Question

Giải PT: \(x\left(x+2\right)\left(x^2+2x+2\right)+1=0\)

Answer 1

$x\left(x+2\right)\left(x^2+2x+2\right)+1=0$

$\iff \left(x^2+2x\right)\left(x^2+2x+2\right)=-1$

Đặt $x^2+2x+1=t$.

Ta có: $\left(t-1\right)\left(t+1\right)=-1$

$\iff t^2-1=-1$

$\iff t^2=0\iff t=0$

Do đó ${\left(x+1\right)}^2=0$

$\iff x+1=0\iff x=-1$

Vậy x = -1.