Question

(x^2+2x)^2-(x+1)^2-55=0
Решите уравнение

Answer 1

(x^2+2x)^2-(x+1)^2-55=0
(x²+2x)²-(x²+2x+1)²-55=0
x²+2x=t
t²-(t+1)²-55=0
t²-t²-2t-1-55=0
2t=-56
t=-28
x²+2x=28
x²+2x-28=0
D=4+112=116
x1=(-2-2√29)/2=-1-√29 U x2=-1+√29