a) Để A nguyên thì \(3⋮\left(x+2\right)\)\(\Rightarrow x+2\inƯ\left(3\right)\)\(\Rightarrow x+2\in\left\{-3;-1;1;3\right\}\)\(\Rightarrow x\in\left\{-5;-3;-1;1\right\}\)
b) Để B nguyên thì \(2x⋮\left(x-1\right)\)\(\Rightarrow\left(2x-2+2\right)⋮\left(x-1\right)\)\(\Rightarrow2\left(x-1\right)+2⋮x-1\)
Vì \(\hept{\begin{cases}2\left(x-1\right)+2⋮x-1\\2\left(x-1\right)⋮x-1\end{cases}}\)\(\Rightarrow2⋮x-1\)\(x-1\inƯ\left(2\right)\)\(\Rightarrow x-1\in\left\{-2;-1;1;2\right\}\)\(\Rightarrow x\in\left\{-1;0;2;3\right\}\)
bn kia lm dai qa
ta co : 2x/x-1
=2(x-1)+1/x-1
=2+1
=3
