Đáp án:
\(\left[ \begin{array}{l}
y = \dfrac{{1 + \sqrt 5 }}{2}\\
y = \dfrac{{1 - \sqrt 5 }}{2}
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{1 - \sqrt 5 }}{2}\\
x = \dfrac{{1 + \sqrt 5 }}{2}
\end{array} \right.\)
Giải thích các bước giải:
Đặt:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + y = a\\
xy = b
\end{array} \right.\\
Hpt \to \left\{ \begin{array}{l}
2a - 3b = 5\\
a - 2b = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a = 2b + 3\\
2\left( {2b + 3} \right) - 3b = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
b = - 1\\
a = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + y = 1\\
xy = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - y\\
\left( {1 - y} \right)y = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - y\\
y - {y^2} = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - y\\
\left[ \begin{array}{l}
y = \dfrac{{1 + \sqrt 5 }}{2}\\
y = \dfrac{{1 - \sqrt 5 }}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 - \sqrt 5 }}{2}\\
x = \dfrac{{1 + \sqrt 5 }}{2}
\end{array} \right.
\end{array}\)
