ĐK: \(x\ne-5\)
\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
\(\Leftrightarrow x^2+\dfrac{25x^2}{\left(x+5\right)^2}-\dfrac{10x^2}{x+5}+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\left(x-\dfrac{5x}{x+5}\right)^2+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\dfrac{x^4}{\left(x+5\right)^2}+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow y^2+10y-11=0\left(y=\dfrac{x^2}{x+5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-11\end{matrix}\right.\)
TH1: \(y=1\)
\(\Leftrightarrow\dfrac{x^2}{x+5}=1\)
\(\Leftrightarrow x^2=x+5\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{21}}{2}\left(tm\right)\)
TH2: \(y=-11\)
\(\Leftrightarrow\dfrac{x^2}{x+5}=-11\)
\(\Leftrightarrow x^2=-11x-55\)
\(\Rightarrow\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{1\pm\sqrt{21}}{2}\)
