Bài tập 2.65 trang 133 SBT Toán 12
Bài tập 2.65 trang 133 SBT Toán 12
Tìm tập xác định của các hàm số sau :
a) \(y = \frac{2}{{\sqrt {{4^x} - 2} }}\)
b) \(y = {\log _6}\frac{{3x + 2}}{{1 - x}}\)
c) \(y = \sqrt {\log x + \log \left( {x + 2} \right)} \)
d) \(y = \sqrt {\log \left( {x - 1} \right) + \log \left( {x + 1} \right)} \)
a) \({4^x} - 2 > 0 \Leftrightarrow {4^x} > 2 \Leftrightarrow x > \frac{1}{2}\)
TXĐ: \(D = \left( {\frac{1}{2}; + \infty } \right)\)
b) \({ \Leftrightarrow \frac{{ - 2}}{3} < x < 1}\)
TXĐ: \({D = \left( {\frac{{ - 2}}{3};1} \right)}\)
c)
\(\begin{array}{l}
\left\{ \begin{array}{l}
\log x + \log \left( {x + 2} \right) \ge 0\\
x > 0\\
x + 2 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\log \left[ {x\left( {x + 2} \right)} \right] \ge 0\\
x > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {x + 2} \right) \ge 1\\
x > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} + 2x - 1 \ge 0\\
x > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x < - 1 - \sqrt 2 \\
x > - 1 + \sqrt 2
\end{array} \right.\\
x > 0
\end{array} \right. \Leftrightarrow x > - 1 + \sqrt 2
\end{array}\)
TXĐ: \(D = \left( { - 1 + \sqrt 2 ; + \infty } \right)\)
d)
\(\begin{array}{l}
\left\{ \begin{array}{l}
\log \left( {x - 1} \right) + \log \left( {x + 1} \right) \ge 0\\
x - 1 > 0\\
x + 1 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {x + 1} \right) \ge 1\\
x > 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 2 \ge 0\\
x > 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \le - \sqrt 2 \\
x \ge \sqrt 2
\end{array} \right.\\
x > 1
\end{array} \right. \Leftrightarrow x \ge \sqrt 2
\end{array}\)
-- Mod Toán 12
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