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Câu hỏi: Cho x, y là hai số thực dương thỏa mãn ${{\log }_{2}}\dfrac{1-xy}{x+y}-x-y=2xy-3$. Tìm giá trị nhỏ nhất của S biết $S=x+2y$
A. ${{S}_{\min }}=\dfrac{2\sqrt{10}-1}{2}$
B. ${{S}_{\min }}=\dfrac{2\sqrt{10}-5}{2}$
C. ${{S}_{\min }}=\dfrac{3\sqrt{10}-7}{2}$
D. ${{S}_{\min }}=\dfrac{2\sqrt{10}-3}{2}$
A. ${{S}_{\min }}=\dfrac{2\sqrt{10}-1}{2}$
B. ${{S}_{\min }}=\dfrac{2\sqrt{10}-5}{2}$
C. ${{S}_{\min }}=\dfrac{3\sqrt{10}-7}{2}$
D. ${{S}_{\min }}=\dfrac{2\sqrt{10}-3}{2}$
Cách giải:
Ta có:
$\begin{aligned}
& {{\log }_{2}}\dfrac{1-xy}{x+y}-x-y=2xy-3 \\
& \Leftrightarrow {{\log }_{2}}\left( 1-xy \right)-{{\log }_{2}}\left( x+y \right)=2xy-3+\left( x+y \right) \\
& \Leftrightarrow {{\log }_{2}}\left( 1-xy \right)+1-{{\log }_{2}}\left( x+y \right)=\left( 2xy-2 \right)+\left( x+y \right) \\
& \Leftrightarrow {{\log }_{2}}\left( 2-2xy \right)+\left( 2xy-2 \right)={{\log }_{2}}\left( x+y \right)+\left( x+y \right) \\
\end{aligned}$
Xét hàm số $f\left( t \right)=lo{{g}_{2}}t+t\left( t> \right)0$ ta có: $f'\left( t \right)=\dfrac{1}{t\ln 2}+1>0\forall t>0.~$
Do đó hàm số đồng biến trên $\left( 0;+\infty \right)$, suy ra $2-2xy=x+y\Leftrightarrow x\left( 2y+1 \right)=2-y\Leftrightarrow x=\dfrac{2-y}{2y+1}$
Ta có:
$S=x+2y=\dfrac{2-y}{2y+1}+2y$
$S'=-\dfrac{5}{{{\left( 2y+1 \right)}^{2}}}+2=\dfrac{2{{\left( 2y+1 \right)}^{2}}-5}{{{\left( 2y+1 \right)}^{2}}}$
$S'=0\Leftrightarrow {{\left( 2y+1 \right)}^{2}}=\dfrac{5}{2}\Leftrightarrow \left[ \begin{aligned}
& 2y+1=\dfrac{\sqrt{10}}{2} \\
& 2y+1=-\dfrac{\sqrt{10}}{2} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& y=\dfrac{-2+\sqrt{10}}{4} \\
& y=\dfrac{-2-\sqrt{10}}{4} \\
\end{aligned} \right.$
BBT:
Dựa vào BBT ta có: $\min S=S\left( \dfrac{-2+\sqrt{10}}{4} \right)=\dfrac{2\sqrt{10}-3}{2}$
Ta có:
$\begin{aligned}
& {{\log }_{2}}\dfrac{1-xy}{x+y}-x-y=2xy-3 \\
& \Leftrightarrow {{\log }_{2}}\left( 1-xy \right)-{{\log }_{2}}\left( x+y \right)=2xy-3+\left( x+y \right) \\
& \Leftrightarrow {{\log }_{2}}\left( 1-xy \right)+1-{{\log }_{2}}\left( x+y \right)=\left( 2xy-2 \right)+\left( x+y \right) \\
& \Leftrightarrow {{\log }_{2}}\left( 2-2xy \right)+\left( 2xy-2 \right)={{\log }_{2}}\left( x+y \right)+\left( x+y \right) \\
\end{aligned}$
Xét hàm số $f\left( t \right)=lo{{g}_{2}}t+t\left( t> \right)0$ ta có: $f'\left( t \right)=\dfrac{1}{t\ln 2}+1>0\forall t>0.~$
Do đó hàm số đồng biến trên $\left( 0;+\infty \right)$, suy ra $2-2xy=x+y\Leftrightarrow x\left( 2y+1 \right)=2-y\Leftrightarrow x=\dfrac{2-y}{2y+1}$
Ta có:
$S=x+2y=\dfrac{2-y}{2y+1}+2y$
$S'=-\dfrac{5}{{{\left( 2y+1 \right)}^{2}}}+2=\dfrac{2{{\left( 2y+1 \right)}^{2}}-5}{{{\left( 2y+1 \right)}^{2}}}$
$S'=0\Leftrightarrow {{\left( 2y+1 \right)}^{2}}=\dfrac{5}{2}\Leftrightarrow \left[ \begin{aligned}
& 2y+1=\dfrac{\sqrt{10}}{2} \\
& 2y+1=-\dfrac{\sqrt{10}}{2} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& y=\dfrac{-2+\sqrt{10}}{4} \\
& y=\dfrac{-2-\sqrt{10}}{4} \\
\end{aligned} \right.$
BBT:
Dựa vào BBT ta có: $\min S=S\left( \dfrac{-2+\sqrt{10}}{4} \right)=\dfrac{2\sqrt{10}-3}{2}$
Đáp án D.