Đặt \(\sqrt{x^2+1}=y\ge1\) pt trở thành \(\left(4x-1\right)y=2y^2-2x\)
\(4xy-y=2y^2-2x\Leftrightarrow2y^2-2x-4xy+y=0\)\(\Leftrightarrow y\left(2y+1\right)-2x\left(2y+1\right)=0\Leftrightarrow\left(2y+1\right)\left(y-2x\right)=0\Leftrightarrow y=2x\)(vì y=-1/2(loại))
\(\Leftrightarrow\sqrt{x^2+1}=2x\Leftrightarrow x=\sqrt{\frac{1}{3}}\)