Google
Câu hỏi: Gọi $a$ là giá trị nhỏ nhất của $f\left( n \right)=\dfrac{\left( {{\log }_{5}}2 \right)\left( {{\log }_{5}}3 \right)\left( {{\log }_{5}}4 \right)...\left( {{\log }_{5}}n \right)}{{{3}^{n}}},$ với $n\in \mathbb{N},n\ge 2.$ Có bao nhiêu số $n$ để $f\left( n \right)=a?$
A. 4.
B. Vô số.
C. 2.
D. 1.
A. 4.
B. Vô số.
C. 2.
D. 1.
Ta có $\forall x\in \mathbb{N},n\ge 2$ ta có: $f\left( n \right)>0.$
Mặt khác: $f\left( n+1 \right)=\dfrac{\left( {{\log }_{5}}2 \right)\left( {{\log }_{5}}3 \right)\left( {{\log }_{5}}4 \right)...\left( {{\log }_{5}}n \right)\left( {{\log }_{5}}\left( n+1 \right) \right)}{{{3}^{n+1}}}=f\left( n \right)\dfrac{{{\log }_{5}}\left( n+1 \right)}{3}.$
$f\left( n-1 \right)=\dfrac{\left( {{\log }_{5}}2 \right)\left( {{\log }_{5}}3 \right)\left( {{\log }_{5}}4 \right)...\left( {{\log }_{5}}\left( n-1 \right) \right)}{{{3}^{n-1}}}=f\left( n \right)\dfrac{3}{{{\log }_{5}}n}.$
Vì $a$ là giá trị nhỏ nhất nên: $\left\{ \begin{aligned}
& f\left( n+1 \right)\ge a \\
& f\left( n-1 \right)\ge a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( n \right)\dfrac{{{\log }_{5}}\left( n+1 \right)}{3}\ge a \\
& f\left( n \right)\dfrac{3}{{{\log }_{5}}n}\ge a \\
\end{aligned} \right.$.
Để $f\left( n \right)=a.$
Suy ra: $\left\{ \begin{aligned}
& f\left( n \right)\dfrac{{{\log }_{5}}\left( n+1 \right)}{3}\ge f\left( n \right) \\
& f\left( n \right)\dfrac{3}{{{\log }_{5}}n}\ge f\left( n \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{{{\log }_{5}}\left( n+1 \right)}{3}\ge 1 \\
& \dfrac{3}{{{\log }_{5}}n}\ge 1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{5}}\left( n+1 \right)\ge 3 \\
& 3\ge {{\log }_{5}}n \\
\end{aligned} \right.$
$\Leftrightarrow {{5}^{3}}-1\le n\le {{5}^{3}}.$
Vậy có 2 số $n$ nguyên thỏa mãn.
Mặt khác: $f\left( n+1 \right)=\dfrac{\left( {{\log }_{5}}2 \right)\left( {{\log }_{5}}3 \right)\left( {{\log }_{5}}4 \right)...\left( {{\log }_{5}}n \right)\left( {{\log }_{5}}\left( n+1 \right) \right)}{{{3}^{n+1}}}=f\left( n \right)\dfrac{{{\log }_{5}}\left( n+1 \right)}{3}.$
$f\left( n-1 \right)=\dfrac{\left( {{\log }_{5}}2 \right)\left( {{\log }_{5}}3 \right)\left( {{\log }_{5}}4 \right)...\left( {{\log }_{5}}\left( n-1 \right) \right)}{{{3}^{n-1}}}=f\left( n \right)\dfrac{3}{{{\log }_{5}}n}.$
Vì $a$ là giá trị nhỏ nhất nên: $\left\{ \begin{aligned}
& f\left( n+1 \right)\ge a \\
& f\left( n-1 \right)\ge a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( n \right)\dfrac{{{\log }_{5}}\left( n+1 \right)}{3}\ge a \\
& f\left( n \right)\dfrac{3}{{{\log }_{5}}n}\ge a \\
\end{aligned} \right.$.
Để $f\left( n \right)=a.$
Suy ra: $\left\{ \begin{aligned}
& f\left( n \right)\dfrac{{{\log }_{5}}\left( n+1 \right)}{3}\ge f\left( n \right) \\
& f\left( n \right)\dfrac{3}{{{\log }_{5}}n}\ge f\left( n \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{{{\log }_{5}}\left( n+1 \right)}{3}\ge 1 \\
& \dfrac{3}{{{\log }_{5}}n}\ge 1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{5}}\left( n+1 \right)\ge 3 \\
& 3\ge {{\log }_{5}}n \\
\end{aligned} \right.$
$\Leftrightarrow {{5}^{3}}-1\le n\le {{5}^{3}}.$
Vậy có 2 số $n$ nguyên thỏa mãn.
Đáp án C.