a: Để (d1) cắt (d2) thì \(2m-1\ne m+1\)=>\(m\ne2\)=>\(m\notin\left\{-1;\dfrac{1}{2};2\right\}\)(1)b: Phương trình hoành độ giao điểm là:\(\left(2m-1\right)x+m-1=\left(m+1\right)x+2m\)=>\(\left(2m-1\right)x-\left(m+1\right)x=2m-m+1\)=>(2m-1-m-1)x=m+1=>(m-2)x=m+1=>\(x=\dfrac{m+1}{m-2}\)=>\(y=\left(m+1\right)\cdot\dfrac{\left(m+1\right)}{m-2}+2m=\dfrac{\left(m+1\right)^2+2m\left(m-2\right)}{m-2}=\dfrac{3m^2-2m+1}{m-2}\)Để x,y nguyên thì \(\left\{{}\begin{matrix}m+1⋮m-2\\3m^2-2m+1⋮m-2\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m-2+3⋮m-2\\3m^2-6m+4m-8+9⋮m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3⋮m-2\\9⋮m-2\end{matrix}\right.\)=>\(3⋮m-2\)=>\(m-2\in\left\{1;-1;3;-3\right\}\)=>\(m\in\left\{3;1;5;-1\right\}\)Kết hợp (1), ta được: \(m\in\left\{3;1;5\right\}\)Câu trả lời: a: Để (d1) cắt (d2) thì \(2m-1\ne m+1\)=>\(m\ne2\)=>\(m\notin\left\{-1;\dfrac{1}{2};2\right\}\)(1)b: Phương trình hoành độ giao điểm là:\(\left(2m-1\right)x+m-1=\left(m+1\right)x+2m\)=>\(\left(2m-1\right)x-\left(m+1\right)x=2m-m+1\)=>(2m-1-m-1)x=m+1=>(m-2)x=m+1=>\(x=\dfrac{m+1}{m-2}\)=>\(y=\left(m+1\right)\cdot\dfrac{\left(m+1\right)}{m-2}+2m=\dfrac{\left(m+1\right)^2+2m\left(m-2\right)}{m-2}=\dfrac{3m^2-2m+1}{m-2}\)Để x,y nguyên thì \(\left\{{}\begin{matrix}m+1⋮m-2\\3m^2-2m+1⋮m-2\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m-2+3⋮m-2\\3m^2-6m+4m-8+9⋮m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3⋮m-2\\9⋮m-2\end{matrix}\right.\)=>\(3⋮m-2\)=>\(m-2\in\left\{1;-1;3;-3\right\}\)=>\(m\in\left\{3;1;5;-1\right\}\)Kết hợp (1), ta được: \(m\in\left\{3;1;5\right\}\)