Giải bài 2.65 trang 133 - SBT Giải tích lớp 12
Tìm tập xác định của các hàm số sau :
\(\begin{align} & a)\,y=\dfrac{2}{\sqrt{{{4}^{x}}-2}};\, \\ & b)\,y={{\log }_{6}}\dfrac{3x+2}{1-x}; \\ & c)\,y=\sqrt{\log x+\log \left( x+2 \right)}; \\ & d)\,y=\sqrt{\log \left( x-1 \right)+\log \left( x+1 \right)}. \\ \end{align} \)
a)
\({{4}^{x}}-2>0\Leftrightarrow {{4}^{x}}>2\Leftrightarrow x>\dfrac{1}{2} \)
\(D=\left( \dfrac{1}{2};+\infty \right) \)
b)
\(\begin{aligned} & \dfrac{3x+2}{1-x}>0 \\ & \Leftrightarrow \dfrac{-2}{3}< x<1 \\ \end{aligned} \)
\(D=\left( \dfrac{-2}{3};1 \right) \)
c)
\(\begin{aligned} & \left\{ \begin{aligned} & \log x+\log \left( x+2 \right)\ge 0 \\ & x>0 \\ & x+2>0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & \log \left[ x\left( x+2 \right) \right]\ge 0 \\ & x>0 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x\left( x+2 \right)\ge 1 \\ & x>0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}+2x-1\ge 0 \\ & x>0 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \left[ \begin{aligned} & x<-1-\sqrt{2} \\ & x>-1+\sqrt{2} \\ \end{aligned} \right. \\ & x>0 \\ \end{aligned} \right.\Leftrightarrow x>-1+\sqrt{2} \\ \end{aligned} \)
\(D=\left( -1+\sqrt{2};+\infty \right) \)
d)
\(\begin{aligned} & \left\{ \begin{aligned} & \log \left( x-1 \right)+\log \left( x+1 \right)\ge 0 \\ & x-1>0 \\ & x+1>0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & \left( x-1 \right)\left( x+1 \right)\ge 1 \\ & x>1 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}-2\ge 0 \\ & x>1 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & \left[ \begin{aligned} & x\le -\sqrt{2} \\ & x\ge \sqrt{2} \\ \end{aligned} \right. \\ & x>1 \\ \end{aligned} \right.\Leftrightarrow x\ge \sqrt{2} \\ \end{aligned} \)