Tìm stn n để:
10 - 2n chia hết cho n-2
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a) \(4\left(n-1\right)-3⋮\left(n-1\right)\)
\(\Rightarrow\left(n-1\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Do \(n\in N\Rightarrow n\in\left\{0;2;4\right\}\)
b) \(-5\left(4-n\right)+12⋮\left(4-n\right)\)
\(\Rightarrow\left(4-n\right)\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do \(n\in N\Rightarrow n\in\left\{16;10;8;7;6;5;3;2;1;0\right\}\)
c) \(-2\left(n-2\right)+6⋮\left(n-2\right)\)
\(\Rightarrow\left(n-2\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Do \(n\in N\Rightarrow n\in\left\{0;1;3;4;5;8\right\}\)
d) \(n\left(n+3\right)+6⋮\left(n+3\right)\)
\(\Rightarrow\left(n+3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Do \(n\in N\Rightarrow n\in\left\{0;3\right\}\)
a) \(2n+7⋮2n+1\)
\(\Rightarrow\left(2n+1\right)+6⋮2n+1\)
\(\Rightarrow6⋮2n+1\)(vì \(2n+1⋮2n+1\))
\(\Rightarrow2n+1\inƯ\left(6\right)\)
\(\Rightarrow2n+1\in\left\{1;2;3;6\right\}\)
\(\Rightarrow\)\(2n\in\left\{0;1;2;5\right\}\)
\(\Rightarrow n\in\left\{0;1\right\}\)
b) \(3m-9⋮3m-1\)
\(\Rightarrow\left(3m-1\right)-8⋮3m-1\)
\(\Rightarrow8⋮3m-1\)(vì \(3m-1⋮3m-1\))
\(\Rightarrow3m-1\inƯ\left(8\right)\)
\(\Rightarrow3m-1\in\left\{1;2;4;8\right\}\)
\(\Rightarrow3m\in\left\{2;3;5;9\right\}\)
\(\Rightarrow m\in\left\{1;3\right\}\)
Hok "tuốt" nha^^
Gợi ý :
10 - 2n ⋮ n - 2
-2n + 4 + 6 ⋮ n - 2
-2 ( n - 2 ) + 6 ⋮ n - 2
Vì -2 ( n - 2 ) ⋮ n - 2
=> 6 ⋮ n - 2
=> n - 2 thuộc Ư(6) = { 1; 2; 3; 6; -1; -2; -3; -6 }
Đến đây tìm n nốt nhé bạn :))
Ta có: ( n - 2) \(⋮\)n - 2
=> 2( n - 2) \(⋮\)n - 2
=> 2n -4 \(⋮\)n - 2
Mà 10 - 2n \(⋮\)n - 2 ( gt )
Nên ta có : \([(\)10 - 2n \()\)\(-\)\((\)2n - 4\()]\)\(⋮\)n -2
=> \([\)10 - 2n - 2n -4\(]\)\(⋮\)n - 2
=> 6 \(⋮\)n - 2
=> n - 2 \(\in\){ 1; 2; 3; 6 }
=> n\(\in\){ 3; 4; 5; 8 }
Vậy n\(\in\){ 3; 4; 5; 8 }