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\(\left(3n+2\right)⋮\left(n-1\right)\)
\(\Rightarrow\left(3n-3+5\right)⋮\left(n-1\right)\)
\(\Rightarrow5⋮\left(n-1\right)\)
\(\Rightarrow n-1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow n\in\left\{-4;0;2;6\right\}\)
ban Nguyen Chau Tuan Kiet tra loi dung nhung ban quen y n thuoc N roi
\(a,n+6⋮n\)
\(\Rightarrow6⋮n\)
\(\Rightarrow n\inƯ\left(6\right)\)
\(\Rightarrow n\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
\(b,n+9⋮n+1\)
\(\Rightarrow n+1+8⋮n+1\)
\(\Rightarrow8⋮n+1\)
\(\Rightarrow n+1\inƯ\left(8\right)\)
\(\Rightarrow n+1\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
\(\Rightarrow n\in\left\{-2;0;-3;1;-5;3;-9;7\right\}\)
\(c,n-5⋮n+1\)
\(\Rightarrow n+1-6⋮n+1\)
\(\Rightarrow6⋮n+1\)
\(\Rightarrow n+1\inƯ\left(6\right)\)
\(\Rightarrow n+1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
\(\Rightarrow n\in\left\{-2;0;-3;0;-4;2;-7;5\right\}\)
\(d,2n+7⋮n-2\)
\(\Rightarrow2n-4+11⋮n-2\)
\(\Rightarrow2\left(n-2\right)+11⋮n-2\)
\(\Rightarrow11⋮n-2\)
\(\Rightarrow n-2\inƯ\left(11\right)\)
\(\Rightarrow n-2\in\left\{-1;1;-11;11\right\}\)
\(\Rightarrow n\in\left\{1;3;-9;13\right\}\)
Ta có :
\(3n+1⋮2n-3\Rightarrow2\left(3n+1\right)⋮2n-3\Rightarrow6n+2⋮2n-3\) (1)
Ta lại có:
\(2n-3⋮2n-3\Rightarrow3\left(2n-3\right)⋮2n-3\Rightarrow6n-9⋮2n-3\) (2)
Trừ (1) cho (2), ta được :
\(\left(6n+2\right)-\left(6n-9\right)⋮2n-3\)
\(\Rightarrow11⋮2n-3\)
\(\Rightarrow2n-3\inƯ\left(11\right)\)
\(\Rightarrow2n-3\in\left\{1;-1;11;-11\right\}\)
\(\Rightarrow2n\in\left\{4;2;14;-8\right\}\)
\(\Rightarrow n\in\left\{2;1;7;-4\right\}\)
Vậy \(n\in\left\{2;1;7;-4\right\}\)
2n - 7 chia hết cho n + 4
=> 2n + 8 - 15 chia hết cho n + 4
=> 2.(n + 4) - 15 chia hết cho n + 4
=> 15 chia hết cho n + 4
=> n + 4 \(\in\)Ư(15) = {-15; -5; -3; -1; 1; 3; 5; 15}
=> n \(\in\){-19; -9; -8; -5; -3; -1; 1; 11}.
a)\(10^{2k}-1=\left(10^k-1\right)\left(10^k+1\right)\)
Dễ thấy: \(10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+1\right)⋮19\)
\(\Rightarrow10^{2k}-1⋮19\)
b)\(10^{3k}-1=\left(10^k-1\right)\left(10^k+10^{2k}+1\right)\)
Dễ thấy: \(10^k-1⋮19\Rightarrow\left(10^k-1\right)\left(10^k+10^{2k}+1\right)⋮19\)
\(\Rightarrow10^{3k}-1⋮19\)
Thắng xem mà học tập đây :v
Vì 10k - 1 \(⋮\) 19 => 10k - 1\(\equiv\) 0 (mod 19)
=> 10k \(\equiv\) 1 (mod 19)
a) 10k \(\equiv\) 1 (mod 19)
=> (10k)2 \(\equiv\) 12 (mod 19)
=> 102k \(\equiv\) 1 (mod 19)
=> 102k - 1 \(⋮\) 19
b) 10k \(\equiv\) 1 (mod 19)
=> (10k)3 \(\equiv\) 13 (mod 19)
=> 103k = 1 (mod 19)
=> 103k - 1 \(⋮\) 19
a, n + 2 \(⋮n-3\)
<=> n - 3 + 5 \(⋮n-3\)
<=> 5 \(⋮n-3\)
=> n - 3 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
=> n = 4; 2; 8; -2 (thỏa mãn)
b, 3n + 15 \(⋮n-4\)
Có 3(n - 4) \(⋮n-4\)
=> (3n + 15) - (3n - 12) \(⋮n-4\)
<=> 27 \(⋮n-4\)
=> n - 4 \(\inƯ\left(27\right)=\left\{\pm1;\pm3;\pm9;\pm27\right\}\)
=> n = 5; 3; 7; 1; 13; -5; 31; -23 (thỏa mãn)
@hoang thuy an
c, 2n - 3 \(⋮3n+2\)
<=> 3(2n - 3) \(⋮3n+2\)
<=> 6n - 9 \(⋮3n+2\)
Có 2(3n + 2) \(⋮3n+2\)
=> (6n - 9) - (6n + 4) \(⋮3n+2\)
<=> -13 \(⋮3n+2\)
=> 3n + 2 \(\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
=> 3n = -1; -3; 11; -15
=> n = -\(\dfrac{1}{3};-1;\dfrac{11}{3};-5\)
Mà n \(\in Z\Rightarrow n=-1;-5\)
d, 4n + 7 \(⋮3n+1\)
<=> 3(4n + 7) \(⋮3n+1\)
<=> 12n + 21 \(⋮3n+1\)
Có 4(3n + 1) \(⋮3n+1\)
=> (12n + 21) - (12n + 4) \(⋮3n+1\)
<=> 17 \(⋮3n+1\)
=> 3n + 1 \(\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
=> 3n = 0; -2; 16; -18
=> n = 0; -\(\dfrac{2}{3};\dfrac{16}{3};-6\)
Mà n \(\in Z\Rightarrow n=0;-6\)
@hoang thuy an
Ta có:\(n^2-3⋮n+3\)
\(\Leftrightarrow n^2+3n-3n-9+6⋮n+3\)
\(\Leftrightarrow\left(n^2+3n\right)-\left(3n+9\right)+6⋮n+3\)
\(\Leftrightarrow n\left(n+3\right)-3\left(n+3\right)+6⋮n+3\)
\(\Leftrightarrow6⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(6\right)\)
Mà \(n\in N\)*\(\Rightarrow n+3\ge4\)
\(\Leftrightarrow n+3=6\)
\(\Leftrightarrow n=3\)
\(n^2-3⋮n+3\\ \Rightarrow\left(n-3\right)\left(n+3\right)+6⋮n+3\\ \Rightarrow6⋮n+3\Rightarrow n+3\in\text{Ư}\left(6\right)\)
Tới đây dễ rồi nha!