\(\left(x+3\right)\sqrt{2x^2+1}-\left(x+3\right)=x^2\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{2x^2+1}-1\right)=x^2\)
\(\Leftrightarrow\dfrac{2x^2.\left(x+3\right)}{\sqrt{2x^2+1}+1}=x^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\Rightarrow x=0\\\dfrac{2\left(x+3\right)}{\sqrt{2x^2+1}+1}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+6=\sqrt{2x^2+1}+1\Leftrightarrow2x+5=\sqrt{2x^2+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+5\ge0\\\left(2x+5\right)^2=2x^2+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-5}{2}\\2x^2+20x+24=0\end{matrix}\right.\) \(\Rightarrow x=-5+\sqrt{13}\)
Vậy pt có 2 nghiệm: \(\left[{}\begin{matrix}x=0\\x=-5+\sqrt{13}\end{matrix}\right.\)
