\(B=\dfrac{x-5+3}{x-5}=1+\dfrac{3}{x-5}\)
Để B nguyên
\(3\text{ }⋮\text{ }\left(x-5\right)\)
=>\(x\in\left\{2;4;6;8\right\}\)
B=x-2/x-5
=x-2/x-2-3
=1- x-2/3
vậy x-2 là ước của 3
Ư(3)={1;-1;3;-3}
-> x thuộc {3;1;5;-1}
\(x\in Z\Rightarrow B\in Z\Rightarrow\dfrac{x-2}{x-5}\in Z\\ \Rightarrow x-2⋮x-5\)
\(\Rightarrow x-5+3⋮x-5\)
Mà \(x-5⋮x-5\Rightarrow3⋮x-5\Rightarrow x-5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\Rightarrow x=\left\{2;4;6;8\right\}\)
Để B nguyên thì \(x-2⋮x-5\)
\(\Leftrightarrow x-5\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{6;4;8;2\right\}\)