\(\lim\limits_{x\rightarrow1}\dfrac{x^2-x}{x^2-3x+2}\) bằng
\(\dfrac{1}{3}\).\(-1\).\(-\dfrac{1}{4}\).\(\dfrac{1}{5}\).Hướng dẫn giải:\(\lim\limits_{x\rightarrow1}\dfrac{x^2-x}{x^2-3x+2}\) \(=\lim\limits_{x\rightarrow1}\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)\(=\lim\limits_{x\rightarrow1}\dfrac{x}{x-2}=-1\).
