Ta có: \(\hept{\begin{cases}x-my=2\\mx+2y=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}-mx+m^2y=-2m\\mx+2y=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-my=2\\\left(m^2+2\right)y=1-2m\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=my+2\\y=\frac{1-2m}{m^2+2}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=m\left(\frac{1-2m}{m^2+2}\right)\\y=\frac{1-2m}{m^2+2}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{m-2m^2}{m^2+2}\\y=\frac{1-2m}{m^2+2}\end{cases}}\)
Để \(3x+2y-1\ge0\)thì \(3\left(\frac{m-2m^2}{m^2+2}\right)+2\left(\frac{1-2m}{m^2+2}\right)\ge1\)\(\Leftrightarrow\frac{3m-6m^2}{m^2+2}+\frac{2-4m}{m^2+2}\ge1\)
\(\Leftrightarrow\frac{-6m^2-m+2}{m^2+2}\ge1\)\(\Leftrightarrow-6m^2-m+2\ge m^2+2\)\(\Leftrightarrow-7m^2-m\ge0\)\(\Leftrightarrow-m\left(7m+1\right)\ge0\)\(\Leftrightarrow m\left(7m+1\right)\le0\)Có hai trường hợp xảy ra:
TH1: \(\hept{\begin{cases}m\ge0\\7m+1\le0\end{cases}\Leftrightarrow\hept{\begin{cases}m\ge0\\m\le-\frac{1}{7}\end{cases}}}\)(loại)
TH2: \(\hept{\begin{cases}m\le0\\7m+1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}m\le0\\m\ge-\frac{1}{7}\end{cases}}\)
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