Nhận thấy khi \(x\le0\Rightarrow VT=x\left(1+\frac{2\sqrt{2}}{\sqrt{1+x^2}}\right)\le0< VP\Rightarrow ptvn\)
Xét \(x>0\):
Đặt \(\frac{1}{\sqrt{1+x^2}}=a\); do \(\sqrt{1+x^2}\ge1\Rightarrow0< \frac{1}{\sqrt{1+x^2}}\le1\Rightarrow0< a\le1\)
\(\Rightarrow1+x^2=\frac{1}{a^2}\Rightarrow x^2=\frac{1-a^2}{a^2}\Rightarrow x=\frac{\sqrt{1-a^2}}{a}\) ta được:
\(\frac{\sqrt{1-a^2}}{a}+\frac{2\sqrt{2}.\sqrt{1-a^2}}{a}.a=1\)
\(\Leftrightarrow\sqrt{1-a^2}+2\sqrt{2}a.\sqrt{1-a^2}=a\)
\(\Leftrightarrow\sqrt{1-a^2}-a+2\sqrt{2}a.\sqrt{1-a^2}=0\)
Đặt \(\sqrt{1-a^2}-a=t\Rightarrow t^2=1-2a\sqrt{1-a^2}\Rightarrow2a\sqrt{1-a^2}=1-t^2\)
\(\Rightarrow t+\sqrt{2}\left(1-t^2\right)=0\Rightarrow-\sqrt{2}t^2+t+\sqrt{2}=0\) \(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\\t=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{1-a^2}-a=\sqrt{2}\\\sqrt{1-a^2}-a=-\frac{\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{1-a^2}=a+\sqrt{2}\\2\sqrt{1-a^2}=2a-\sqrt{2}\left(a\ge\frac{\sqrt{2}}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2a^2+2\sqrt{2}a+1=0\\4a^2-2\sqrt{2}a-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=-\frac{\sqrt{2}}{2}< 0\left(l\right)\\a=\frac{\sqrt{6}+\sqrt{2}}{4}\\a=\frac{-\sqrt{6}+\sqrt{2}}{4}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{\sqrt{1-a^2}}{a}=2-\sqrt{3}\)
Vậy pt có nghiệm duy nhất \(x=2-\sqrt{3}\)
