·4 ≤ (3 + 5х) · 3
- 24х - 32 ≤ 9 + 15х
- 24х - 15х ≤ 9 + 32
-39х≤ 41
39х≥ - 41
х ≥ - 41/39
Ответ: х∈ [-41/39; + ∞)