Chắc là bạn ghi nhầm đề, đề đúng là
\(x^2+\left(3-\sqrt{x^2+2}\right)x=1+2\sqrt{x^2+2}\)
Đặt \(\sqrt{x^2+2}=a>0\Rightarrow x^2=a^2-2\) ta được:
\(a^2-2+\left(3-a\right)x-2a-1=0\)
\(\Leftrightarrow a^2-\left(x+2\right)a+3x-3=0\)
\(\Delta=\left(x+2\right)^2-4\left(3x-3\right)=x^2-8x+16=\left(x-4\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}a=\frac{x+2+x-4}{2}=x-1\\a=\frac{x+2-x+4}{2}=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=x-1\left(x\ge1\right)\\\sqrt{x^2+2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2=x^2-2x+1\\x^2+2=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\left(l\right)\\x=\pm\sqrt{7}\end{matrix}\right.\)
