\(12⋮2n+1\Rightarrow2n+1\inƯ\left(12\right)\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Vì 2n +1 chia 2 dư 1 nên \(2n+1\in\left\{\pm1;\pm3\right\}\)

\(\Rightarrow n\in\left\{0;-1;1;-2\right\}\)

làm tiếp 

\(3n+5⋮n+2\Rightarrow3\left(n+2\right)+3⋮n+2\)

\(\Rightarrow3⋮n+2\Rightarrow n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow n\in\left\{-1;-3;1;-5\right\}\)

a. 12 chia hết cho 2n+1

\(\Rightarrow2n+1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

\(\Rightarrow n\in\left\{0;\pm1;\pm2\right\}\)

b. n-1 chia hết cho n+1

\(\Rightarrow\left(n+1\right)-2⋮n+1\)

\(\Rightarrow n+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow n\in\left\{0;1;-2;-4\right\}\)

c. 3n+5 chia hết cho n+2

\(\Rightarrow3\left(n+2\right)-1⋮n+2\)

\(\Rightarrow n+2\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Rightarrow n\in\left\{-3;-1\right\}\)

d. 3n+2 chia hết cho 2n+3

\(\Rightarrow6n+4⋮2n+3\)

\(\Rightarrow3\left(2n+3\right)-5⋮2n+3\)

\(\Rightarrow2n+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow n\in\left\{\pm1;-2;-4\right\}\)

\(2n-4⋮2n+1\)

\(\Rightarrow2n+1-5⋮2n+1\)

=> \(5⋮2n+1\)

=> \(2n+1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

=> \(2n\in\left\{0;-2;4;-6\right\}\)

=> \(n\in\left\{0;-1;2;-3\right\}\) (TM)

Ta co : x/10=y/6=z/21=>5x/50=y/6=2z/42 va 5x+y-2z=-56

Ap dung tinh chat day ti so bang nhau ta co : 

5x/50=y/6=2z/42=5x+y-2z/50+6-42=-56/14

Suy ra :5x/50=-4=>x=-4.50:5=-40

y/6=-4=>y=-4.6=-24

2z/42=-4=>z=-4.42:2=-84

**** nhe

a: Thay x=2 và y=-3 vào (d), ta được:

10-4m=-3

hay m=13/4

ấn vào Đúng 0 sẽ hiện ra kết quả

nhớ 10 lần

a)-1

b)11

c)11

d)-1