ĐKXĐ: ...
\(x^2+3x+2-y^2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)-y^2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1-y^2\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=y^2-1\end{matrix}\right.\)
- Với \(x=-2\Rightarrow y-1-3\sqrt{y-1}-1=0\)
- Với \(x=y^2-1\Rightarrow y^2+y-1-3\sqrt{y-1}=0\)
Đặt \(\sqrt{y-1}=t\ge0\Rightarrow y=t^2+1\)
\(\Leftrightarrow\left(t^2+1\right)^2+t^2-3t=0\)
\(\Leftrightarrow t^4+3t^2-3t+1=0\)
\(\Leftrightarrow t^4+3\left(t-\frac{1}{2}\right)^2+\frac{1}{4}=0\left(vn\right)\)
