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\(n⋮n-2\\ \Rightarrow n-\left(n-2\right)⋮n-2\\ \Rightarrow2⋮n-2\\ \Rightarrow n-2\in\left\{1;2\right\}\\ \Rightarrow n\in\left\{3;4\right\}\)Vậy \(n\in\left\{3;4\right\}\)
\(n+7⋮n+1\\ \Rightarrow n+7-\left(n+1\right)⋮n+1\\ \Rightarrow6⋮n+1\\ \Rightarrow n+1\in\left\{1;2;3;6\right\}\\ \Rightarrow n\in\left\{0;1;2;5\right\}\)Vậy \(n\in\left\{0;1;2;5\right\}\)
\(21⋮2n+5\\ \Rightarrow2n+5\in\left\{1;3;7;21\right\}\\ \Rightarrow2n\in\left\{2;16\right\}\\ \Rightarrow n\in\left\{1;8\right\}\)Vậy \(n\in\left\{1;8\right\}\)
\(2n+7⋮2n+1\\ \Rightarrow2n+7-\left(2n+1\right)⋮2n+1\\ \Rightarrow6⋮2n+1\\ \Rightarrow2n+1\in\left\{1;2;3;6\right\}\\ \Rightarrow2n\in\left\{0;1;2;5\right\}\\ \Rightarrow n\in\left\{0;1\right\}\)Vậy \(n\in\left\{0;1\right\}\)
2n+ 18 \(⋮\) 2n+5
=> \(\left(2n+18\right)-\left(2n+5\right)⋮\left(2n+5\right)\)
=> \(\left(2n+18-2n-5\right)⋮\left(2n+5\right)\)
=> \(13⋮\left(2n+5\right)\)
=> \(\left(2n+5\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
ta có bảng sau
2n+5 | -13 | -1 | 1 | 13 |
2n
|
-18 | -6 | -4 | 8 |
n | -9 | -3 | -2 | 4 |
vây n \(\in\left\{-9;-3;-2;4\right\}\)
a) n \(\in\text{ }\text{ }\left\{2;7;12;17;22;27;...\right\}\)
b) \(n\in\left\{3;10;17;24;31;39;46;...\right\}\)
c) \(n\in\left\{14;27;40;53;66;79;...\right\}\)
a) \(n\in\text{ }\left\{2;7;12;17;22;27;...\right\}\)
b) \(n\in\text{ }\left\{3;10;17;24;31;39;...\right\}\)
c) \(n\in\text{ }\left\{14;27;40;53;66;79;...\right\}\)
Chúc bạn học tốt !! Mik nhanh nhất nha
a)
Ta có: 2n+21 chia hết cho 5
=> 2n+21 = Ư(5)={-1;1;-5;5}
=> 2n = {-22;-20;-26;-16}
=> n ={-11;-10;-13;-8}
Ta có: 5n-8 chia hết cho 7
====> 5n-8 = Ư(7)={-1;1;-7;7}
=> 5n = {7;9;1;15)
=> n = {3}
a) 2;7;12;17;22;27;...
b) 3;10;17;24;31;39;46
c) 14;27;40;53;66;79;...
2n+21 chia hết cho 5
nên 2n+21EB(5)={0;5;10;15;20;25;30;35;...}
=>2nE{4;9;14;...}
=>nE{2;7;...}
\(21⋮\left(2n-5\right)\Leftrightarrow2n-5\inƯ\left(21\right)=\left\{-21,-7,-3,-1,1,3,7,21\right\}\)
\(\Leftrightarrow n\in\left\{-8,-1,1,2,3,4,6,13\right\}\).