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Câu hỏi: Có bao nhiêu số nguyên $x\in \left[ -10;10 \right]$ thỏa mãn
$\left[ {{\log }_{4}}\left( x+5 \right)-{{\log }_{16}}\left( 5{{x}^{2}}+12 \right) \right].\left( {{5}^{4x-1}}-{{125}^{x}} \right)\ge 0?$
A. 10.
B. 13.
C. 3.
D. 17.
$\left[ {{\log }_{4}}\left( x+5 \right)-{{\log }_{16}}\left( 5{{x}^{2}}+12 \right) \right].\left( {{5}^{4x-1}}-{{125}^{x}} \right)\ge 0?$
A. 10.
B. 13.
C. 3.
D. 17.
Điều kiện $x>-5.$
Trường hợp 1:
$\left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)-{{\log }_{16}}\left( 5{{x}^{2}}+12 \right)\ge 0 \\
& {{5}^{4x-1}}-{{125}^{x}}\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)\ge {{\log }_{16}}\left( 5{{x}^{2}}+12 \right) \\
& {{5}^{4x-1}}\ge {{5}^{3x}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x+5\ge \sqrt{5{{x}^{2}}+12} \\
& 4x-1\ge 3x \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -4{{x}^{2}}+10x+13\ge 0 \\
& x\ge 1 \\
\end{aligned} \right.\Leftrightarrow 1\le x\le \dfrac{5+\sqrt{77}}{4}.$
Mà $x$ nguyên nên $x\in \left\{ 1;2;3 \right\}.$
Trường hợp 2:
$\left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)-{{\log }_{16}}\left( 5{{x}^{2}}+12 \right)\le 0 \\
& {{5}^{4x-1}}-{{125}^{x}}\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)\le {{\log }_{16}}\left( 5{{x}^{2}}+12 \right) \\
& {{5}^{4x-1}}\le {{5}^{3x}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x+5\le \sqrt{5{{x}^{2}}+12} \\
& 4x-1\le 3x \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -4{{x}^{2}}+10x+13\le 0 \\
& x\le 1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x\ge \dfrac{5+\sqrt{77}}{4} \\
& x\le \dfrac{5-\sqrt{77}}{4} \\
\end{aligned} \right. \\
& x\le 1 \\
\end{aligned} \right.\Leftrightarrow x\le \dfrac{5-\sqrt{77}}{4}.$
Mà $x$ nguyên và $x\in \left[ -10;10 \right]$ nên $x\in \left\{ -10;-9;...;-1 \right\}.$
Vậy có tất cả 13 nghiệm nguyên $x.$
Trường hợp 1:
$\left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)-{{\log }_{16}}\left( 5{{x}^{2}}+12 \right)\ge 0 \\
& {{5}^{4x-1}}-{{125}^{x}}\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)\ge {{\log }_{16}}\left( 5{{x}^{2}}+12 \right) \\
& {{5}^{4x-1}}\ge {{5}^{3x}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x+5\ge \sqrt{5{{x}^{2}}+12} \\
& 4x-1\ge 3x \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -4{{x}^{2}}+10x+13\ge 0 \\
& x\ge 1 \\
\end{aligned} \right.\Leftrightarrow 1\le x\le \dfrac{5+\sqrt{77}}{4}.$
Mà $x$ nguyên nên $x\in \left\{ 1;2;3 \right\}.$
Trường hợp 2:
$\left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)-{{\log }_{16}}\left( 5{{x}^{2}}+12 \right)\le 0 \\
& {{5}^{4x-1}}-{{125}^{x}}\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{4}}\left( x+5 \right)\le {{\log }_{16}}\left( 5{{x}^{2}}+12 \right) \\
& {{5}^{4x-1}}\le {{5}^{3x}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x+5\le \sqrt{5{{x}^{2}}+12} \\
& 4x-1\le 3x \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -4{{x}^{2}}+10x+13\le 0 \\
& x\le 1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x\ge \dfrac{5+\sqrt{77}}{4} \\
& x\le \dfrac{5-\sqrt{77}}{4} \\
\end{aligned} \right. \\
& x\le 1 \\
\end{aligned} \right.\Leftrightarrow x\le \dfrac{5-\sqrt{77}}{4}.$
Mà $x$ nguyên và $x\in \left[ -10;10 \right]$ nên $x\in \left\{ -10;-9;...;-1 \right\}.$
Vậy có tất cả 13 nghiệm nguyên $x.$
Đáp án B.