题目内容

17.不等式$\frac{x-1}{{{x^2}-x-6}}$≥0的解集为(  )
A.(-∞,-2)∪(3,+∞)B.(-∞,-2)∪[1,3)C.(-2,1]∪(3,+∞)D.(-2,1)∪[1,3)

分析 原不等式等价于$\left\{\begin{array}{l}{x-1≥0}\\{(x+2)(x-3)>0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1≤0}\\{(x+2)(x-3)<0}\end{array}\right.$,解得即可.

解答 解:$\frac{x-1}{{{x^2}-x-6}}$≥0等价于$\frac{x-1}{(x+2)(x-3)}$≥0等价于$\left\{\begin{array}{l}{x-1≥0}\\{(x+2)(x-3)>0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1≤0}\\{(x+2)(x-3)<0}\end{array}\right.$,
解得x>3或-2<x<1,
故不等式的解集为(-2,1]∪(3,+∞),
故选:C

点评 本题考查了分式不等式的解法,关键是分类讨论,属于基础题.

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