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Câu hỏi: Tất cả các giá trị nguyên của $x$ thoả mãn bất phương trình ${{\log }_{\dfrac{\pi }{4}}}\left( {{x}^{2}}-3x \right)<{{\log }_{\dfrac{\pi }{4}}}\left( x+4 \right)$
A. $2-2\sqrt{2}<x<2+2\sqrt{2}$.
B. $\left[ \begin{aligned}
& x<2-2\sqrt{2} \\
& x>2+2\sqrt{2} \\
\end{aligned} \right.$.
C. $\left[ \begin{aligned}
& -4<x<2-2\sqrt{2} \\
& x>2+2\sqrt{2} \\
\end{aligned} \right. $.
D. $ 2-2\sqrt{2}<x<0$.
A. $2-2\sqrt{2}<x<2+2\sqrt{2}$.
B. $\left[ \begin{aligned}
& x<2-2\sqrt{2} \\
& x>2+2\sqrt{2} \\
\end{aligned} \right.$.
C. $\left[ \begin{aligned}
& -4<x<2-2\sqrt{2} \\
& x>2+2\sqrt{2} \\
\end{aligned} \right. $.
D. $ 2-2\sqrt{2}<x<0$.
${{\log }_{\dfrac{\pi }{4}}}\left( {{x}^{2}}-3x \right)<{{\log }_{\dfrac{\pi }{4}}}\left( x+4 \right)$ $\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-3x>x+4 \\
& x+4>0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-4x-4>0 \\
& x>-4 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>2+2\sqrt{2} \\
& x<2-2\sqrt{2} \\
\end{aligned} \right. \\
& x>-4 \\
\end{aligned} \right. $ $ \Leftrightarrow \left[ \begin{aligned}
& -4<x<2-2\sqrt{2} \\
& x>2+2\sqrt{2} \\
\end{aligned} \right.$.
& {{x}^{2}}-3x>x+4 \\
& x+4>0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-4x-4>0 \\
& x>-4 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>2+2\sqrt{2} \\
& x<2-2\sqrt{2} \\
\end{aligned} \right. \\
& x>-4 \\
\end{aligned} \right. $ $ \Leftrightarrow \left[ \begin{aligned}
& -4<x<2-2\sqrt{2} \\
& x>2+2\sqrt{2} \\
\end{aligned} \right.$.
Đáp án C.