(х+1)⁴+ (x+1)² -6 =0
(x+1)² = t
t² + t -6 =0
D= 1² -4 *1 *(-6) = 1+24= 25 = 5²
t₁= (-1-5)/2 = -6/2=-3
t₂= (-1+5)/2 =4/2 =2
(x+1)² =-3
х²+2х+1+3=0
х²+2х+4=0
D= 4-4*1*4= 4-16 =-12
D<0 - нет корней<br>
(x+1)² = 2
x²+2x+1-2=0
x²+2x-1=0
D= 4-4*1*(-1) = 4+4=8 ⇒√D= √8= √(4*2)= 2√2
x₁= (-2-2√2) /2 = (2 (-1-√2)/2 = -1-√2= - (√2+1)
x₂= (-2+2√2)/2 =(2(-1+√2)/2)= (√2 - 1)
Ответ: х₁= -(√2+1) ; х₂= √2-1