Cho hệ phương trình

- Hướng dẫn giải

Ta có

$$\begin{gathered} \left\{\begin{array}{l}x\sqrt{2}-y\sqrt{3}=1\\ x+y\sqrt{3}=\sqrt{2}\end{array}\right.\iff \left\{\begin{array}{l}x\sqrt{2}-y\sqrt{3}=1\\ x\sqrt{2}+y\sqrt{6}=2\end{array}\right. \\ \iff \left\{\begin{array}{l}x\sqrt{2}-y\sqrt{3}=1\\ \left(\sqrt{6}+\sqrt{3}\right)y=1\end{array}\right.\iff \left\{\begin{array}{l}x\sqrt{2}-y\sqrt{3}=1\\ y=\frac{1}{\sqrt{6}+\sqrt{3}}\end{array}\right. \\ \iff \left\{\begin{array}{l}y=\frac{\sqrt{6}-\sqrt{3}}{3}\\ x\sqrt{2}-\sqrt{3}.\frac{\sqrt{6}-\sqrt{3}}{3}=1\end{array}\right.\iff \left\{\begin{array}{l}y=\frac{\sqrt{6}-\sqrt{3}}{3}\\ x=1\end{array}\right. \end{gathered}$$

Vậy hệ đã cho có nghiệm duy nhất $(x; y) =\left(1;\frac{\sqrt{6}-\sqrt{3}}{3}\right)$  

Ta có: $x+3\sqrt{3}y = 1 + 3\sqrt{3}.\frac{\sqrt{6}-\sqrt{3}}{3}=1+\sqrt{3}.\left(\sqrt{6}-\sqrt{3}\right)$

$=1+\sqrt{18}-3=\sqrt{18}-2=3\sqrt{2}-2$

Đáp án: D