ĐKXĐ:...
\(\Leftrightarrow10x^2+8x-2\left(3x+1\right)\sqrt{2x^2-1}-1=0\)
\(\Leftrightarrow2x^2-1-2\left(3x+1\right)\sqrt{2x^2-1}+8x^2+8x=0\)
Đặt \(\sqrt{2x^2-1}=t\ge0\)
\(\Rightarrow t^2-2\left(3x+1\right)t+8x^2+8x=0\)
\(\Delta'=\left(3x+1\right)^2-\left(8x^2+8x\right)=x^2-2x+1=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=3x+1-x+1=2x+2\\t=3x+1+x-1=4x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2-1}=2x+2\left(x\ge-1\right)\\\sqrt{2x^2-1}=4x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-1=4x^2+8x+4\left(x\ge-1\right)\\2x^2-1=16x^2\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{\sqrt{6}-4}{2}\)
