Lời giải:
Đặt \(\sqrt{2x^2+1}=a(a>0)\)
Khi đó pt trở thành:
\((2x^2+1)+6x=(2x+3)\sqrt{2x^2+1}\)
\(\Leftrightarrow a^2+6x=(2x+3).a\)
\(\Leftrightarrow (a^2-3a)+6x-2ax=0\)
\(\Leftrightarrow a(a-3)-2x(a-3)=0\Leftrightarrow (a-3)(a-2x)=0\)
\(\Rightarrow \left[\begin{matrix} a-3=0\\ a-2x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} a=\sqrt{2x^2+1}=3\\ a=\sqrt{2x^2+1}=2x\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x^2=4\\ 2x^2+1=4x^2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\pm 2\\ 2x^2=1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\pm 2\\ x=\pm \sqrt{\frac{1}{2}}\end{matrix}\right.\) (loại TH \(x=-\sqrt{\frac{1}{2}}\) do không t/m)
Vậy \(x\in \left\{\sqrt{\frac{1}{2}}; \pm 2\right\}\)
