ĐKXĐ: \(-\sqrt{10}\le x\le\sqrt{10}\)
Đặt \(\sqrt{10-x^2}=t\ge0\Rightarrow x^2=10-t^2\)
Pt trở thành:
\(\left(x+3\right)t=10-t^2-x-12\)
\(\Leftrightarrow t^2+\left(x+3\right)t+x+2=0\)
\(\Leftrightarrow\left(t^2-1\right)+\left(x+3\right)t+x+3=0\)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)+\left(x+3\right)\left(t+1\right)=0\)
\(\Leftrightarrow\left(t+1\right)\left(t+x+2\right)=0\)
\(\Leftrightarrow t+x+2=0\) (do \(t+1>0\) ; \(\forall t\ge0\))
\(\Leftrightarrow\sqrt{10-x^2}+x+2=0\)
\(\Leftrightarrow\sqrt{10-x^2}=-x-2\) (\(x\le-2\))
\(\Leftrightarrow10-x^2=x^2+4x+4\)
\(\Leftrightarrow x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=-3\end{matrix}\right.\)
Ta có: \(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}-\left(x^2-x-12\right)=0\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\10-x^2=x^2-8x+16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2-x^2+8x-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\-2x^2+8x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\-2\left(x^2-4x+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)
