1. Để hệ phương trình có nghiệm duy nhất thì:
\(\dfrac{2}{4}\ne\dfrac{1}{m}\)
\(\Leftrightarrow m\ne\dfrac{4\cdot1}{2}\)
\(\Leftrightarrow m\ne2\)
2. Ta có: \(\left\{{}\begin{matrix}2x+y=8\\4x+my=2m+18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=16\\4x+my=2m+18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=8\\\left(m-2\right)y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=8-\dfrac{2m+2}{m-2}\\y=\dfrac{2m+2}{m-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m-7}{m-2}\\y=\dfrac{2m+2}{m-2}\end{matrix}\right.\)
a) Để: 2x - 3y > 0 thì:
\(\Leftrightarrow2\cdot\dfrac{3m-7}{m-2}-3\cdot\dfrac{2m+2}{m-2}>0\)
\(\Leftrightarrow\dfrac{6m-14}{m-2}-\dfrac{6m-6}{m-2}>0\)
\(\Leftrightarrow\dfrac{-8}{m-2}>0\)
\(\Leftrightarrow m-2< 0\)
\(\Leftrightarrow m< 2\)
b) Để: \(\left\{{}\begin{matrix}x\in Z\\y\in Z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3m-7}{m-2}\in Z\\\dfrac{2m+2}{m-2}\in Z\end{matrix}\right.\)
Mà: \(\dfrac{3m-7}{m-2}=\dfrac{3m-6-1}{m-2}=3-\dfrac{1}{m-2}\)
Để bt nguyên thì \(m-2\inƯ\left(1\right)=\left\{1;-1\right\}\Rightarrow m\in\left\{3;1\right\}\) (1)
\(\dfrac{2m+2}{m-2}=\dfrac{2m-4+6}{m-2}=2+\dfrac{6}{m-2}\)
Để bt nguyên thì \(m-2\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\Rightarrow m\in\left\{3;1;4;0;5;-1;8;-4\right\}\) (2)
Từ (1) và (2) ta có để hpt có nghiệm x,y ∈ Z thì \(m\in\left\{1;3\right\}\)
