Question

tìm số nguyên n

a,\(\left(3n-2n\right)⋮\left(n+1\right)\)

b, \(\left(2n-4\right)⋮\left(n+2\right)\)

Answer 1

\(------huongdan-----\)

\(Taco:\)

\(\left(3n-2n\right)⋮n+1\Leftrightarrow n⋮n+1\Leftrightarrow\left(n+1\right)-n⋮n+1\Leftrightarrow1⋮n+1\)

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

\(b,2n-4⋮n+2\Leftrightarrow2n+4-2n+4⋮2n+4\Leftrightarrow8⋮2n+4\)

dễ thấy: 2n+4 chẵn => 2n+4 là ước chẵn của 8

\(\Rightarrow2n+4\in\left\{2;4;8;-2;-4;-8\right\}\Rightarrow2n\in\left\{-2;0;4;-6;-8;-12\right\}\)

\(\Rightarrow n\in\left\{-1;0;2;-3;-4;-6\right\}\)

Answer 2

\(2n-4⋮n+2\)

\(\Rightarrow2n+4-8⋮n+2\)

\(\Rightarrow2\left(n+2\right)+8⋮n+2\)

\(\Rightarrow n+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

bn tụ lập bảng ha ~ 

Answer 3

a)\(3n-2n⋮n+1\)

\(\Rightarrow n⋮n+1\)

\(\Rightarrow n+1-1⋮n+1\Rightarrow1⋮n+1\orbr{\begin{cases}n+1=1\\n+1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}n=0\\n=-2\end{cases}}\)

b)\(2n-4⋮n+2\Rightarrow2n+4-8⋮n+2\)

\(\Rightarrow2\left(n+2\right)-8⋮n+2\Rightarrow8⋮n+2\)

\(+,n+2=1\Rightarrow n=-2\)

\(+,n+2=-1\Rightarrow n=-3\)

\(+,n+2=2\Rightarrow n=0\)

\(+,n+2=-2\Rightarrow n=-4\)

\(+,n+2=4\Rightarrow n=2\)

\(+,n+2=-4\Rightarrow n=-6\)