Решить 3 задания по математике. Быстрее пожалуйста)

1) x•(x+5)² = 14•(x+5)
x=-5 - один корень найден
x•(х+5) = х²+5х = 14
х²+5х-14 = 0 => х = (-5±9)/2 = -7 и 2
Ответ: х₁=-7, х₂=-5, х₃=2.

2) Непонятно что написано.

3) (6x-1)²-(x-6)² ≤ 0
(6х-1-х+6)(6х-1+х-6) ≤ 0
5•7•(х+1)(х-1) ≤ 0
-1 ≤ x ≤ 1
Ответ: х∈[-1;1].

1) x * (x^2 + 10x + 25) = 14*(x+5)
x * (x+5)^2 - 14*(x+5) = 0
(x+5) * [x*(x+5) - 14) = 0
(x+5) * (x^2 + 5x - 14) = 0
(x+5) * ((x^2 + 5x + 25) - 9) = 0
(x+5) * ((x+5)^2 - 3^2) = 0
(x+5) * ((x+5) + 3) * ((x+5) - 3) = 0
(x+5) * (x+8) * (x+2) = 0
x1 = - 5
x2 = - 8
x3 = - 2

2)
{ (3x+7y)^2 = 10y
{ (3x+7y)^2 = 10x
=>
10x = 10y ------> x = y =>
{ (3x+7x)^2 = 10x
100x^2 = 10x
100x^2 - 10x = 0
10 * (10x^2 - x) = 0
x * (10x - 1) = 0
x * (x - 1/10) = 0
x1 = y1 = 0;
x2 = y2 = 1/10

3) (x-6)^2 >= (6x-1)^2
(x-6)^2 - (6x-1)^2 >= 0
[(x-6) + (6x-1)] * [(x-6) - (6x-1)] >= 0
(7x-7) * (-5x-5) >= 0
7*(-5) * (x-1) * (x+1) >= 0
(x-1)(x+1) =< 1
- 1 =< x =< 1