Прошу помогите с алгеброй

Нет

Ответ

Пусть [(x-3)/(x+2)]² = y
=> y -15=16/y; y² -15y -16 =0;
y₁ = -1; y₂ =16.

Остаётся решить [(x-3)/(x+2)]² = -1 -- реш.нет
и [(x-3)/(x+2)]² = 16.
т.е одно квадр ур-ие.
Дерзай!

ПС для контроля
x = -1;
x = -11/3.

[(x-3)/(x+2)]^2 - 15 = 16*[(x+2)/(x-3)]^2
ОДЗ: x не= - 2; x не= 3
[(x-3)/(x+2)]^2 = t ---> [(x+2)/(x-3)]^2 = 1/t
t - 15 = 16*1/t
t^2 - 15t - 16 = 0
t(1,2) = [15 + - V(15^2 + 4*16)]/2 = (15 + - 17)/2
t1 = (15-17)/2 < 0 - не подходит
t2 = (15+17)/2 = 32/2 = 16
=>
[(x-3)/(x+2)]^2 = 16
[(x-3)/(x+2)]^2 - 4^2 = 0
[(x-3)/(x+2) + 4] * [(x-3)/(x+2) - 4] = 0
[{(x+3) + 4x + 8} / (x+2)] * [{(x-3) - 4x - 8} / (x+2)] = 0
(5x + 11) * (- 3x - 11) = 0
(5x + 11) * (3x + 11) = 0
(x + 11/5) * (x + 11/3) = 0
x1 = 11/5
x2 = - 11/3