Question

14．試求下列函數(shù)的定義域：
（1）y=log_（x+2）（32-4^x）；
（2）y=$\frac{\sqrt{lo{g}_{0.8}x-1}}{2x-1}$；
（3）y=$\frac{1}{\sqrt{1-lo{g}_{a}（x+a）}}$．

Answer 1

分析 根據(jù)函數(shù)成立的條件即可求函數(shù)的定義域．

解答 解：（1）要使函數(shù)有意義，則$\left\{\begin{array}{l}{32-{4}^{x}＞0}\\{x+2＞0}\\{x+2≠1}\end{array}\right.$，
即$\left\{\begin{array}{l}{{2}^{2x}＜{2}^{5}}\\{x＞-2}\\{x≠-1}\end{array}\right.$，即$\left\{\begin{array}{l}{2x＜5}\\{x＞-2}\\{x≠1}\end{array}\right.$，則$\left\{\begin{array}{l}{x＜\frac{5}{2}}\\{x＞-2}\\{x≠-1}\end{array}\right.$，即-2＜x＜-1或-1＜x＜$\frac{5}{2}$，故函數(shù)的定義域?yàn)閧x|-2＜x＜-1或-1＜x＜$\frac{5}{2}$}．
（2）要使函數(shù)有意義，則$\left\{\begin{array}{l}{2x-1≠0}\\{lo{g}_{0.8}x-1≥0}\end{array}\right.$，即$\left\{\begin{array}{l}{x≠\frac{1}{2}}\\{lo{g}_{0.8}x≥1}\end{array}\right.$，即$\left\{\begin{array}{l}{x≠\frac{1}{2}}\\{0＜x≤0.8}\end{array}\right.$，解得0＜x≤$\frac{4}{5}$且x≠$\frac{1}{2}$，即函數(shù)的定義域?yàn)閧x|0＜x≤$\frac{4}{5}$且x≠$\frac{1}{2}$}．
（3）要使函數(shù)有意義，則$\left\{\begin{array}{l}{x+a＞0}\\{1-lo{g}_{a}（x+a）＞0}\end{array}\right.$，
即$\left\{\begin{array}{l}{x＞-a}\\{lo{g}_{a}（x+a）＜1}\end{array}\right.$，
則$\left\{\begin{array}{l}{a＞1}\\{x＞-a}\\{x+a＜a}\end{array}\right.$或$\left\{\begin{array}{l}{0＜a＜1}\\{x＞-a}\\{x+a＞a}\end{array}\right.$，
即$\left\{\begin{array}{l}{a＞1}\\{x＞-a}\\{x＜0}\end{array}\right.$或$\left\{\begin{array}{l}{0＜a＜1}\\{x＞-a}\\{x＞0}\end{array}\right.$，
即a＞1時(shí)，-a＜x＜0，此時(shí)函數(shù)的定義域?yàn)椋?a，0），
0＜a＜1時(shí)，x＞0，此時(shí)函數(shù)的定義域?yàn)椋?，+∞）．

點(diǎn)評(píng) 本題主要考查函數(shù)定義域的求解，要求熟練掌握常見函數(shù)成立的條件