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Câu hỏi: Cho các số thực $a, b$ thỏa mãn $\dfrac{1}{3}<b<a<1$. Tìm giá trị nhỏ nhất của biểu thức $P={{\log }_{a}}\dfrac{4\left( 3b-1 \right)}{9}+8\log _{\dfrac{b}{a}}^{2}a$.
A. $7$.
B. $8$.
C. $6$.
D. $9$.
A. $7$.
B. $8$.
C. $6$.
D. $9$.
Vì $\dfrac{1}{3}<b<a<1$ nên ${{\left( 3b-2 \right)}^{2}}\ge 0\Leftrightarrow {{b}^{2}}\ge \dfrac{4\left( 3b-1 \right)}{9}\Rightarrow {{\log }_{a}}\dfrac{4\left( 3b-1 \right)}{9}\le {{\log }_{a}}{{b}^{2}}$
Ta có $8\log _{\dfrac{b}{a}}^{2}a=8{{\left( \dfrac{1}{{{\log }_{a}}b-1} \right)}^{2}}$
Đặt ${{\log }_{a}}b=x$. Vì $\dfrac{1}{3}<b<a<1$ nên $x={{\log }_{a}}b>1$. Khi đó
$P={{\log }_{a}}\dfrac{4\left( 3b-1 \right)}{9}+8\log _{\dfrac{b}{a}}^{2}a\ge {{\log }_{a}}{{b}^{2}}+8{{\left( \dfrac{1}{{{\log }_{a}}b-1} \right)}^{2}}\Rightarrow P\ge 2x+\dfrac{8}{{{\left( x-1 \right)}^{2}}}$
Mà $2x+\dfrac{8}{{{\left( x-1 \right)}^{2}}}=\left( x-1 \right)+\left( x-1 \right)+\dfrac{8}{{{\left( x-1 \right)}^{2}}}+2\ge 3.\sqrt[3]{\left( x-1 \right).\left( x-1 \right).\dfrac{8}{{{\left( x-1 \right)}^{2}}}}+2=8$
Suy ra $P\ge 8$
Dấu $''=''$ xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{2}{3} \\
& a=\sqrt[3]{\dfrac{2}{3}} \\
\end{aligned} \right.$
Vậy $\underset{{}}{\mathop{\min }} P=8$.
Ta có $8\log _{\dfrac{b}{a}}^{2}a=8{{\left( \dfrac{1}{{{\log }_{a}}b-1} \right)}^{2}}$
Đặt ${{\log }_{a}}b=x$. Vì $\dfrac{1}{3}<b<a<1$ nên $x={{\log }_{a}}b>1$. Khi đó
$P={{\log }_{a}}\dfrac{4\left( 3b-1 \right)}{9}+8\log _{\dfrac{b}{a}}^{2}a\ge {{\log }_{a}}{{b}^{2}}+8{{\left( \dfrac{1}{{{\log }_{a}}b-1} \right)}^{2}}\Rightarrow P\ge 2x+\dfrac{8}{{{\left( x-1 \right)}^{2}}}$
Mà $2x+\dfrac{8}{{{\left( x-1 \right)}^{2}}}=\left( x-1 \right)+\left( x-1 \right)+\dfrac{8}{{{\left( x-1 \right)}^{2}}}+2\ge 3.\sqrt[3]{\left( x-1 \right).\left( x-1 \right).\dfrac{8}{{{\left( x-1 \right)}^{2}}}}+2=8$
Suy ra $P\ge 8$
Dấu $''=''$ xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{2}{3} \\
& a=\sqrt[3]{\dfrac{2}{3}} \\
\end{aligned} \right.$
Vậy $\underset{{}}{\mathop{\min }} P=8$.
Đáp án B.