题目内容
18.解不等式:(1)$\frac{{x}^{2}-2x-1}{x-2}$<0
(2)$\frac{(x-1)^{3}({x}^{2}+x+6)}{(x+3)^{2}}$≤0.
分析 (1)原不等式即为$\left\{\begin{array}{l}{x-2>0}\\{{x}^{2}-2x-1<0}\end{array}\right.$或$\left\{\begin{array}{l}{x-2<0}\\{{x}^{2}-2x-1>0}\end{array}\right.$,由二次不等式的解法,即可得到所求解集;
(2)由x2+x+6>0恒成立,可得原不等式即为x-1≤0且x+3≠0,即可得到所求解集.
解答 解:(1)原不等式即为$\left\{\begin{array}{l}{x-2>0}\\{{x}^{2}-2x-1<0}\end{array}\right.$或$\left\{\begin{array}{l}{x-2<0}\\{{x}^{2}-2x-1>0}\end{array}\right.$,
即有$\left\{\begin{array}{l}{x>2}\\{1-\sqrt{2}<x<1+\sqrt{2}}\end{array}\right.$或$\left\{\begin{array}{l}{x<2}\\{x>1+\sqrt{2}或x<1-\sqrt{2}}\end{array}\right.$,
即为2<x<1+$\sqrt{2}$或x<1-$\sqrt{2}$.
则解集为(2,1+$\sqrt{2}$)∪(-∞,1-$\sqrt{2}$);
(2)由x2+x+6=(x+$\frac{1}{2}$)2+$\frac{23}{4}$>0,
原不等式等价为x-1≤0且x+3≠0,
解得x≤1且x≠-3,
则解集为{x|x≤1且x≠-3}.
点评 本题考查分式不等式的解法,注意运用等价变形为整式不等式,考查运算能力,属于基础题和易错题.
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