\(A=\frac{3n-5}{n+4}=\frac{3\left(n+4\right)-17}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{17}{n+4}=3-\frac{17}{n+4}\in Z\)
\(\Rightarrow17⋮n+4\)
\(\Rightarrow n+4\inƯ\left(17\right)=\left\{1;-1;17;-17\right\}\)
\(\Rightarrow n\in\left\{-3;-5;13;-21\right\}\)
Ta có:\(A\in Z\Leftrightarrow\frac{3n-5}{n+4}\in Z\Leftrightarrow\frac{3n+12-17}{n+4}\in Z\Leftrightarrow\frac{3\left(n+4\right)}{n+4}-\frac{17}{n+4}\in Z\Leftrightarrow3-\frac{17}{n+4}\in Z\Leftrightarrow\frac{-17}{n+4}\in Z\)
\(\Leftrightarrow n+4\inƯ17\Leftrightarrow n+4\in\left\{-1;-17;1;17\right\}\)
Thay \(n+4=-1\Rightarrow n=-5\) (TM)
\(n+4=-17\Rightarrow n=-21\) (TM)
\(n+4=1\Rightarrow n=-3\) (TM)
\(n+4=17\Rightarrow n=13\) (TM)
Vậy \(n\in\left\{-21;-5;-3;13\right\}\) thì \(A\in Z\)
Để A nguyên thì:
3n - 5 chia hết cho n + 4
=> 3n + 12 - 17 chia hết cho n + 4 => 3.(n + 4) - 17 chia hết cho n + 4
=> 17 chia hết cho n + 4
=> n + 4 thuộc Ư(17) = {-17; -1; 1; 17}
=> n thuộc {-21; -5; -3; 13}.
tìm nϵ Z để H có giá trị nguyên.H=\(\dfrac{3}{n-5}\)
