Đáp án:$(x;y)=(x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}};\dfrac{-2}{5+\sqrt{6}} )$
Giải thích các bước giải:
\(\begin{cases}\sqrt{3}x-y=1\\5x+\sqrt{2}y=\sqrt{3}\end{cases}\)
\(\begin{cases}y=\sqrt{3}x-1\\5x+\sqrt{2}(\sqrt{3}x-1)=\sqrt{3}\end{cases}\)
\(\begin{cases}y=\sqrt{3}x-1\\5x+\sqrt{6}x-\sqrt{2}=\sqrt{3}\end{cases}\)
\(\begin{cases}y=\sqrt{3}x-1\\5x+\sqrt{6}x=\sqrt{3}+\sqrt{2}\end{cases}\)
\(\begin{cases}y=\sqrt{3}x-1\\(5+\sqrt{6})x=\sqrt{3}+\sqrt{2}\end{cases}\)
\(\begin{cases}y=\sqrt{3}x-1\\x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}}\end{cases}\)
\(\begin{cases}y=\sqrt{3}(\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}})-1\\x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}}\end{cases}\)
\(\begin{cases}y=(\dfrac{3+\sqrt{6}}{5+\sqrt{6}})-1\\x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}}\end{cases}\)
\(\begin{cases}y=\dfrac{3+\sqrt{6}-\sqrt{5}-\sqrt{6}}{5+\sqrt{6}}\\x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}}\end{cases}\)
\(\begin{cases}y=\dfrac{-2}{5+\sqrt{6}}\\x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}}\end{cases}\)
Vậy $(x;y)=(x=\dfrac{\sqrt{3}+\sqrt{2}}{5+\sqrt{6}};\dfrac{-2}{5+\sqrt{6}} )$
