để \(\frac{n^2+n-5}{n+2}\)nguyên \(\Leftrightarrow n^2+n-5⋮n+2\)
\(\Leftrightarrow n^2+2n-n-5⋮n+2\)
\(\Leftrightarrow n.\left(n+2\right)-n-5⋮n+2\)
mà \(n.\left(n+2\right)⋮n+2\)
\(\Rightarrow n-5⋮n+2\)
\(\Rightarrow n+2-7⋮n+2\)
mà \(n+2⋮n+2\)
\(\Rightarrow7⋮n+2\)
em tự làm típ nhé
Điều kiện xác định:\(n\ne-2\)
Ta có:
\(\frac{n^2+n-5}{n+2}=\left(n-1\right)-\frac{3}{n+2}\) (chia đa thức, có \(-\frac{3}{n+2}\)vì chia dư -3)
Để \(\frac{n^2+n-5}{n+2}\) là số nguyên
=> \(3⋮n+2\)
=>\(\left(n+2\right)\inƯ\left(3\right)\)
=>\(\left(n+2\right)\in\left\{-3;-1;1;3\right\};n\in Z\)
=>\(n\in\left\{-5;-3;-1;1\right\}\)
ĐKXĐ : \(n\ne-2\)
Để \(\frac{n^2+n-5}{n+2}\)nguyên thì :
\(\left(n^2+n-5\right)⋮\left(n+2\right)\)
\(\left(n^2+2n-n-5\right)⋮\left(n+2\right)\)
\(\left[n\left(n+2\right)-\left(n+5\right)\right]⋮\left(n+2\right)\)
Vì \(n\left(n+2\right)⋮\left(n+2\right)\)
\(\Rightarrow\left(n+5\right)⋮\left(n+2\right)\)
\(\left(n+2+3\right)⋮\left(n+2\right)\)
Vì \(\left(n+2\right)⋮\left(n+2\right)\Rightarrow3⋮\left(n+2\right)\)
\(\Rightarrow\left(n+2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow n\in\left\{-1;-3;1;-5\right\}\)( thỏa mãn )
Vậy....
