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Đề bài là tìm n chứ:
a) Ta có:
\(n+5⋮n+2\)
\(\Rightarrow\left(n+2\right)+3⋮n+2\)
\(\Rightarrow3⋮n+2\)
\(\Rightarrow n+2\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n+2=-1\Rightarrow n=-3\\n+2=1\Rightarrow n=-1\\n+2=-3\Rightarrow n=-5\\n+2=3\Rightarrow n=1\end{matrix}\right.\)
Vậy \(n\in\left\{-3;-1;-5;1\right\}\)
b) Ta có:
\(2n+1⋮n-5\)
\(\Rightarrow\left(2n-10\right)+11⋮n-5\)
\(\Rightarrow2\left(n-5\right)+11⋮n-5\)
\(\Rightarrow11⋮n-5\)
\(\Rightarrow n-5\in U\left(11\right)=\left\{-1;1;-11;11\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n-5=-1\Rightarrow n=4\\n-5=1\Rightarrow n=6\\n-5=-11\Rightarrow n=-6\\n-5=11\Rightarrow n=16\end{matrix}\right.\)
Vậy \(n\in\left\{4;6;-6;16\right\}\)
c) Ta có:
\(n^2+3n-13⋮n+3\)
\(\Rightarrow n\left(n+3\right)-13⋮n+3\)
\(\Rightarrow-13⋮n+3\)
\(\Rightarrow n+3\in U\left(13\right)=\left\{-1;1;-13;13\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n+3=-1\Rightarrow n=-4\\n+3=1\Rightarrow n=-2\\n+3=-13\Rightarrow n=-16\\n+3=13\Rightarrow n=10\end{matrix}\right.\)
Vậy \(n\in\left\{-4;-2;-16;10\right\}\)
a: \(\Leftrightarrow n+2+5⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
b: \(\Leftrightarrow n-3-6⋮n-3\)
\(\Leftrightarrow n-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(n\in\left\{4;2;5;1;6;0;9;-3\right\}\)
c: \(\Leftrightarrow17⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1;17;-17\right\}\)
hay \(n\in\left\{0;-2;16;-18\right\}\)
a, 2n+1 chia hết cho 21=>21 thuộc Ư(2n+1)
=>2n+1 thuộc {1,3,7,21}
| 2n+1 | 1 | 3 | 7 | 21 |
| n | 0 | 1 | 3 | 10 |
Vậy n thuộc{0,1,3,10}
a) n + 5 ⋮ n - 2
n - 2 + 7 ⋮ n - 2
Vì n - 2 ⋮ n - 2
=> 7 ⋮ n - 2
=> n - 2 thuộc Ư(7) = { 1; -1; 7; -7 }
=> n thuộc { 3; 1; 9; -5 }
Vậy..........
b) 2n + 1 ⋮ n - 5
2n - 10 + 11 ⋮ n - 5
2( n - 5 ) + 11 ⋮ n - 5
Vì 2( n - 5 ) ⋮ n - 5
=> 11 ⋮ n - 5
=> n - 5 thuộc Ư(11) = { 1; -1; 11; -11 }
=> n thuộc { 6; 4; 16; -6 }
Vậy...........
c) n2 + 3 ⋮ n - 1
n2 - n + n + 3 ⋮ n - 1
n( n - 1 ) + n + 3 ⋮ n - 1
Vì n( n - 1 ) ⋮ n - 1
=> n + 3 ⋮ n - 1
n - 1 + 4 ⋮ n - 1
Vì n - 1 ⋮ n - 1
=> 4 ⋮ n - 1
=> n - 1 thuộc Ư(4) = { 1; 2; 4; -1; -2; -4 }
=> n thuộc { 2; 3; 5; 0; -1; -3 }
Vậy..........
\(a,n+5⋮n-2\Leftrightarrow n-2+7⋮n-2\)
\(\Leftrightarrow7⋮n-2\left(n-2⋮n-2\right)\)
\(\Leftrightarrow n-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow n=1;3;-5;9\)
\(b,2n+1⋮n-5\Leftrightarrow2n-10+11⋮n-5\)
\(\Leftrightarrow2\left(n-5\right)+11⋮n-5\)
\(\Leftrightarrow11⋮n-5\) ( vì \(2\left(n-5\right)⋮n-5\))
\(\Leftrightarrow n-5\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(\Leftrightarrow n=4;6;-6;16\)
d) n2 + 3n - 13 ⋮ n + 3
n( n + 3 ) - 13 ⋮ n + 3
Vì n( n + 3 ) ⋮ n + 3
=> 13 ⋮ n + 3
=> n + 3 thuộc Ư(13) = { 1; 13; -1; -13 }
=> n thuộc { -2; 10; -4; -16 }
Vậy.......
\(c,n^2+3⋮n-1\Leftrightarrow n^2-1+4⋮n-1\)
\(\Leftrightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)
\(\Leftrightarrow4⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Bn tự lập bảng để xét các giá trị của n
\(\left(n+5\right)⋮\left(n-2\right)\)
\(\Rightarrow\left(n-2+7\right)⋮\left(n-2\right)\)
\(\Rightarrow7⋮\left(n-2\right)\left[\text{vì }\left(n-2\right)⋮\left(n-2\right)\right]\)
\(\Rightarrow n-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\text{Trường hợp : }n-2=-7\)
\(\Rightarrow n=-7+2\)
\(\Rightarrow n=-5\)
\(\text{Trường hợp : }n-2=-1\)
\(\Rightarrow n=-1+2\)
\(\Rightarrow n=1\)
\(\text{Trường hợp : }n-2=1\)
\(\Rightarrow n=1+2\)
\(\Rightarrow n=3\)
\(\text{Trường hợp : }n-2=7\)
\(\Rightarrow n=7+2\)
\(\Rightarrow n=9\)
\(\text{Vậy }n\in\left\{-5;1;3;9\right\}\)
\(\left(2n+1\right)⋮\left(n-5\right)\)
\(\Rightarrow\left(2n-10+11\right)⋮\left(n-5\right)\)
\(\Rightarrow11⋮\left(n-5\right)\left[\text{vì }\left(2n-10\right)⋮\left(n-5\right)\right]\)
\(\Rightarrow n-5\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\text{Trường hợp : }n-5=-11\)
\(\Rightarrow n=-11+5\)
\(\Rightarrow n=-6\)
\(\text{Trường hợp : }n-5=-1\)
\(\Rightarrow n=-1+5\)
\(\Rightarrow n=4\)
\(\text{Trường hợp : }n-5=1\)
\(\Rightarrow n=1+5\)
\(\Rightarrow n=6\)
\(\text{Trường hợp : }n-5=11\)
\(\Rightarrow n=11+5\)
\(\Rightarrow n=16\)
\(\text{Vậy }n\in\left\{-6;4;6;16\right\}\)
\(\left(n^2+3\right)⋮\left(n-1\right)\)
\(\Rightarrow\left(n^2-n+n+3\right)⋮\left(n-1\right)\)
\(\Rightarrow\left[n\left(n-1\right)+n+3\right]⋮\left(n-1\right)\)
\(\Rightarrow\left(n+3\right)⋮\left(n-1\right)\left\{\text{vì }\left[n\left(n-1\right)⋮\left(n-1\right)\right]\right\}\)
\(\Rightarrow\left(n-1+4\right)⋮\left(n-1\right)\)
\(\Rightarrow4⋮\left(n-1\right)\left[\text{vì }\left(n-1\right)⋮\left(n-1\right)\right]\)
\(\Rightarrow n-1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow n\in\left\{-3;-1;0;2;3;5\right\}\)( tự giải từng trường hợp như các bài trên nhé )