Question

12．材料：分母中含有未知数的不等式叫分式不等式，如：$\frac{2x}{x+1}$＞0；$\frac{x+3}{x-1}$＜0等．那么如何求出它们的解集呢？
根据我们学过的有理数除法法则可知：两数相除，同号得正，异号得负．其字母表达式为：
（1）若a＞0，b＞0，则$\frac{a}{b}$＞0；若a＜0，b＜0，则$\frac{a}{b}$＞0；
（2）若a＞0，b＜0，则$\frac{a}{b}$＜0；若a＜0，b＞0，则$\frac{a}{b}$＜0．
反之：（1）若$\frac{a}{b}$＞0，则$\left\{\begin{array}{l}{a＞0}\\{b＞0}\end{array}\right.$或$\left\{\begin{array}{l}{a＜0}\\{b＜0}\end{array}\right.$
（2）若$\frac{a}{b}$＜0，则$\left\{\begin{array}{l}{a＞0}\\{b＜0}\end{array}\right.$或$\left\{\begin{array}{l}{a＜0}\\{b＞0}\end{array}\right.$．
根据上述规律，求不等式$\frac{x+1}{x-3}$＞0的解集．

Answer 1

分析 由题意知若$\frac{a}{b}$＜0，则$\left\{\begin{array}{l}{a＞0}\\{b＜0}\end{array}\right.$或$\left\{\begin{array}{l}{a＜0}\\{b＞0}\end{array}\right.$；将不等式$\frac{x+1}{x-3}$＞0转化为$\left\{\begin{array}{l}{x+1＞0}\\{x-3＞0}\end{array}\right.$或$\left\{\begin{array}{l}{x+1＜0}\\{x-3＜0}\end{array}\right.$，分别求每个不等式组的解集即可．

解答 解：根据题意若$\frac{a}{b}$＜0，则$\left\{\begin{array}{l}{a＞0}\\{b＜0}\end{array}\right.$或$\left\{\begin{array}{l}{a＜0}\\{b＞0}\end{array}\right.$，
由不等式$\frac{x+1}{x-3}$＞0得：$\left\{\begin{array}{l}{x+1＞0}\\{x-3＞0}\end{array}\right.$或$\left\{\begin{array}{l}{x+1＜0}\\{x-3＜0}\end{array}\right.$，
解不等式组$\left\{\begin{array}{l}{x+1＞0}\\{x-3＞0}\end{array}\right.$得：x＞3，
解不等式组$\left\{\begin{array}{l}{x+1＜0}\\{x-3＜0}\end{array}\right.$得：x＜-1，
故不等式$\frac{x+1}{x-3}$＞0的解集为：x＜-1或x＞3，
故答案为：$\left\{\begin{array}{l}{a＞0}\\{b＜0}\end{array}\right.$，$\left\{\begin{array}{l}{a＜0}\\{b＞0}\end{array}\right.$．

点评 本题主要考查解不等式、不等式组的能力，将原不等式转化为两个不等式组是解题的关键．