\(A=\frac{2n+1}{n+5}\inℤ\Leftrightarrow2n+1⋮n+5\)
\(\Rightarrow2n+10-9⋮n+5\)
\(\Rightarrow2\left(n+5\right)-9⋮n+5\)
\(2\left(n+5\right)⋮n+5\)
\(\Rightarrow9⋮n+5\)
\(\Rightarrow n+5\inƯ\left(9\right)\)
\(n\inℤ\Rightarrow n+5\inℤ\)
\(\Rightarrow n+5\in\left\{-1;1;-3;3;-9;9\right\}\)
\(\Rightarrow n\in\left\{-6;-4;-8;-2;-14;4\right\}\)
gọi d\(\in\)uc(2n+1,n+5)
\(\Rightarrow1\left(2n+1\right)-2\left(n+5\right)⋮d\)
\(\Rightarrow2n+1-2n-10⋮d\)
\(\Rightarrow-9⋮d\Rightarrow d\in u\left(-9\right)=\left\{1;-1;9;-9\right\}\)
Lập bảng:
| \(2n+1\) | \(1\) | \(-1\) | \(9\) | \(-9\) |
| \(n\) | \(0\) | \(-1\) | \(4\) | \(-5\) |
\(n\inℤ\Rightarrow n\in\left\{0;-1;4;-5\right\}\)
\(A=\frac{2n+1}{n+5}\in Z\)
\(\Rightarrow2n+1⋮n+5\left(1\right)\)
+)Ta có:\(n+5⋮n+5\)
\(\Rightarrow2.\left(n+5\right)⋮n+5\)
\(\Rightarrow2n+10⋮n+5\left(2\right)\)
+)Từ (1) và (2)
\(\Rightarrow\left(2n+10\right)-\left(2n+1\right)⋮n+5\)
\(\Rightarrow2n+10-2n-1⋮n+5\)
\(\Rightarrow9⋮n+5\)
\(\Rightarrow n+5\in\left\{\pm1;................;\pm9\right\}\)
+)Ta có bảng sau:
| n+5 | -1 | 1 | -2 | 2 | -3 | 3 | -4 | 4 | -5 | 5 | -6 | 6 | -7 | 7 | -8 | 8 | -9 | 9 |
| n | -6\(\in Z\) | -4\(\in Z\) | -7\(\in Z\) | -3\(\in Z\) | -8\(\in Z\) | -2\(\in Z\) | -9\(\in Z\) | -1\(\in Z\) | -10\(\in Z\) | 0\(\in Z\) | -11\(\in Z\) | 1\(\in Z\) | -12\(\in Z\) | 2\(\in Z\) | -13\(\in Z\) | 3\(\in Z\) | -14\(\in Z\) | 4\(\in Z\) |
Vậy \(n\in\left\{\right\}\)bạn tự liệt kê các n trên xuống nha dài quá nên mk k viết
Chúc bạn học tốt
biến đổi A =\(\frac{2n+1}{n+5}=\frac{2n+10-10+1}{n+5}=\frac{2\left(n+5\right)-9}{n+5}\)
A =\(\frac{2\left(n+5\right)}{n+5}+\frac{9}{n+5}=2+\frac{9}{n+5}\)
để A=\(2+\frac{9}{n+5}\inℤ\Rightarrow\frac{9}{n+5}\inℤ\)
\(\Rightarrow n+5\inư\left(9\right)\in\left\{\pm9;\pm3;\pm1\right\}\)
| n+5 | 9 | 3 | 1 | -1 | -3 | -9 |
| n | 4 | -2 | -4 | -6 | -8 | -14 |
Vây x\(\in\left(4;-2;-4;-6;-8;-14\right)\)
