Đặt \(\sqrt{x^2+1}=a\ge1\Rightarrow x^2=a^2-1\)
\(\left(4x-1\right)a=2\left(a^2-1\right)+2x+1\)
\(\Leftrightarrow2a^2-\left(4x-1\right)a+2x-1=0\)
\(\Leftrightarrow2a^2-a-2\left(2x-1\right)a+2x-1=0\)
\(\Leftrightarrow a\left(2a-1\right)-\left(2x-1\right)\left(2a-1\right)=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a-2x+1\right)=0\)