Câu hỏi: Có bao nhiêu số nguyên $x$ thỏa mãn $\left( \log _{5}^{2}x-{{\log }_{5}}{{x}^{3}}+2 \right)\sqrt{6561-{{3}^{x}}}\ge 0$ là
A. $8$.
B. $5$.
C. $6$.
D. $7$.
A. $8$.
B. $5$.
C. $6$.
D. $7$.
Điều kiện xác định: $\left\{ \begin{aligned}
& x>0 \\
& 6561-{{3}^{x}}\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0 \\
& x\le {{\log }_{3}}6561 \\
\end{aligned} \right.\Leftrightarrow 0<x\le 8.$.
Với điều kiện trên ta có:
$\begin{aligned}
& \left( \log _{5}^{2}x-{{\log }_{5}}{{x}^{3}}+2 \right)\sqrt{6561-{{3}^{x}}}\ge 0\Leftrightarrow \left( \log _{5}^{2}x-3{{\log }_{5}}x+2 \right)\sqrt{6561-{{3}^{x}}}\ge 0 \\
& \Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left( \log _{5}^{2}x-3{{\log }_{5}}x+2 \right)\sqrt{6561-{{3}^{x}}}=0 \\
\end{aligned} \right. \\
& \left( \log _{5}^{2}x-3{{\log }_{5}}x+2 \right)\sqrt{6561-{{3}^{x}}}>0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left[ \begin{aligned}
& x=8 \\
& \log _{5}^{2}x-3{{\log }_{5}}x+2=0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 0<x<8 \\
& \log _{5}^{2}x-3{{\log }_{5}}x+2>0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned}$
$\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left[ \begin{aligned}
& x=8 \\
& {{\log }_{5}}x=1 \\
& {{\log }_{5}}x=2 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 0<x<8 \\
& \left[ \begin{aligned}
& {{\log }_{5}}x>2 \\
& {{\log }_{5}}x<1 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left[ \begin{aligned}
& x=8\left( tm \right) \\
& x=5\ \left( tm \right) \\
& x=25\ \left( ktm \right) \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 0<x<8 \\
& \left[ \begin{aligned}
& x>25 \\
& x<5 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=8 \\
& 0<x\le 5 \\
\end{aligned} \right.$.
Mà $x\in Z\Rightarrow x\in \left\{ 1;2;3;4;5;8 \right\}\Rightarrow $ có $6$ số nguyên $x$ thỏa mãn.
& x>0 \\
& 6561-{{3}^{x}}\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0 \\
& x\le {{\log }_{3}}6561 \\
\end{aligned} \right.\Leftrightarrow 0<x\le 8.$.
Với điều kiện trên ta có:
$\begin{aligned}
& \left( \log _{5}^{2}x-{{\log }_{5}}{{x}^{3}}+2 \right)\sqrt{6561-{{3}^{x}}}\ge 0\Leftrightarrow \left( \log _{5}^{2}x-3{{\log }_{5}}x+2 \right)\sqrt{6561-{{3}^{x}}}\ge 0 \\
& \Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left( \log _{5}^{2}x-3{{\log }_{5}}x+2 \right)\sqrt{6561-{{3}^{x}}}=0 \\
\end{aligned} \right. \\
& \left( \log _{5}^{2}x-3{{\log }_{5}}x+2 \right)\sqrt{6561-{{3}^{x}}}>0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left[ \begin{aligned}
& x=8 \\
& \log _{5}^{2}x-3{{\log }_{5}}x+2=0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 0<x<8 \\
& \log _{5}^{2}x-3{{\log }_{5}}x+2>0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned}$
$\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left[ \begin{aligned}
& x=8 \\
& {{\log }_{5}}x=1 \\
& {{\log }_{5}}x=2 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 0<x<8 \\
& \left[ \begin{aligned}
& {{\log }_{5}}x>2 \\
& {{\log }_{5}}x<1 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 0<x\le 8 \\
& \left[ \begin{aligned}
& x=8\left( tm \right) \\
& x=5\ \left( tm \right) \\
& x=25\ \left( ktm \right) \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 0<x<8 \\
& \left[ \begin{aligned}
& x>25 \\
& x<5 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=8 \\
& 0<x\le 5 \\
\end{aligned} \right.$.
Mà $x\in Z\Rightarrow x\in \left\{ 1;2;3;4;5;8 \right\}\Rightarrow $ có $6$ số nguyên $x$ thỏa mãn.
Đáp án C.