Помогите решить алгебру 8 класс!Хотя-бы с 1 по 4 задание.

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

2.
3x² - 2x - 5 = 0
D = (-2)² - 4*3*(-5) = 4 + 60 = 64 = 8² ; D>0
x₁ = ( - (-2) - 8)/(2*1) = (2 - 8)/2 = ( - 6)/2 = - 3
x₂ = ( - (-2) + 8)/(2*1) = (2 +8)/2 = 10/2 = 5

x² - 49 = 0
x² - 7² = 0
(x - 7)(x + 7) = 0
x - 7 = 0
x₁ = 7
x + 7 = 0
x₂ = - 7

3.
(√ 5 + √8)² - √90 = (√5)² + 2 √5 *√8 + (√8)² - √(3²*10) =
= 5 + 2√(2²*10) + 8 - 3√10 = 13 + 4√10 - 3√10 =
= 13 + √10

4.
5x - 2(x -4) < 9x + 20
5x - 2x + 8 < 9x + 20
3x + 8 < 9x + 20<br>3x - 9x < 20 - 8<br>- 6x < 12 |*(-1)<br>6x > - 12
x > - 2
x∈ ( - 2; +∞)

х² - 3х - 10 ≤ 0
Вспомогательное уравнение
x² - 3x - 10 = 0
D = ( - 3)² - 4*1*(-10) = 9 + 40 = 49 = 7²
D> 0
x₁ = ( - (-3) - 7)/(2*1) = (3 - 7)/2 = -4/2 = -2
x₂ = ( - (-3) + 7)/(2*1) = (3+7)/2 = 10/2 = 5
+ - +
-----.----------. ------
-2 5
- 2 ≤ x ≤ 5
x∈ [ - 2 ; 5 ]