题目先化简，再求值：(x2−xx2−2x+1+21−x)÷x−2x2−1(\dfrac{x^{2}-x}{x^{2}-2x+1}+\dfrac{2}{1-x})÷\dfrac{x-2}{x^{2}-1}(x2−2x+1x2−x+1−x2)÷x2−1x−2，其中xxx是不等式组{12(x+1)⩽2x+23⩾x+34\begin{cases}\dfrac{1}{2}(x+1)\leqslant 2\\ \dfrac {x+2}{3}\geqslant\dfrac{x+3}{4}\end{cases}⎩⎨⎧21(x+1)⩽23x+2⩾4x+3的整数解．知识点：一元一次不等式组的整数解章节：方程与不等式 / 不等式与不等式组
