tìm n sao cho
a/30<2^n<300
b/20<6^n<4300
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\(3n+14⋮n+2\)
=>\(3n+6+8⋮n+2\)
=>\(8⋮n+2\)
=>\(n+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>\(n\in\left\{-1;-3;0;-4;2;-6;6;-10\right\}\)
mà n>=0
nên \(n\in\left\{0;2;6\right\}\)
a: \(\Leftrightarrow n-5\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{6;4;8;2\right\}\)
b: \(\Leftrightarrow3n-18+18⋮n-6\)
\(\Leftrightarrow n-6\in\left\{1;-1;2;-2;3;-3;6;-6;9;-9;18;-18\right\}\)
hay \(n\in\left\{7;5;8;4;9;3;12;0;15;-3;24;-12\right\}\)
a) \(-3⋮n-5\)
\(\Rightarrow n-5\inƯ\left(-3\right)=\left\{-1;1;-3;3\right\}\)
Có bảng sau:
n-5 | -1 | 1 | -3 | 3 |
n | 4 | 6 | 2 | 8 |
Vậy...
b)
\(\begin{matrix}3n⋮n-6\\n-6⋮n-6\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}3n⋮n-6\\3n-18⋮n-6\end{matrix}\right.\){18\(⋮\) n-6
\(\Leftrightarrow n-6\inƯ\left(18\right)=\left\{1,2,3,6,9,18,-1,-2,-3,-6,-9,-18\right\}\)
Có bảng sau:
n-6 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 | 9 | -9 | 18 | -18 |
n | 7 | 5 | 8 | 4 | 9 | 3 | 12 | 0 | 15 | -3 | 24 | -12 |
Vậy...
a: \(\Leftrightarrow n+8-11⋮n+8\)
\(\Leftrightarrow n+8\in\left\{1;-1;11;-11\right\}\)
hay \(n\in\left\{-7;-9;3;-19\right\}\)
b: Đề thiếu rồi bạn
a, \(\dfrac{n-3}{n+8}=\dfrac{n+8-11}{n+8}=1-\dfrac{11}{n+8}\)
\(\Rightarrow n+8\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
n+8 | 1 | -1 | 11 | -11 |
n | -7 | -9 | 3 | -19 |
b, bạn bổ sung đề nhé
a) \(\Rightarrow2\left(n+3\right)-38⋮\left(n+3\right)\)
Do \(n\in N\)
\(\Rightarrow\left(n+3\right)\inƯ\left(38\right)=\left\{19;38\right\}\)
\(\Rightarrow n\in\left\{16;35\right\}\)
b) \(\Rightarrow5\left(n+5\right)-74⋮\left(n+5\right)\)
Do \(n\in N\)
\(\Rightarrow\left(n+5\right)\inƯ\left(74\right)=\left\{37;74\right\}\)
\(\Rightarrow N\in\left\{32;69\right\}\)
a) \(\Rightarrow2\left(n+3\right)-38⋮\left(n+3\right)\)
Mà \(n\in N\Rightarrow n+3\ge3\)
\(\Rightarrow\left(n+3\right)\inƯ\left(38\right)=\left\{19;38\right\}\)
\(\Rightarrow n\in\left\{16;35\right\}\)
b) \(\Rightarrow5\left(n+5\right)-74⋮\left(n+5\right)\)
Do \(n\in N\Rightarrow n+5\ge5\)
\(\Rightarrow\left(n+5\right)\inƯ\left(74\right)=\left\{37;74\right\}\)
\(\Rightarrow n\in\left\{32;69\right\}\)
\(a,2n-32⋮n+3\Rightarrow2\left(n+3\right)-38⋮n+3\\ \Rightarrow n+3\inƯ\left(38\right)=\left\{1;2;19;38\right\}\\ \Rightarrow n\in\left\{16;35\right\}\\ b,5n-49⋮n+5\Rightarrow5\left(n+5\right)-74⋮n+5\\ \Rightarrow n+5\inƯ\left(74\right)=\left\{1;2;37;74\right\}\\ \Rightarrow n\in\left\{32;69\right\}\)
a) \(4n-5⋮13\)
\(\Rightarrow4n-5+13⋮13\Rightarrow4n+8⋮13\Rightarrow4\left(n+2\right)⋮13\)
Vì (4;13) = 1 nên n+2 chia hết cho 13
=> n=13k-2 ( \(k\in N\)*)
b) \(5n+1⋮7\Rightarrow5n+1+14⋮7\Rightarrow5n+15⋮7\Rightarrow5\left(n+3\right)⋮7\)
Vì 5 không chia hết cho 7 nên để 5(n+3) chia hết cho 7 thì n+3 chia hết cho 7
=> n = 7k-3 ( \(k\in N\)*)
c) \(25n+3⋮53\Rightarrow25n+3-53⋮53\Rightarrow25n-50⋮53\Rightarrow25\left(n-2\right)⋮53\Rightarrow n-2⋮53\)
=> n = 53k+2 ( k thuộc N*)
\(a,\Rightarrow n-1+7⋮n-1\)
Mà \(n-1⋮n-1\Rightarrow7⋮n-1\)
\(\Rightarrow n-1\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow n\in\left\{2;8\right\}\)
\(b,\Rightarrow3\left(n+1\right)+2⋮n+1\)
Mà \(3\left(n+1\right)⋮n+1\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;2\right\}\\ \Rightarrow n=1\left(n\ne0\right)\)
a,Ta có:n+2 chia hết cho n-3
=>n-3+5 chia hết cho n-3
Mà n-3 chia hết cho n-3
=>5 chia hết cho n-3
=>n-3\(\in\)Ư(5)={-5,-1,1,5}
=>n\(\in\){-2,2,4,8}
b,Ta có:2n-7 chia hết cho n-1
=>2n-2-5 chia hết cho n-1
=>2(n-1)-5 chia hết cho n-1
Mà 2(n-1) chia hết cho n-1
=>5 chia hết cho n-1
=>n-1\(\in\)Ư(5)={-5,-1,1,5}
=>n\(\in\){-4,0,2,6}
A = 6 + 62 + 63 + 64 + ... + 62016
6A = 62 + 63 + 64 + 65 + ... + 62017
6A - A = (62 + 63 + 64 + 65 + ... + 62017) - (6 + 62 + 63 + 64 + ... + 62016)
5A = 62017 - 6
6n = 5A + 6
6n = (62017 - 6) + 6
6n = 62017
=>n = 2017
a) x Î Ư(6) = {-6; -3; -2; -l; l; 2; 3; 6}.
b) x + l Î Ư (8) = {- 8; -4; -2; -1; 1; 2; 4; 8}. Từ đó tìm được
x Î{-9; -5; -3; -2; 0; 1; 3; 7}.
c) x - 2 Î Ư(10) = {-10; -5; - 2; -1; 1; 2; 5; 10). Từ đó tìm được
x Î {-8; -3; 0; l; 3; 5; 7; 12}.