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@MinhAnh
`a)`
`25 \vdots x + 2`
`=> x + 2 \in Ư_((25))`
Ta có bảng:
\begin{array}{|c|c|c|}\hline \text{x+2}&\text{-1}&\text{-25}&\text{-5}&\text{5}\\\hline \text{x}&\text{-3}&\text{-27}&\text{-7}&\text{3}\\\hline\end{array}
`b)`
`2x + 4 \vdots x`
`=>{(2x+4 \vdots x),( 2(x) \vdots x ):}`
`=>{(2x+4 \vdots x),( 2x \vdots x):}`
`=> (2x+4)-2x \vdots x`
`=> 2x + 4 - 2x \vdots x`
`=> 4 \vdots x`
`=> x \in Ư_((4))`
\begin{array}{|c|c|c|}\hline \text{x}&\text{-2}&\text{2}&\text{-1}&\text{1}&\text{-4}&\text{4}\\\hline\end{array}
`c)`
`x + 7 \vdots x + 3`
`=> (x+7)-(x+3) \vdots x+3`
`=> x + 7 - x - 3 \vdots x + 3`
`=> 4 \vdots x + 3`
`=> x+3 \in Ư_((4))`
Ta có bảng:\begin{array}{|c|c|c|}\hline \text{x+3}&\text{-2}&\text{2}&\text{-4}&\text{4}&\text{-1}&\text{1}\\\hline \text{x}&\text{-5}&\text{-1}&\text{-7}&\text{1}&\text{-4}&\text{-2}\\\hline\end{array}
`d)`
`3x - 5 \vdots x - 1`
`=>{(3x-5 \vdots x - 1),( 3(x-1) \vdots x-1):}`
`=>{(3x-5 \vdots x - 1),( 3x-3 \vdots x -1):}`
`=>(3x-5)-(3x-3) \vdots x - 1`
`=>3x - 5 - 3x + 3 \vdots x - 1`
`=>-2 \vdots x - 1`
`=> x - 1 \in Ư_((-2))`
Ta có bảng:
\begin{array}{|c|c|c|}\hline \text{x-1}&\text{-1}&\text{1}&\text{-2}&\text{2}\\\hline \text{x}&\text{0}&\text{2}&\text{-1}&\text{3}\\\hline\end{array}
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