Question

5．函數(shù)y=$\frac{\sqrt{{x}^{2}-4}}{lg（{x}^{2}+2x-3）}$定義域為（-∞，-1-$\sqrt{5}$）∪（-1-$\sqrt{5}$，-3）∪[2，+∞）．

Answer 1

分析 根據(jù)函數(shù)y的解析式，列出不等式組$\left\{\begin{array}{l}{{x}^{2}-4≥0}\\{{x}^{2}+2x-3＞0}\\{{x}^{2}+2x-3≠1}\end{array}\right.$，求出解集即可．

解答 解：∵函數(shù)y=$\frac{\sqrt{{x}^{2}-4}}{lg（{x}^{2}+2x-3）}$，
∴$\left\{\begin{array}{l}{{x}^{2}-4≥0}\\{{x}^{2}+2x-3＞0}\\{{x}^{2}+2x-3≠1}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x≤-2或x≥2}\\{x＜-3或x＞1}\\{x≠-1-\sqrt{5}且x≠-1+\sqrt{5}}\end{array}\right.$
即x＜-1-$\sqrt{5}$或-1-$\sqrt{5}$＜x＜-3或x≥2，
∴y的定義域為（-∞，-1-$\sqrt{5}$）∪（-1-$\sqrt{5}$，-3）∪[2，+∞）．
故答案為：（-∞，-1-$\sqrt{5}$）∪（-1-$\sqrt{5}$，-3）∪[2，+∞）．

點(diǎn)評 本題考查了根據(jù)函數(shù)的解析式求定義域的應(yīng)用問題，解題的關(guān)鍵是求不等式組的解集．