Câu hỏi: Có bao nhiêu số nguyên $x\in \left[ -2022;2022 \right]$ thỏa mãn $\left( {{3}^{{{x}^{2}}}}-{{27}^{x}} \right)\sqrt{{{\log }_{2}}\left( 4x \right)-2}\ge 0$ ?
A. $2021$
B. $2020$
C. $2023$
D. $2022$
A. $2021$
B. $2020$
C. $2023$
D. $2022$
$\left( {{3}^{{{x}^{2}}}}-{{27}^{x}} \right)\sqrt{{{\log }_{2}}\left( 4x \right)-2}\ge 0$ $\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}\left( 4x \right)-2=0 \\
& \left\{ \begin{aligned}
& {{\log }_{2}}\left( 4x \right)-2>0 \\
& {{3}^{{{x}^{2}}}}-{{27}^{x}}\ge 0 \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& \left\{ \begin{aligned}
& x>1 \\
& {{x}^{2}}\ge 3x \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x\ge 3 \\
\end{aligned} \right.$.
Mà $x$ nguyên thuộc $\left[ -2022;2022 \right]$ nên có 2021 số.
& {{\log }_{2}}\left( 4x \right)-2=0 \\
& \left\{ \begin{aligned}
& {{\log }_{2}}\left( 4x \right)-2>0 \\
& {{3}^{{{x}^{2}}}}-{{27}^{x}}\ge 0 \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& \left\{ \begin{aligned}
& x>1 \\
& {{x}^{2}}\ge 3x \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x\ge 3 \\
\end{aligned} \right.$.
Mà $x$ nguyên thuộc $\left[ -2022;2022 \right]$ nên có 2021 số.
Đáp án A.