题目(1)(1)(1)解不等式组：{2x−3>12−x3>x3−2\begin{cases}2x-3 > 1 \\ \dfrac{2-x}{3} > \dfrac{x}{3}-2\end{cases}⎩⎨⎧2x−3>132−x>3x−2． (2)(2)(2)解方程：x+1x−1+41−x2=1\dfrac{x+1}{x-1}+\dfrac{4}{1-{{x}^{2}}}=1x−1x+1+1−x24=1．知识点：解分式方程,一元一次不等式组的解法章节：未标注
