Giải hệ: \(\left\{\begin{array}{l} \sqrt{2} x-\sqrt{5} y=1 \\ x+\sqrt{5} y=\sqrt{2} \end{array}\right.\)

* Hướng dẫn giải

Ta có:

\(\left\{\begin{array}{l} \sqrt{2} x-\sqrt{5} y=1 \\ x+\sqrt{5} y=\sqrt{2} \end{array} \Leftrightarrow\left\{\begin{array}{l} x=-\sqrt{5} y+\sqrt{2} \\ \sqrt{2} x-\sqrt{5} y=1 \end{array}\right.\right.\)

\(\begin{array}{l} \Leftrightarrow\left\{\begin{array}{l} x=-\sqrt{5} y+\sqrt{2} \\ \sqrt{2} x-\sqrt{5} y=1 \end{array} \Leftrightarrow\left\{\begin{array}{l} x=-\sqrt{5} y+\sqrt{2} \\ \sqrt{2}(-\sqrt{5} y+\sqrt{2})-\sqrt{5} y=1 \end{array}\right.\right. \\ \Leftrightarrow\left\{\begin{array}{l} x=-\sqrt{5} y+\sqrt{2} \\ \sqrt{2}(-\sqrt{5} y+\sqrt{2})-\sqrt{5} y=1 \end{array} \Leftrightarrow\left\{\begin{array}{l} x=-\sqrt{5} y+\sqrt{2} \\ -\sqrt{5}(\sqrt{2}+1) y=-1 \end{array}\right.\right. \end{array}\)

\(\Leftrightarrow\left\{\begin{array}{l} x=-\sqrt{5}\left(\frac{1}{\sqrt{5}(\sqrt{2}+1)}\right) \\ y=\frac{1}{\sqrt{5}(\sqrt{2}+1)} \end{array} \Leftrightarrow\left\{\begin{array}{l} x=1 \\ y=\frac{\sqrt{2}-1}{\sqrt{5}} \end{array}\right.\right.\)

Vậy, nghiệm của hệ phương trình là \(\left(1 ; \frac{\sqrt{2}-1}{\sqrt{5}}\right)\)