очень нужно выручайте. решение должно быть записано подробно, со всеми промежуточными вычислениями.​

-4х² - 4х = 0
- 4х (х  + 1) = 0
произведение =0, если один из множителей = 0
-4х = 0
х₁ = 0
х + 1 = 0
х₂ = -1

5х²  - 9х  = 0
х(5х - 9) = 0
х₁= 0
5х - 9 = 0
5х = 9
х = 9/5
х ₂= 1,8

-6х = 0
х = 0

¹/₅ х² - 5 = 0
¹/₅ х²  = 5
х²  =  5 :  ¹/₅  =  5 * ⁵/₁ 
х² = 25
х₁ = √25
х₁ = 5
х₂ = -√25
х₂ = - 5

¹/₄ х² -  1 = 0
(¹/₂ х)²  - 1²  = 0
(0,5х  - 1)(0,5х + 1) = 0
0,5х  - 1 = 0
0,5 х = 1
х = 1/0,5
х₁ = 2
0,5х   + 1 = 0
0,5х = -1
х = - 1/0,5
х₂ = - 2

2х² + 3х = 0
х (2х + 3) =0
х₁ = 0
2х + 3 = 0
х₂ = - 1,5

⁴/₅ х² - 45 = 0
0,8х²  = 45
х²  = 45/0,8
х² = 56,25
х₁ = 7,5
х₂ = - 7,5

х² + х =0
х(х+1) =0
х₁ = 0
х + 1=0
х₂ = -1

2(5х -7)(1+х) = 0
5х - 7 =0
5х = 7
х = 7/5
х₁ = 1,4
1 + х = 0
х₂ = -1

25 - х² = 0
(5-х)(5+х) =0
5 - х = 0
-х = -5
х₁ = 5
5 +х = 0
х₂ = - 5

18  - х² = 14
18  - х² -14 = 0
4 - х²  = 0
2²  - х² = 0
(2 -х)(2+х) = 0
2 - х = 0
х₁ = 2
2 + х = 0
х₂ = -2

4 - 36х² = 0
2²  - (6х)² = 0
(2-6х)(2+6х)=0
2 - 6х = 0
- 6х = -2
6х = 2
х =2/6
х₁ = 1/3
2 + 6х =0
6х = -2
х₂ = - 1/3

х² + 3 = 3 - х
х²  + 3 - 3 +х = 0
х²  + х =0
х(х+1) =0
х₁=0
х+1=0
х₂ = -1

2х² - 14=0
2(х²  - 7) = 0     |÷2
х² - 7 =0
x² = 7
x₁ = √7
x₂ = - √7

А)cos²α-cos²β=(cosα-cosβ)(cosα+cosβ)=-2sin(α-β)/2sin(α+β)/2·2cos(α+β)/2cos(α-β)/2=-2sin(α-β)/2cos(α-β)/2·2sin(α+β)/2cos(α+β)/2=-sin(α-β)sin(α+β)
б)3/4-sin²x=(√3/2)²-sin²x=(√3/2-sinx)(√3/2+sinx)=(sinπ/3-sinx)(sinπ/3+sinx)=2sin(π/3+x)/2cos(π/3-x)/2·2sin(π/3-x)/2cos(π/3+x)/2=2sin(π/3+x)/2cos(π/3+x)/2·2sin(π/3-x)/2cos(π/3-x)/2=sin(π/3+x)sin(π/3-x)
в)cos²x-1/2=cos²x-(√2/2)²=(cosx-√2/2)(cosx+√2/2)=(cosx-cosπ/4)(cosx+cosπ/4)=-2sin(x-π/4)/2sin(x+π/4)/2·2cos(x+π/4)/2cos(x-π/4)/2=-2sin(x-π/4)/2cos(x-π/4)/2·2sin(x+π/4)/2cos(x+π/4)/2=-sin(x-π/4)sin(x+π/4)
г)sin²α-cos²π/3=(sinα-cosπ/3)(sinα+cosπ/3)=(sinα-1/2)(sinα+1/2)=(sinα-sinπ/6)(sinα+sinπ/6)=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α-π/6)/2=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α+π/6)/2=sin(α-π/6)sin(α+π/6)