\(x^2-2x+\sqrt{2x^2+1}=\sqrt{4x+1}\) (ĐKXĐ : \(x\ge-\frac{1}{4}\) )
\(\Leftrightarrow2x^2-4x+2\sqrt{2x^2+1}-2\sqrt{4x+1}=0\)
\(\Leftrightarrow\left[\left(2x^2+1\right)+2\sqrt{2x^2+1}+1\right]-\left[\left(4x+1\right)+2\sqrt{4x+1}+1\right]=0\)
\(\Leftrightarrow\left(\sqrt{2x^2+1}+1\right)^2-\left(\sqrt{4x+1}+1\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{2x^2+1}+1-\sqrt{4x+1}-1\right)\left(\sqrt{2x^2+1}+1+\sqrt{4x+1}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x^2+1}-\sqrt{4x+1}\right)\left(\sqrt{2x^2+1}+\sqrt{4x+1}+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{2x^2+1}-\sqrt{4x+1}=0\\\sqrt{2x^2+1}+\sqrt{4x+1}+2=0\end{array}\right.\)
Vì \(\sqrt{2x^2+1}+\sqrt{4x+1}+2>0\) với mọi \(x\ge-\frac{1}{4}\) nên vô nghiệm.
Do đó ta xét \(\sqrt{2x^2+1}-\sqrt{4x+1}=0\Leftrightarrow2x^2+1=4x+1\Leftrightarrow2x\left(x-2\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\) (thoả mãn)
Vậy tập nghiệm của phương trình : \(S=\left\{0;2\right\}\)
