Đáp án:
\(\left\{ \begin{array}{l}
x = \dfrac{{\sqrt 5 + \sqrt {15} }}{{\sqrt {15} + 1}}\\
y = \dfrac{{\sqrt {15} - \sqrt 3 }}{{\sqrt {15} + 1}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + \sqrt 5 y = \sqrt 5 \\
\sqrt 3 x - y = \sqrt 3
\end{array} \right. \to \left\{ \begin{array}{l}
x + \sqrt 5 y = \sqrt 5 \\
\sqrt {15} x - \sqrt 5 y = \sqrt {15}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {\sqrt {15} + 1} \right)x = \sqrt 5 + \sqrt {15} \\
\sqrt 3 x - y = \sqrt 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 5 + \sqrt {15} }}{{\sqrt {15} + 1}}\\
y = \sqrt 3 x - \sqrt 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 5 + \sqrt {15} }}{{\sqrt {15} + 1}}\\
y = \sqrt 3 .\dfrac{{\sqrt 5 + \sqrt {15} }}{{\sqrt {15} + 1}} - \sqrt 3 = \dfrac{{\sqrt {15} + 3\sqrt 5 - 3\sqrt 5 - \sqrt 3 }}{{\sqrt {15} + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 5 + \sqrt {15} }}{{\sqrt {15} + 1}}\\
y = \dfrac{{\sqrt {15} - \sqrt 3 }}{{\sqrt {15} + 1}}
\end{array} \right.
\end{array}\)
( đề là \(\left\{ \begin{array}{l}
x + \sqrt 5 y = \sqrt 5 \\
\sqrt 3 x - y = \sqrt 3
\end{array} \right.\) t nghĩ hợp lý hơn bạn nha )
