Đáp án:
a) \(\left\{ \begin{array}{l}
x = 1\\
y = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:a = 1\\
Hpt \to \left\{ \begin{array}{l}
x - 2y = 1\\
2x + y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - 2y = 1\\
4x + 2y = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
5x = 5\\
y = 2 - 2x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 0
\end{array} \right.\\
b)\left\{ \begin{array}{l}
ax - 2y = a\\
2x + y = a + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
ax - 2y = a\\
4x + 2y = 2a + 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {a + 4} \right)x = 3a + 2\\
y = a + 1 - 2x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{3a + 2}}{{a + 4}}\\
y = a + 1 - 2.\dfrac{{3a + 2}}{{a + 4}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{3a + 2}}{{a + 4}}\\
y = \dfrac{{{a^2} + 4a + a + 4 - 6a - 4}}{{a + 4}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{3a + 2}}{{a + 4}}\\
y = \dfrac{{{a^2} - a}}{{a + 4}}
\end{array} \right.\\
DK:a \ne - 4\\
Do:x + y = - 3\\
\to \dfrac{{3a + 2}}{{a + 4}} + \dfrac{{{a^2} - a}}{{a + 4}} = - 3\\
\to {a^2} + 2a + 2 = - 3a - 12\\
\to {a^2} + 5a + 14 = 0\\
Do:\Delta = 25 - 4.14 = - 31 < 0\\
\to a \in \emptyset
\end{array}\)
