\(\Leftrightarrow\left(x^2+2x\right)^2+5\left(x^2+2x\right)+6-2=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+5\left(x^2+2x\right)+4=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2+2x+4\right)=0\)
=>x+1=0
hay x=-1
\(\left(x^2+2x+2\right)\left(x^2+2x+3\right)-2=0\\ \Leftrightarrow\left(x^2+2x+2\right)\left[\left(x^2+2x+2\right)+1\right]-2=0\\ \Leftrightarrow\left(x^2+2x+2\right)^2+\left(x^2+2x+2\right)-2=0\\ \Leftrightarrow\left[\left(x^2+2x+2\right)^2+2\left(x^2+2x+2\right)\right]-\left[\left(x^2+2x+2\right)+2\right]=0\\ \Leftrightarrow\left(x^2+2x+2\right)\left(x^2+2x+2+2\right)-\left(x^2+2x+2+2\right)=0\\ \Leftrightarrow\left(x^2+2x+2-1\right)\left(x^2+2x+4\right)=0\\ \Leftrightarrow\left(x^2+2x+1\right)\left(x^2+2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2+2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x+1\right)^2+3=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\)
Đặt \(x^2+2x+2=t\)đk t > 0
\(t\left(t+1\right)-2=0\Leftrightarrow t^2+t-2=0\Leftrightarrow t=1;t=2\left(ktm\right)\)
Với t = 1 \(x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
