Câu hỏi: Tập xác định của hàm số
A.
B.
C.
D.
A.
B.
C.
D.
Ta có điều kiện:
$\left\{ \begin{aligned}
& {{x}^{2}}+7x>0 \\
& {{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}+7x \right)+3\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& {{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}+7x \right)\ge -3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& {{x}^{2}}+7x\le 8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& {{x}^{2}}+7x-8\le 0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& -8\le x\le 1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& 0<x\le 1 \\
& -8\le x<-7 \\
\end{aligned} \right.
$\left\{ \begin{aligned}
& {{x}^{2}}+7x>0 \\
& {{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}+7x \right)+3\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& {{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}+7x \right)\ge -3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& {{x}^{2}}+7x\le 8 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& {{x}^{2}}+7x-8\le 0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>0 \\
& x<-7 \\
\end{aligned} \right. \\
& -8\le x\le 1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& 0<x\le 1 \\
& -8\le x<-7 \\
\end{aligned} \right.
Đáp án D.