Question

Решите уравнение (x^2 + 2x)^2 - 2(x + 1)^2 = 1

Answer 1

(x^2 + 2x)^2 - 2(x + 1)^2 - 1 = 0
(x^2 + 2x)^2 - 2(x^2 + 2x + 1) - 1 = 0
Замена x^2 + 2x = y
y^2 - 2(y + 1) - 1 = 0
y^2 - 2y - 3 = 0
(y + 1)(y - 3) = 0
Обратная замена
y1 = x^2 + 2x = -1
x^2 + 2x + 1 = (x + 1)^2 = 0
x1 = x2 = -1
y2 = x^2 + 2x = 3
x^2 + 2x - 3 = (x - 1)(x + 3) = 0
x3 = -3, x4 = 1

Answer 2

(x^2+2x)^2-2*(x^2+2x+1)=1
Пусть x^2+2x=y:
y^2-2y-2=1
y^2-2y-3=0
[y1=-1
[y2=3
Подставим y1=-1 и y2=3 в 1)x^2+2x=y:
х^2+2х+1=0
х1=-1
2)х^2+2х-3=0
[х2=1
[х3=-3
Ответ: х1=-1, х2=1, х3=-3.