a) Ta có: \(\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{x+3}\)
\(\Leftrightarrow\frac{x\left(x^2-9\right)}{5\left(3-x\right)}=\frac{-x\left(x+3\right)}{x+3}\)
\(\Leftrightarrow\frac{-x\left(x+3\right)\left(3-x\right)}{5\left(3-x\right)}=-x\)
\(\Leftrightarrow\frac{-x\left(x+3\right)}{5}=-x\)
\(\Leftrightarrow-x\left(x+3\right)=-5x\)
\(\Leftrightarrow-x^2-3x+5x=0\)
\(\Leftrightarrow-x^2+2x=0\)
\(\Leftrightarrow x\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)
