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\(\left(m-1\right)x=2-3m\) (với \(m\ne1\))
\(\Rightarrow x=\dfrac{2-3m}{m-1}\)
\(x\ge1\Rightarrow\dfrac{2-3m}{m-1}\ge1\)
\(\Rightarrow\dfrac{2-3m}{m-1}-1\ge0\Rightarrow\dfrac{3-4m}{m-1}\ge0\)
\(\Rightarrow\dfrac{3}{4}\le m< 1\)
\( (m-1)x+3m-2 =0 \\ \Leftrightarrow x= \dfrac{2-3m}{m-1} \\ \Rightarrow \) PT có nghiệm \(\Leftrightarrow m-1 \ne 0 \Leftrightarrow m \ne 1\)
\(x ≥ 1 \Leftrightarrow 2-3m ≥ m-1 \Leftrightarrow m ≤ \dfrac{3}{4}\)
Vậy \(m ≤ \dfrac{3}{4}\).
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m+1}{m^2}\ne\dfrac{-2}{-1}=2\)
=>\(2m^2\ne m+1\)
=>\(2m^2-m-1\ne0\)
=>\(\left(m-1\right)\left(2m+1\right)\ne0\)
=>\(m\notin\left\{1;-\dfrac{1}{2}\right\}\)
\(\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\m^2x-y=m^2+2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\2m^2\cdot x-2y=2m^2+4m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(2m^2-m-1\right)=2m^2+4m-m+1\\\left(m+1\right)x-2y=m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\cdot\left(m-1\right)\left(2m+1\right)=2m^2+3m+1=\left(m+1\right)\left(2m+1\right)\\\left(m+1\right)x-2y=m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\2y=\left(m+1\right)x-\left(m-1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\2y=\dfrac{m^2+2m+1-\left(m-1\right)^2}{m-1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m+1}{m-1}\\y=\dfrac{m^2+2m+1-m^2+2m-1}{2m-2}=\dfrac{4m}{2m-2}=\dfrac{2m}{m-1}\end{matrix}\right.\)
Để x,y đều nguyên thì \(\left\{{}\begin{matrix}m+1⋮m-1\\2m⋮m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m-1+2⋮m-1\\2m-2+2⋮m-1\end{matrix}\right.\)
=>\(2⋮m-1\)
=>\(m-1\in\left\{1;-1;2;-2\right\}\)
=>\(m\in\left\{2;0;3;-1\right\}\)
\(\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\m^2x-y=m^2+2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+1\right)x-2y=m-1\\2m^2x-2y=2m^2+4m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2m^2-m-1\right)x=2m^2+3m+1\\y=m^2x-m^2-2m\end{matrix}\right.\)
Pt có nghiệm duy nhất khi \(2m^2-m-1\ne0\Rightarrow m\ne\left\{1;-\dfrac{1}{2}\right\}\)
Khi đó: \(\left\{{}\begin{matrix}x=\dfrac{2m^2-2m-1}{2m^2+3m+1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{\left(m+1\right)\left(2m+1\right)}=\dfrac{m-1}{m+1}\\y=m^2x-m^2-2m=\dfrac{-4m^2-2m}{m+1}\end{matrix}\right.\)
Để x nguyên \(\Rightarrow\dfrac{m-1}{m+1}\in Z\Rightarrow1-\dfrac{2}{m+1}\in Z\)
\(\Rightarrow\dfrac{2}{m+1}\in Z\)
\(\Rightarrow m+1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow m=\left\{-3;-2;0;1\right\}\)
Thay vào y thấy đều thỏa mãn y nguyên.
Vậy ...
PT có nghiệm duy nhất khi và chỉ khi m - 1 khác 0, tức m khác 1.
Khi đó \(x=\dfrac{2-3m}{m-1}\).
\(x\ge1\Leftrightarrow\dfrac{2-3m}{m-1}\ge1\Leftrightarrow\dfrac{2-3m-m+1}{m-1}\ge0\Leftrightarrow\dfrac{3-4m}{m-1}\ge0\Leftrightarrow\dfrac{4}{3}\ge m>1\).
Vậy ....
\(\left\{{}\begin{matrix}x+my=3\\m^2x+my=2m^2+m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=3\\\left(m^2-1\right)x=2m^2+m-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+my=3\\x=\dfrac{2m+3}{m+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2m+3}{m+1}\\y=\dfrac{1}{m+1}\end{matrix}\right.\)
\(P=\left(\dfrac{2m+3}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}=\left(2+\dfrac{1}{m+1}\right)^2+\dfrac{3}{\left(m+1\right)^2}\)
\(=4+\dfrac{4}{m+1}+\dfrac{4}{\left(m+1\right)^2}=\left(\dfrac{2}{m+1}+1\right)^2+3\ge3\)
\(P_{min}=3\) khi \(m=-3\)