Question

tìm n thuộc N để:

A=\(\dfrac{2n+7}{n-2}\) là số nguyên

Answer 1

Ta có : \(A=\dfrac{2n+7}{n-2}\) \(=\dfrac{2n-4+11}{n-2}\) \(=\dfrac{2n-4}{n-2}\) \(+\dfrac{11}{n-2}\)

\(=\dfrac{2\left(n-2\right)}{n-2}+\dfrac{11}{n-2}\)\(=2+\dfrac{11}{n-2}\)

Vậy để A nguyên thì \(n-2\inƯ\left(11\right)\)\(=\left\{-11;-11;1;11\right\}\)

* \(n-2=-11\Rightarrow n=-11+2=-9\)

* \(n-2=-1\Rightarrow n=-1+2=1\)

* \(n-2=1\Rightarrow n=1+2=3\)

* \(n-2=11\Rightarrow n=11+2=13\)

Vì \(n\in N\) \(\Rightarrow\) \(n=\left\{1;3;13\right\}\)

Vậy để A là số nguyên thì \(n=\left\{1;3;13\right\}\)