Đáp án:
\(\left[ \begin{array}{l}
a = 2\\
a = - 2
\end{array} \right. \to \left[ \begin{array}{l}
b = 1\\
b = - 1
\end{array} \right.\)
Giải thích các bước giải:
Để hệ có vô số nghiệm
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{2}{a} = \dfrac{a}{2}\\
\dfrac{a}{2} = \dfrac{b}{1}
\end{array} \right.\left( {DK:a;b \ne 0} \right)\\
\to \left\{ \begin{array}{l}
{a^2} = 4\\
a = 2b
\end{array} \right.\\
\to \left[ \begin{array}{l}
a = 2\\
a = - 2
\end{array} \right. \to \left[ \begin{array}{l}
b = 1\\
b = - 1
\end{array} \right.\\
Có:\left\{ \begin{array}{l}
2x + ay = b\\
ax + 2y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{b - ay}}{2}\\
\dfrac{{ab - {a^2}y}}{2} + 2y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{b - ay}}{2}\\
\dfrac{{ab - {a^2}y + 4y}}{2} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{b - ay}}{2}\\
\left( {4 - {a^2}} \right)y = ab - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{ab - 2}}{{\left( {4 - {a^2}} \right)}}\\
x = \dfrac{{b - a.\dfrac{{ab - 2}}{{\left( {4 - {a^2}} \right)}}}}{2} = \dfrac{{4 - {a^2} - {a^2}b + 2a}}{{2\left( {4 - {a^2}} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{ab - 2}}{{\left( {4 - {a^2}} \right)}}\\
x = \dfrac{{4 - {a^2} - {a^2}b + 2a}}{{2\left( {4 - {a^2}} \right)}}
\end{array} \right.
\end{array}\)
