ĐKXĐ: ...
\(y^2\left(x+2\right)-\left(x^2+3x+2\right)=0\)
\(\Leftrightarrow y^2\left(x+2\right)-\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(y^2-x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=y^2-1\end{matrix}\right.\)
TH1: \(x=-2\Rightarrow y-2=3\sqrt{y-1}\)
Đặt \(\sqrt{y-1}=t\ge0\Rightarrow t^2-1=3t\Rightarrow t^2-3t-1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{3+\sqrt{13}}{2}\\t=\frac{3-\sqrt{13}}{2}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow y-1=\left(\frac{3+\sqrt{13}}{2}\right)^2\Rightarrow y=...\)
Th2: \(x=y^2-1\)
\(\Rightarrow y^2-1+y=3\sqrt{y-1}\)
Đặt \(\sqrt{y-1}=t\ge0\Rightarrow y=t^2+1\)
\(\Rightarrow\left(t^2+1\right)^2+t^2=3t\)
\(\Leftrightarrow t^4+3t^2-3t+1=0\)
\(\Leftrightarrow t^4+3\left(t-\frac{1}{2}\right)^2+\frac{1}{4}=0\) (vô nghiệm)
