1,tìm các số nguyên n để

a,5 chia hết n+1

b,n+2chia hết n-1

c,3n+1 chia hết n+2

d, phân số 4n-1/n+1 nhận giá trị nguyên

`a.`

`5/(n+1)`          `(n\ne-1)`

Để `5` chia hết cho `n+1`

Thì `n+1∈Ư(5)={-5;-1;1;5}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{n+1}&\text{-5}&\text{-1}&\text{1}&\text{5}\\\hline \text{n}&\text{-6(tm)}&\text{-2(tm)}&\text{0(tm)}&\text{4(tm)}\\\hline\end{array}

Vậy `n∈{-6;-2;0;4}`

`b.`

`(n+2)/(n-1)`   `(n\ne1)`

`=(n-1+3)/(n-1)=1+3/(n-1)`

Để `n+2` chia hết cho `n-1` thì `n-1∈Ư(3)={-3;-1;1;3}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{n-1}&\text{-3}&\text{-1}&\text{1}&\text{3}\\\hline \text{n}&\text{-2(tm)}&\text{0(tm)}&\text{2(tm)}&\text{4(tm)}\\\hline\end{array}

Vậy `n∈{-2;0;2;4}`

`c.`

`(3n+1)/(n+2)`         `(n\ne-2)`

`=(3n+6-5)/(n+2)=3-5/(n+2)`

Để `3n+1` chia hết cho `n+2` thì `n+2∈Ư(5)={-5;-1;1;5}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{n+2}&\text{-5}&\text{-1}&\text{1}&\text{5}\\\hline \text{n}&\text{-7(tm)}&\text{-3(tm)}&\text{-1(tm)}&\text{3(tm)}\\\hline\end{array}

Vậy `n∈{-7;-3;-1;3}`

`d.`

`(4n-1)/(n+1)`       `(n\ne-1)`

`=(4n+4-5)/(n+1)=1-5/(n+1)`

Để `(4n-1)/(n+1)` nguyên `⇔n+1∈Ư(5)={-5;-1;1;5}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{n+1}&\text{-5}&\text{-1}&\text{1}&\text{5}\\\hline \text{n}&\text{-6(tm)}&\text{-2(tm)}&\text{0(tm)}&\text{4(tm)}\\\hline\end{array}

Vậy `n∈{-6;-2;0;4}`