Cho x, y là các số thực thỏa mãn 

B

- Hướng dẫn giải

Đáp án B

Ta có

$\begin{array}{l}\frac{{\log }_{2}x}{{\log }_2\left(2xy\right)}=\frac{{\log }_{2}y}{{\log }_2\left(\frac{xy}{2}\right)}\\ ={\log }_2\left(xy\right)\Rightarrow {\log }_{2xy}x\\ ={\log }_{\frac{xy}{2}}y=\log 2\left(xy\right)=t\end{array}$

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}x={\left(2xy\right)}^t\\ y={\left(\frac{xy}{2}\right)}^t\\ xy=2^t\end{array}\right.\\ \Rightarrow {\left(2xy.\frac{xy}{2}\right)}^t=2^t\\ \Rightarrow \left[\begin{array}{l}t=0\\ xy=\sqrt{2}\end{array}\right.\end{array}$

Với $t=0\Rightarrow x=y=1\Rightarrow x+y=2$

Với

$\begin{array}{l}xy=\sqrt{2}\Rightarrow t=\frac{1}{2}\\ \Rightarrow x+y={\left(2\sqrt{2}\right)}^{\frac{1}{2}}+{\left(\frac{\sqrt{2}}{2}\right)}^{\frac{1}{2}}\\ =\sqrt[4]{8}+\frac{1}{\sqrt[4]{2}}\end{array}$