Đáp án:
a) \(\left\{ \begin{array}{l}
x = \dfrac{{17 + 12\sqrt 3 }}{{13}}\\
y = \dfrac{{ - 7 + 5\sqrt 3 }}{{13}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
ax - y = 2\\
x + ay = 3
\end{array} \right. \to \left\{ \begin{array}{l}
{a^2}x - ay = 2a\\
x + ay = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{a^2} + 1} \right)x = 2a + 3\\
ax - y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2a + 3}}{{{a^2} + 1}}\\
y = ax - 2 = a.\dfrac{{2a + 3}}{{{a^2} + 1}} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2a + 3}}{{{a^2} + 1}}\\
y = \dfrac{{2{a^2} + 3a - 2{a^2} - 2}}{{{a^2} + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2a + 3}}{{{a^2} + 1}}\\
y = \dfrac{{3a - 2}}{{{a^2} + 1}}
\end{array} \right.\\
a)Thay:a = \sqrt 3 - 1\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2\left( {\sqrt 3 - 1} \right) + 3}}{{{{\left( {\sqrt 3 - 1} \right)}^2} + 1}} = \dfrac{{17 + 12\sqrt 3 }}{{13}}\\
y = \dfrac{{3\left( {\sqrt 3 - 1} \right) - 2}}{{{{\left( {\sqrt 3 - 1} \right)}^2} + 1}} = \dfrac{{ - 7 + 5\sqrt 3 }}{{13}}
\end{array} \right.\\
b)x - y = 1\\
\to \dfrac{{2a + 3}}{{{a^2} + 1}} - \dfrac{{3a - 2}}{{{a^2} + 1}} = 1\\
\to \dfrac{{ - a + 5}}{{{a^2} + 1}} = 1\\
\to 5 - a = {a^2} + 1\\
\to {a^2} + a - 4 = 0\\
\to \left[ \begin{array}{l}
a = \dfrac{{ - 1 + \sqrt {17} }}{2}\\
a = \dfrac{{ - 1 - \sqrt {17} }}{2}
\end{array} \right.
\end{array}\)
( bạn xem lại đề có nhầm dấu hay số ở đâu không nhé )
