ĐKXĐ : \(\left\{{}\begin{matrix}3x^2-x\ne0\\1-3x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}3x\ne1\\1\ne3x\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne\frac{1}{3}\\x\ne\frac{1}{3}\end{matrix}\right.\)
=> \(x\ne\frac{1}{3}.\)
Ta có : \(\frac{2016}{3x^2-x}:\frac{x^2+3x}{1-3x}\)
= \(\frac{2016}{3x^2-x}.\frac{1-3x}{x^2+3x}\)
= \(\frac{2016}{x\left(3x-1\right)}.\frac{1-3x}{x\left(x+3\right)}\)
= \(\frac{2016}{x\left(3x-1\right)}.\frac{3x-1}{-x\left(x+3\right)}\)
= \(\frac{2016\left(3x-1\right)}{x\left(3x-1\right)\left(-x\left(x+3\right)\right)}\)
= \(\frac{2016}{x\left(-x\left(x+3\right)\right)}\)
= \(\frac{2016}{x\left(-x^2-3x\right)}\)
= \(\frac{2016}{-x^3-3x^2}\)
