(x^2 + 2x)^2 - 2(x + 1)^2=1Помогите :3

Question

(x^2 + 2x)^2 - 2(x + 1)^2=1
Помогите :3

Answer 1

(x^2 + 2x)^2 - 2(x + 1)^2=1
x^2+2x=t
t^2-2(t+1)=1
t^2-2t-3=0
t12=(2+-корень(4+12))/2=3 -1
x^2+2x=-1
x^2+2x+1=0
(x+1)^2=0
x=-1
x^2+2x=3
x^2+2x-3=0
x1=-3
x2=1

Answer 2

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

Пусть  x^2 + 2x = t, тогда имеем:
t^2 - 2(t+1) = 1
t^2 - 2t - 3 = 0
Δ = 4 + 4*3 = 4 +12 = 16 = 4 ^2
t1 = (2+4)/2 = 6/2 = 3
t2 = (2-4)/2 = - 2/2 = - 1

Обратная замена:
1) x^2 + 2x = 3
x^2 + 2x - 3 = 0
Δ = 4 + 4*3 = 16 = 4^2
x1 = ( - 2 + 4)/2 = 2/2 = 1
x2 = (- 2 -4)/2 = - 6/2 = - 3

2) x^2 + 2x = - 1
x^2 + 2x +  1 = 0
(x + 1)^2 = 0
x3 = - 1

ОТВЕТ:
-3; - 1 ; 1