Cho HPT \(
\begin{cases}x+ay=3 \\ax+4y= 6\end{cases}
\)
a) TÌm a để HPT có nghiệm nguyên và dương
b) Tìm hệ thức giữa x và y độc lập với a.
\(
a)
D=4-a^{2}
\)
Để  HPT có nghiệm thì \( 4-a^{2}\neq 0\Leftrightarrow a\neq \pm 2\)
\(
D_{x}=12-6a, D_{y}=6-3a
\)
\(
\Rightarrow \begin{cases}x= \frac{6\left ( 2-a \right )}{4-a^{2}}=\frac{6}{2+a}\\ y=\frac{3\left ( 2-a \right )}{4-a^{2}}=\frac{3}{2+a} \end{cases}
\)
\(\Rightarrow \)Để HPT có nghiệm dương thì \(2+a>0\Leftrightarrow a>-2\)
Để HPT có nghiệm nguyên thì \(2+a=\pm 1, 2+a=\pm 3\Leftrightarrow a=-1, a=-3, a=1, a=-5\)
Chọn \( a=-1, a=1 \)
b) Thấy ngay \( x=2y \)
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