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Câu hỏi: Cho các số thực $a,b>1$ thỏa mãn ${{a}^{{{\log }_{b}}a}}+16{{b}^{{{\log }_{a}}\left( \dfrac{{{b}^{8}}}{{{a}^{3}}} \right)}}=12{{b}^{2}}$. Giá trị của biểu thức $P={{a}^{3}}+{{b}^{3}}$ là
A. $P=20$
B. $P=39$
C. $P=125$
D. $P=72$
A. $P=20$
B. $P=39$
C. $P=125$
D. $P=72$
Ta có: ${{a}^{{{\log }_{b}}a}}+16{{b}^{{{\log }_{a}}\left( \dfrac{{{b}^{8}}}{{{a}^{3}}} \right)}}=12{{b}^{2}}\Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( {{\log }_{a}}{{b}^{8}}-{{\log }_{a}}{{a}^{3}} \right)}}=12{{b}^{2}}$
$\begin{aligned}
& \Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( {{\log }_{a}}{{b}^{8}}-{{\log }_{a}}{{a}^{3}} \right)}}=12{{b}^{2}}\Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( 8{{\log }_{a}}b-3 \right)}}=12{{b}^{2}} \\
& \Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( \dfrac{8}{{{\log }_{b}}a}-3 \right)}}=12{{b}^{2}}\left( * \right) \\
\end{aligned}$
Đặt ${{\log }_{a}}b=t\Rightarrow a={{b}^{t}}$
Lại có vì $a,b>1\Rightarrow {{\log }_{a}}b>0$ hay $t>0$
Khi đó ta có: $VT\left( * \right)={{a}^{{{\log }_{b}}a}}+16{{b}^{\left( \dfrac{8}{{{\log }_{b}}a}-3 \right)}}={{\left( {{b}^{t}} \right)}^{t}}+16.{{b}^{\dfrac{8}{t}-3}}={{\left( {{b}^{t}} \right)}^{t}}+8.{{b}^{\dfrac{8}{t}-3}}+8.{{b}^{\dfrac{8}{t}-3}}$
$\overset{C\hat{o}-si}{\mathop{\ge }} 3\sqrt[3]{{{b}^{{{t}^{2}}}}.8.{{b}^{\dfrac{8}{t}-3}}.8.{{b}^{\dfrac{8}{t}-3}}}=12\sqrt[3]{{{b}^{{{t}^{2}}}}{{b}^{\dfrac{8}{t}-3}}{{b}^{\dfrac{8}{t}-3}}}12\sqrt[3]{{{b}^{{{t}^{2}}+\dfrac{8}{t}+\dfrac{8}{t}-6}}}$
$\overset{C\hat{o}-si}{\mathop{\ge }} 12\sqrt[3]{{{b}^{3}}^{\sqrt[3]{{{t}^{2}}+\dfrac{8}{t}+\dfrac{8}{t}-6}}}=12\sqrt[3]{{{b}^{6}}}=12{{b}^{2}}$ $\left( v\grave{i}{{t}^{2}}+\dfrac{8}{t}+\dfrac{8}{t}\ge 3\sqrt[3]{{{t}^{2}}.\dfrac{8}{t}.\dfrac{8}{t}}=3 \right)$
Hay $VT\left( * \right)\ge 12{{b}^{2}}$, dấu "=" xảy ra khi
$\left\{ \begin{aligned}
& {{b}^{{{t}^{2}}}}=8{{b}^{\dfrac{8}{t}-3}} \\
& {{t}^{2}}=\dfrac{8}{t} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& t=2 \\
& {{b}^{4}}=8b \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=2 \\
& b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{b}}a=2 \\
& b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2 \\
& a=4 \\
\end{aligned} \right.\left( TM \right)$
Suy ra $P={{a}^{3}}+{{b}^{3}}=64+8=72$
$\begin{aligned}
& \Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( {{\log }_{a}}{{b}^{8}}-{{\log }_{a}}{{a}^{3}} \right)}}=12{{b}^{2}}\Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( 8{{\log }_{a}}b-3 \right)}}=12{{b}^{2}} \\
& \Leftrightarrow {{a}^{{{\log }_{b}}a}}+16{{b}^{\left( \dfrac{8}{{{\log }_{b}}a}-3 \right)}}=12{{b}^{2}}\left( * \right) \\
\end{aligned}$
Đặt ${{\log }_{a}}b=t\Rightarrow a={{b}^{t}}$
Lại có vì $a,b>1\Rightarrow {{\log }_{a}}b>0$ hay $t>0$
Khi đó ta có: $VT\left( * \right)={{a}^{{{\log }_{b}}a}}+16{{b}^{\left( \dfrac{8}{{{\log }_{b}}a}-3 \right)}}={{\left( {{b}^{t}} \right)}^{t}}+16.{{b}^{\dfrac{8}{t}-3}}={{\left( {{b}^{t}} \right)}^{t}}+8.{{b}^{\dfrac{8}{t}-3}}+8.{{b}^{\dfrac{8}{t}-3}}$
$\overset{C\hat{o}-si}{\mathop{\ge }} 3\sqrt[3]{{{b}^{{{t}^{2}}}}.8.{{b}^{\dfrac{8}{t}-3}}.8.{{b}^{\dfrac{8}{t}-3}}}=12\sqrt[3]{{{b}^{{{t}^{2}}}}{{b}^{\dfrac{8}{t}-3}}{{b}^{\dfrac{8}{t}-3}}}12\sqrt[3]{{{b}^{{{t}^{2}}+\dfrac{8}{t}+\dfrac{8}{t}-6}}}$
$\overset{C\hat{o}-si}{\mathop{\ge }} 12\sqrt[3]{{{b}^{3}}^{\sqrt[3]{{{t}^{2}}+\dfrac{8}{t}+\dfrac{8}{t}-6}}}=12\sqrt[3]{{{b}^{6}}}=12{{b}^{2}}$ $\left( v\grave{i}{{t}^{2}}+\dfrac{8}{t}+\dfrac{8}{t}\ge 3\sqrt[3]{{{t}^{2}}.\dfrac{8}{t}.\dfrac{8}{t}}=3 \right)$
Hay $VT\left( * \right)\ge 12{{b}^{2}}$, dấu "=" xảy ra khi
$\left\{ \begin{aligned}
& {{b}^{{{t}^{2}}}}=8{{b}^{\dfrac{8}{t}-3}} \\
& {{t}^{2}}=\dfrac{8}{t} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& t=2 \\
& {{b}^{4}}=8b \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& t=2 \\
& b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{b}}a=2 \\
& b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2 \\
& a=4 \\
\end{aligned} \right.\left( TM \right)$
Suy ra $P={{a}^{3}}+{{b}^{3}}=64+8=72$
Đáp án D.