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Câu hỏi: Cho hai số thực dương x, y thỏa mãn ${{5}^{x+2y}}+\dfrac{3}{{{3}^{xy}}}+x+1=\dfrac{{{5}^{xy}}}{5}+{{3}^{-x-2y}}+y\left( x-2 \right)$. Tìm giá trị nhỏ nhất của biểu thức $P=x+2y$.
A. $P=6-2\sqrt{3}$
B. $P=4+2\sqrt{6}$
C. $P=4-2\sqrt{6}$
D. $P=6+2\sqrt{3}$
A. $P=6-2\sqrt{3}$
B. $P=4+2\sqrt{6}$
C. $P=4-2\sqrt{6}$
D. $P=6+2\sqrt{3}$
Theo giả thiết ta có ${{5}^{x+2y}}+\dfrac{3}{{{3}^{xy}}}+x+1=\dfrac{{{5}^{xy}}}{5}+{{3}^{-x-2y}}+y\left( x-2 \right)$.
$\Leftrightarrow {{5}^{x+2y}}-{{3}^{-x-2y}}+x+2y={{5}^{xy-1}}-{{3}^{1-xy}}+xy-1\Leftrightarrow x+2y=xy-1$.
$\Leftrightarrow 1-xy+x+2y=0\Leftrightarrow y\left( x-2 \right)=x+1>0\Rightarrow \left\{ \begin{aligned}
& x>2 \\
& y=\dfrac{x+1}{x-2} \\
\end{aligned} \right.$
$\Rightarrow P=f\left( x \right)=x+\dfrac{2\left( x+1 \right)}{x-2}\ge \underset{\left( 2;+\infty \right)}{\mathop{\min }} f\left( x \right)=f\left( 2+\sqrt{6} \right)=4+2\sqrt{6}$.
$\Leftrightarrow {{5}^{x+2y}}-{{3}^{-x-2y}}+x+2y={{5}^{xy-1}}-{{3}^{1-xy}}+xy-1\Leftrightarrow x+2y=xy-1$.
$\Leftrightarrow 1-xy+x+2y=0\Leftrightarrow y\left( x-2 \right)=x+1>0\Rightarrow \left\{ \begin{aligned}
& x>2 \\
& y=\dfrac{x+1}{x-2} \\
\end{aligned} \right.$
$\Rightarrow P=f\left( x \right)=x+\dfrac{2\left( x+1 \right)}{x-2}\ge \underset{\left( 2;+\infty \right)}{\mathop{\min }} f\left( x \right)=f\left( 2+\sqrt{6} \right)=4+2\sqrt{6}$.
Đáp án B.