ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm a\end{matrix}\right.\)
Ta có: \(\dfrac{4}{x+a}+\dfrac{8}{x-a}=\dfrac{5a^2}{x\left(x^2-a^2\right)}\)
\(\Leftrightarrow\dfrac{4x\left(x-a\right)}{x\left(x+a\right)\left(x-a\right)}+\dfrac{8x\left(x+a\right)}{x\left(x+a\right)\left(x-a\right)}=\dfrac{5a^2}{x\left(x-a\right)\left(x+a\right)}\)
Suy ra: \(4x^2-4xa+8x^2+8xa-5a^2=0\)
\(\Leftrightarrow12x^2+4xa-5a^2=0\)
\(\Leftrightarrow12x^2+10xa-6xa-5a^2=0\)
\(\Leftrightarrow2x\left(6x+5a\right)-a\left(6x+5a\right)=0\)
\(\Leftrightarrow\left(6x+5a\right)\left(2x-a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+5a=0\\2x-a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-5a\\2x=a\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5a}{6}\\x=\dfrac{a}{2}\end{matrix}\right.\)
