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Câu hỏi: Tìm tập xác định của các hàm số sau:
a) \(\displaystyle y = {\log _8}\left( {{x^2} - 3x - 4} \right)\)
b) \(\displaystyle y = {\log _{\sqrt 3 }}\left( { - {x^2} + 5x + 6} \right)\)
c) \(\displaystyle y = {\log _{0,7}}\dfrac{{{x^2} - 9}}{{x + 5}}\)
d) \(\displaystyle y = {\log _{\frac{1}{3}}}\dfrac{{x - 4}}{{x + 4}}\)
e) \(\displaystyle y = {\log _\pi }\left( {{2^x} - 2} \right)\)
g) \(\displaystyle y = {\log _3}\left( {{3^{x - 1}} - 9} \right)\)
a) \(\displaystyle y = {\log _8}\left( {{x^2} - 3x - 4} \right)\)
b) \(\displaystyle y = {\log _{\sqrt 3 }}\left( { - {x^2} + 5x + 6} \right)\)
c) \(\displaystyle y = {\log _{0,7}}\dfrac{{{x^2} - 9}}{{x + 5}}\)
d) \(\displaystyle y = {\log _{\frac{1}{3}}}\dfrac{{x - 4}}{{x + 4}}\)
e) \(\displaystyle y = {\log _\pi }\left( {{2^x} - 2} \right)\)
g) \(\displaystyle y = {\log _3}\left( {{3^{x - 1}} - 9} \right)\)
Phương pháp giải
Hàm số \(\displaystyle y = {\log _a}f\left( x \right)\) xác định khi \(\displaystyle f\left( x \right)\) xác định và \(\displaystyle f\left( x \right) > 0\).
Lời giải chi tiết
a) ĐKXĐ: \(\displaystyle {x^2} - 3x - 4 > 0\) \(\displaystyle \Leftrightarrow \left( {x + 1} \right)\left({x - 4} \right) > 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x > 4\\x < - 1\end{array} \right.\).
Vậy TXĐ \(\displaystyle D = \left( { - \infty ; - 1} \right) \cup \left({4; + \infty } \right)\).
b) ĐKXĐ: \(\displaystyle - {x^2} + 5x + 6 > 0\) \(\displaystyle \Leftrightarrow \left( {x + 1} \right)\left({6 - x} \right) > 0\) \(\displaystyle \Leftrightarrow - 1 < x < 6\).
Vậy TXĐ \(\displaystyle D = \left( { - 1; 6} \right)\).
c) ĐKXĐ: \(\displaystyle \dfrac{{{x^2} - 9}}{{x + 5}} > 0\) \(\displaystyle \Leftrightarrow \dfrac{{\left( {x - 3} \right)\left({x + 3} \right)}}{{x + 5}} > 0\).
Xét dấu vế trái ta được:
Vậy TXĐ \(\displaystyle D = \left( { - 5; - 3} \right) \cup \left({3; + \infty } \right)\).
d) ĐKXĐ: \(\displaystyle \dfrac{{x - 4}}{{x + 4}} > 0 \Leftrightarrow \left[ \begin{array}{l}x > 4\\x < - 4\end{array} \right.\).
Vậy TXĐ: \(\displaystyle D = \left( { - \infty ; - 4} \right) \cup \left({4; + \infty } \right)\).
e) ĐKXĐ: \(\displaystyle {2^x} - 2 > 0 \Leftrightarrow {2^x} > 2\) \(\displaystyle \Leftrightarrow {2^x} > {2^1} \Leftrightarrow x > 1\).
Vậy TXĐ: \(\displaystyle D = \left( {1; + \infty } \right)\).
g) ĐKXĐ: \(\displaystyle {3^{x - 1}} - 9 > 0 \Leftrightarrow {3^{x - 1}} > 9\) \(\displaystyle \Leftrightarrow {3^{x - 1}} > {3^2} \Leftrightarrow x - 1 > 2\) \(\displaystyle \Leftrightarrow x > 3\).
Vậy TXĐ: \(\displaystyle D = \left( {3; + \infty } \right)\).
Hàm số \(\displaystyle y = {\log _a}f\left( x \right)\) xác định khi \(\displaystyle f\left( x \right)\) xác định và \(\displaystyle f\left( x \right) > 0\).
Lời giải chi tiết
a) ĐKXĐ: \(\displaystyle {x^2} - 3x - 4 > 0\) \(\displaystyle \Leftrightarrow \left( {x + 1} \right)\left({x - 4} \right) > 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x > 4\\x < - 1\end{array} \right.\).
Vậy TXĐ \(\displaystyle D = \left( { - \infty ; - 1} \right) \cup \left({4; + \infty } \right)\).
b) ĐKXĐ: \(\displaystyle - {x^2} + 5x + 6 > 0\) \(\displaystyle \Leftrightarrow \left( {x + 1} \right)\left({6 - x} \right) > 0\) \(\displaystyle \Leftrightarrow - 1 < x < 6\).
Vậy TXĐ \(\displaystyle D = \left( { - 1; 6} \right)\).
c) ĐKXĐ: \(\displaystyle \dfrac{{{x^2} - 9}}{{x + 5}} > 0\) \(\displaystyle \Leftrightarrow \dfrac{{\left( {x - 3} \right)\left({x + 3} \right)}}{{x + 5}} > 0\).
Xét dấu vế trái ta được:
Vậy TXĐ \(\displaystyle D = \left( { - 5; - 3} \right) \cup \left({3; + \infty } \right)\).
d) ĐKXĐ: \(\displaystyle \dfrac{{x - 4}}{{x + 4}} > 0 \Leftrightarrow \left[ \begin{array}{l}x > 4\\x < - 4\end{array} \right.\).
Vậy TXĐ: \(\displaystyle D = \left( { - \infty ; - 4} \right) \cup \left({4; + \infty } \right)\).
e) ĐKXĐ: \(\displaystyle {2^x} - 2 > 0 \Leftrightarrow {2^x} > 2\) \(\displaystyle \Leftrightarrow {2^x} > {2^1} \Leftrightarrow x > 1\).
Vậy TXĐ: \(\displaystyle D = \left( {1; + \infty } \right)\).
g) ĐKXĐ: \(\displaystyle {3^{x - 1}} - 9 > 0 \Leftrightarrow {3^{x - 1}} > 9\) \(\displaystyle \Leftrightarrow {3^{x - 1}} > {3^2} \Leftrightarrow x - 1 > 2\) \(\displaystyle \Leftrightarrow x > 3\).
Vậy TXĐ: \(\displaystyle D = \left( {3; + \infty } \right)\).