Решите уравнение (х²+2х+2)(х²+2х-4)=-5​

X1,2=-1+-sqrt(2)

Объяснение:

x^2+2x=t

(t+2)(t-4)+5=0

t^2-2t-3=0

D=4+12=16

Sqrt(D)=4

T1,2=2+-4/2=1+-2

x^2+2x+3=0. x^2+2x-1=0

D<0. D=8

Sqrt(D) =2sqrt(2)

X1,2=-1+-sqrt(2)

(х²+2х+2)(х²+2х-4)=-5​

х²+2х-1 = t

(t + 3)(t - 3) = - 5

t² - 9 = -5

t² = 4

t = +- 2

1/ t =2

х²+2х-1 = 2

х²+2х-3 = 0

D = 4 + 12 = 16

x12 = (-2 +- 4)/2 = -3  1

2/ t = -2

х²+2х-1 = -2

х²+2х+1 = 0

(x + 1)² = 0

x = -1

ответ x = {-3, -1 , 1 }