\(\Leftrightarrow2\left(x^2+1\right)-2x\sqrt{x^2+1}-5=0\)
\(\Leftrightarrow\left(x^2+1\right)-2x\sqrt{x^2+1}+x^2-4=0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)^2-2^2=0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x-2\right)\left(\sqrt{x^2+1}-x+2=0\right)\)
TH1: \(\sqrt{x^2+1}-x-2=0\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\x^2+1=\left(x+2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\4x=-3\end{matrix}\right.\) \(\Rightarrow x=-\dfrac{3}{4}\)
TH2: \(\sqrt{x^2+1}-x+2=0\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2+1=\left(x-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-4x=-3\end{matrix}\right.\) \(\Rightarrow x=\dfrac{3}{4}< 2\) (loại)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{3}{4}\)
