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Câu hỏi: Cho ${{x}}$ và ${{y}}$ là các số thực dương thỏa mãn ${{\dfrac{1}{2} \log _3 \dfrac{x}{9}+\log _3 y=\dfrac{9-x y^2}{y^2}}}$. Khi ${{P=x+6 y}}$ đạt giá trị nhỏ nhất thì giá trị của ${{\dfrac{x}{y}}}$ bằng
A. ${{\sqrt[3]{3}}}$.
B. ${{\dfrac{3}{2}}}$.
C. ${{\sqrt[3]{9}}}$.
D. 3.
A. ${{\sqrt[3]{3}}}$.
B. ${{\dfrac{3}{2}}}$.
C. ${{\sqrt[3]{9}}}$.
D. 3.
Với $x,y\in {{\mathbb{R}}^{+}}$, ta có: $\dfrac{1}{2}{{\log }_{3}}\dfrac{x}{9}+{{\log }_{3}}y=\dfrac{9-x{{y}^{2}}}{{{y}^{2}}}\Leftrightarrow {{\log }_{3}}\dfrac{x}{9}+2{{\log }_{3}}y=\dfrac{18-2x{{y}^{2}}}{{{y}^{2}}}$
$\Leftrightarrow {{\log }_{3}}\dfrac{x{{y}^{2}}}{9}=\dfrac{18-2x{{y}^{2}}}{{{y}^{2}}}\Leftrightarrow {{\log }_{3}}x{{y}^{2}}+2\cdot \dfrac{x{{y}^{2}}}{{{y}^{2}}}={{\log }_{3}}9+2\cdot \dfrac{9}{{{y}^{2}}}\quad \left( 1 \right)$
Xét hàm: $f\left( t \right)={{\log }_{3}}t+2\cdot \dfrac{t}{{{v}^{2}}}$, $t>0$
Khi đó: ${f}'\left( t \right)=\dfrac{1}{3\ln t}+\dfrac{2}{{{v}^{2}}}>0,\forall t>0,v\in \mathbb{R}$. Suy ra: $\left( 1 \right)\Leftrightarrow x{{y}^{2}}=9\Rightarrow x=\dfrac{9}{{{y}^{2}}}$.
$\Rightarrow P=x+6y=\dfrac{9}{{{y}^{2}}}+6y=\dfrac{9}{{{y}^{2}}}+3y+3y\ge \sqrt[3]{\dfrac{9}{{{y}^{2}}}\cdot 3y\cdot 3y}=\sqrt[3]{81}$
Dấu bằng xảy ra khi $\dfrac{9}{{{y}^{2}}}=3y\Rightarrow y=\sqrt[3]{3}$.
Vậy khi ${{P}_{\min }}$ thì $\dfrac{x}{y}=\dfrac{9}{{{y}^{3}}}=\dfrac{9}{3}=3$.
$\Leftrightarrow {{\log }_{3}}\dfrac{x{{y}^{2}}}{9}=\dfrac{18-2x{{y}^{2}}}{{{y}^{2}}}\Leftrightarrow {{\log }_{3}}x{{y}^{2}}+2\cdot \dfrac{x{{y}^{2}}}{{{y}^{2}}}={{\log }_{3}}9+2\cdot \dfrac{9}{{{y}^{2}}}\quad \left( 1 \right)$
Xét hàm: $f\left( t \right)={{\log }_{3}}t+2\cdot \dfrac{t}{{{v}^{2}}}$, $t>0$
Khi đó: ${f}'\left( t \right)=\dfrac{1}{3\ln t}+\dfrac{2}{{{v}^{2}}}>0,\forall t>0,v\in \mathbb{R}$. Suy ra: $\left( 1 \right)\Leftrightarrow x{{y}^{2}}=9\Rightarrow x=\dfrac{9}{{{y}^{2}}}$.
$\Rightarrow P=x+6y=\dfrac{9}{{{y}^{2}}}+6y=\dfrac{9}{{{y}^{2}}}+3y+3y\ge \sqrt[3]{\dfrac{9}{{{y}^{2}}}\cdot 3y\cdot 3y}=\sqrt[3]{81}$
Dấu bằng xảy ra khi $\dfrac{9}{{{y}^{2}}}=3y\Rightarrow y=\sqrt[3]{3}$.
Vậy khi ${{P}_{\min }}$ thì $\dfrac{x}{y}=\dfrac{9}{{{y}^{3}}}=\dfrac{9}{3}=3$.
Đáp án D.