\(M=\dfrac{2x-7}{x-5}\\ M=\dfrac{2x-10+3}{x-5}\\ M=\dfrac{2x-10}{x-5}+\dfrac{3}{x-5}\\ M=\dfrac{2\left(x-5\right)}{x-5}+\dfrac{3}{x-5}\\ M=2+\dfrac{3}{x-5}\)
Để \(M\in Z\) thì \(\dfrac{3}{x-5}\in Z\left(x\ne5\right)\)
\(\Rightarrow\left(x-5\right)\inƯ_{\left(3\right)}=\left\{-7;1;1;7\right\}\)
Xét các TH:
\(TH1:x-3=-7\Rightarrow x=-7+3=-4\left(tm\right)\\ TH2:x-3=-1\Rightarrow x=-1+3=2\left(tm\right)\\ TH3:x-3=1\Rightarrow x=1+3=4\left(tm\right)\\ TH4:x-3=7\Rightarrow x=7+3=10\left(tm\right)\)
KL: \(M\in Z\) khi \(x\in\left\{-4;2;4;10\right\}\)
\(M=\dfrac{2x-10+3}{x-5}=2+\dfrac{3}{x-5}\)
Để M nguyên \(\Rightarrow\dfrac{3}{x-5}\) nguyên
\(\Rightarrow x-5=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x=\left\{2;4;6;8\right\}\)
với mn ơi
