ĐKXĐ: ...
\(\dfrac{5}{x^2}+1+\dfrac{2x}{\sqrt{5+x^2}}=3\)
\(\Leftrightarrow\dfrac{5+x^2}{x^2}+\dfrac{2x}{\sqrt{5+x^2}}=3\)
Đặt \(\dfrac{x}{\sqrt{5+x^2}}=t\)
\(\Rightarrow\dfrac{1}{t^2}+2t=3\)
\(\Rightarrow2t^3-3t^2+1=0\)
\(\Rightarrow\left(t-1\right)^2\left(2t+1\right)=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{5+x^2}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{5+x^2}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{5+x^2}=1\left(vn\right)\\\dfrac{x^2}{5+x^2}=\dfrac{1}{4}\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow x=-\sqrt{\dfrac{5}{3}}\)
