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Câu hỏi: Gọi ${{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{20}}$ là các số thực thuộc khoảng $\left( \dfrac{1}{4};1 \right)$ và $M$ là giá trị nhỏ nhất của biểu thức sau:
$P={{\log }_{\sqrt{{{a}_{1}}}}}{{\left( {{a}_{2}}-\dfrac{1}{4} \right)}^{3}}+{{\log }_{\sqrt{{{a}_{2}}}}}{{\left( {{a}_{3}}-\dfrac{1}{4} \right)}^{3}}{{\log }_{\sqrt{{{a}_{3}}}}}{{\left( {{a}_{4}}-\dfrac{1}{4} \right)}^{3}}+...+{{\log }_{\sqrt{{{a}_{20}}}}}{{\left( {{a}_{1}}-\dfrac{1}{4} \right)}^{3}}$
Vậy giá trị $M$ thuộc khoảng nào dưới đây?
A. $\left( 235;245 \right)$
B. $\left( 225;235 \right)$
C. $\left( 245;255 \right)$
D. $\left( 215;225 \right)$
$P={{\log }_{\sqrt{{{a}_{1}}}}}{{\left( {{a}_{2}}-\dfrac{1}{4} \right)}^{3}}+{{\log }_{\sqrt{{{a}_{2}}}}}{{\left( {{a}_{3}}-\dfrac{1}{4} \right)}^{3}}{{\log }_{\sqrt{{{a}_{3}}}}}{{\left( {{a}_{4}}-\dfrac{1}{4} \right)}^{3}}+...+{{\log }_{\sqrt{{{a}_{20}}}}}{{\left( {{a}_{1}}-\dfrac{1}{4} \right)}^{3}}$
Vậy giá trị $M$ thuộc khoảng nào dưới đây?
A. $\left( 235;245 \right)$
B. $\left( 225;235 \right)$
C. $\left( 245;255 \right)$
D. $\left( 215;225 \right)$
Ta có: $P=6\left[ {{\log }_{{{a}_{1}}}}\left( {{a}_{2}}-\dfrac{1}{4} \right)+{{\log }_{{{a}_{2}}}}\left( {{a}_{3}}-\dfrac{1}{4} \right)+{{\log }_{{{a}_{3}}}}\left( {{a}_{4}}-\dfrac{1}{4} \right)+...{{\log }_{{{a}_{20}}}}\left( {{a}_{1}}-\dfrac{1}{4} \right) \right]$
Theo bất đẳng thức Cô-si ta có: $\sqrt{{{a}_{2}}-\dfrac{1}{4}}=2\sqrt{\left( {{a}_{2}}-\dfrac{1}{4} \right).\dfrac{1}{4}}\le \left( {{a}_{2}}-\dfrac{1}{4} \right)+\dfrac{1}{4}={{a}_{2}}$
$\Rightarrow \sqrt{{{a}_{2}}-\dfrac{1}{4}}\le {{a}_{2}}\Leftrightarrow {{a}_{2}}-\dfrac{1}{4}\le a_{2}^{2}\Leftrightarrow {{\log }_{{{a}_{1}}}}\left( {{a}_{2}}-\dfrac{1}{4} \right)\ge {{\log }_{{{a}_{1}}}}a_{2}^{2}$ (do ${{a}_{1}}\in \left( \dfrac{1}{4};1 \right)\Rightarrow {{a}_{1}}<1$ )
$\Leftrightarrow {{\log }_{{{a}_{1}}}}\left( {{a}_{2}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{1}}}}{{a}_{2}}$
Tương tự ta cũng có:
$\left. \begin{aligned}
& {{\log }_{{{a}_{2}}}}\left( {{a}_{3}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{2}}}}{{a}_{3}} \\
& {{\log }_{{{a}_{3}}}}\left( {{a}_{4}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{3}}}}{{a}_{4}} \\
& ........................................ \\
& {{\log }_{{{a}_{20}}}}\left( {{a}_{1}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{20}}}}{{a}_{1}} \\
\end{aligned} \right\}\Rightarrow P\ge 12\left[ {{\log }_{{{a}_{1}}}}{{a}_{2}}+{{\log }_{{{a}_{2}}}}{{a}_{3}}+{{\log }_{{{a}_{3}}}}{{a}_{4}}+...+{{\log }_{{{a}_{20}}}}{{a}_{1}} \right]$
$\Rightarrow P\overset{Co-si}{\mathop{\ge }} 12.20.\sqrt[20]{{{\log }_{{{a}_{1}}}}{{a}_{2}}.{{\log }_{{{a}_{2}}}}{{a}_{3}}.{{\log }_{{{a}_{3}}}}{{a}_{4}}...{{\log }_{{{a}_{20}}}}{{a}_{1}}}=240$
Vậy $M=\min P=240\in \left( 235;245 \right)$
Theo bất đẳng thức Cô-si ta có: $\sqrt{{{a}_{2}}-\dfrac{1}{4}}=2\sqrt{\left( {{a}_{2}}-\dfrac{1}{4} \right).\dfrac{1}{4}}\le \left( {{a}_{2}}-\dfrac{1}{4} \right)+\dfrac{1}{4}={{a}_{2}}$
$\Rightarrow \sqrt{{{a}_{2}}-\dfrac{1}{4}}\le {{a}_{2}}\Leftrightarrow {{a}_{2}}-\dfrac{1}{4}\le a_{2}^{2}\Leftrightarrow {{\log }_{{{a}_{1}}}}\left( {{a}_{2}}-\dfrac{1}{4} \right)\ge {{\log }_{{{a}_{1}}}}a_{2}^{2}$ (do ${{a}_{1}}\in \left( \dfrac{1}{4};1 \right)\Rightarrow {{a}_{1}}<1$ )
$\Leftrightarrow {{\log }_{{{a}_{1}}}}\left( {{a}_{2}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{1}}}}{{a}_{2}}$
Tương tự ta cũng có:
$\left. \begin{aligned}
& {{\log }_{{{a}_{2}}}}\left( {{a}_{3}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{2}}}}{{a}_{3}} \\
& {{\log }_{{{a}_{3}}}}\left( {{a}_{4}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{3}}}}{{a}_{4}} \\
& ........................................ \\
& {{\log }_{{{a}_{20}}}}\left( {{a}_{1}}-\dfrac{1}{4} \right)\ge 2{{\log }_{{{a}_{20}}}}{{a}_{1}} \\
\end{aligned} \right\}\Rightarrow P\ge 12\left[ {{\log }_{{{a}_{1}}}}{{a}_{2}}+{{\log }_{{{a}_{2}}}}{{a}_{3}}+{{\log }_{{{a}_{3}}}}{{a}_{4}}+...+{{\log }_{{{a}_{20}}}}{{a}_{1}} \right]$
$\Rightarrow P\overset{Co-si}{\mathop{\ge }} 12.20.\sqrt[20]{{{\log }_{{{a}_{1}}}}{{a}_{2}}.{{\log }_{{{a}_{2}}}}{{a}_{3}}.{{\log }_{{{a}_{3}}}}{{a}_{4}}...{{\log }_{{{a}_{20}}}}{{a}_{1}}}=240$
Vậy $M=\min P=240\in \left( 235;245 \right)$
Đáp án A.