`#tnvt`
`A=\frac{14-x}{4-x}(x\ne4,x\inZZ)`
`=\frac{4-x+10}{4-x}`
`=\frac{4-x}{4-x}+\frac{10}{4-x}`
`=1+\frac{10}{4-x}`
Để `A\inZZ` thì `10\vdots4-x`
`=>4-x\inƯ(10)`
`=>4-x\in{+-1;+-2;+-5;+-10}`
Ta lập bảng giá trị như sau:
\begin{array}{|c|c|c|}\hline 4-x &-1&1&-2&2&-5&5&-10&10\\\hline x&5(tm)&3(tm)&6(tm)&2(tm)&-9(tm)&-1(tm)&14(tm)&-6(tm)\\\hline\end{array}
Vậy `x\in{5;3;6;2;-9;-1;14;-6}` thì `A\inZZ`
