решить:

(x-1)^4-5(x^2-1)^2+4(x+1)^4=0

(x-1)^4-5(x^2-1)^2+4(x+1)^4=0

ну можно в степени возвести

а можно сделать замену

(x-1)^4-5(x^2-1)^2+4(x+1)^4=0

(x-1)^4-5(x-1)^2(x+1)^2+4(x+1)^4=0

(x + 1)^2 = t

(x - 1)^2 = u

u^2 - 5ut + 4t^2 = 0

D = 25t^2 - 16t^2 = 9t^2

u12 = (5t +- 3t)/2 = 4t  t

(u - t)*(u - 4t) = 0

1. u = t

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

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

(x + 1 + x -1)(x + 1 - x + 1) = 0

4x = 0

x = 0

2. (x - 1)^2 = 4(x + 1)^2

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

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

(3x + 1)(x + 3) = 0

x = -1/3

x = -3

x= {-3, -1/3. 0}