ĐKXĐ: ....
Đặt \(\sqrt{y+1}=a\Rightarrow y=a^2-1\)
\(\left\{{}\begin{matrix}x^2a-2x\left(a^2-1\right)-2x=1\\x^3-3x-3x\left(a^2-1\right)=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2a-2a^2x=1\\x^3-3xa^2=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(ax-2a^2\right)=1\\x\left(x^2-3a^2\right)=6\end{matrix}\right.\)
\(\Rightarrow\frac{ax-2a^2}{x^2-3a^2}=\frac{1}{6}\Rightarrow6ax-12a^2=x^2-3a^2\)
\(\Leftrightarrow x^2-6ax+9a^2=0\)
\(\Leftrightarrow\left(x-3a\right)^2=0\Rightarrow x=3a\)
\(\Rightarrow x=3\sqrt{y+1}\Rightarrow y=\frac{x^2-9}{9}\) (\(x>0\))
\(\Rightarrow x^3-3x-\frac{3x\left(x^2-9\right)}{9}=6\)
