\(A=\frac{n+5}{n-2}\)
\(=\frac{n-2+7}{n-2}\)
\(=\frac{n-2}{n-2}+\frac{7}{n-2}\)
\(\Rightarrow\)để \(A\in Z\)\(\Rightarrow\frac{7}{n-2}\in Z\)
\(\Rightarrow n-2\inƯ_7\)
Mà \(Ư_7=\left\{1;-1;7;-7\right\}\)nên ta có
* \(n-2=1\Rightarrow n=3\)
*\(n-2=-1\Rightarrow n=1\)
*\(n-2=7\Rightarrow n=9\)
*\(n-2=-7\Rightarrow n=-5\)
Vậy để \(A\in Z\Leftrightarrow n\in\left\{3;1;9;-5\right\}\)
n + 5 / n - 2
n + 5 : n-2
( n + 5 )-(n-2):n-2
n+5-n+2:n-2
7:n-2
=>n-2=Ư(7)
=>n=-5;1;3;9
\(A=\frac{n+5}{n-2}=\frac{n-2+7}{n-2}=1+\frac{7}{n-2}\)
Để \(A\in Z\Rightarrow7⋮n-2\)
Lập bảng tính
n-2 | 1 7 -1 -7 |
n | 3 9 1 -6 |
\(B=\frac{2n+7}{n+1}=\frac{2.\left(n+1\right)+5}{n+1}=2+\frac{5}{n+1}\)
Để \(B\in Z\Rightarrow5⋮n+1\)
Lập bảng
n+1 | 1 5 -1 -5 |
n | 0 4 -2 -6 |
\(C=\frac{3n-5}{n+2}=\frac{3.\left(n+2\right)-11}{n+2}=3+\frac{11}{n+2}\)
Để \(C\in Z\Rightarrow11⋮n+2\)
n+2 | 1 11 -1 -11 |
n | -1 9 -3 -13 |
Hok tốt :))
\(B=\frac{2n+7}{n+1}\)
\(=\frac{2n+2+5}{n+1}\)
\(=\frac{2\left(n+1\right)}{n+1}+\frac{5}{n+1}\)
Để \(B\in Z\Rightarrow\frac{5}{n+1}\in Z\)\(\Rightarrow n+1\inƯ_5\)
Tương tự câu A mà làm nốt nhé
2n + 7 / n + 1
=>2n +7 : n +1
=>(2n+7)-2(n+1): n+1
=>2n+7-2n-2:n+1
=>5:n+1
=>n+1=Ư(5)
=>n=-6;-2;0;4
\(C=\frac{3n-5}{n+2}\)
\(=\frac{3n+6-11}{n+2}\)
\(=\frac{3\left(n+2\right)}{n+2}-\frac{11}{n+2}\)
\(=3-\frac{11}{n+2}\)
Để \(C\in Z\Rightarrow\frac{11}{n+2}\in Z\)\(\Rightarrow n+2\inƯ_{11}\)
Tự làm nốt nha