Question

tim n thuoc N de (4n+12):(3n+1)

Answer 1

\(4n+12⋮3n+1\)

\(\Leftrightarrow3\left(4n+12\right)⋮\left(3n+1\right)\)

\(\Leftrightarrow\left(12n+36\right)⋮\left(3n+1\right)\)

\(\Leftrightarrow\left[4\left(3n+1\right)+32\right]⋮\left(3n+1\right)\)

\(\Leftrightarrow32⋮\left(3n+1\right)\)

\(\Rightarrow3n+1\inƯ\left(32\right)\)

\(\Rightarrow3n+1\in\left\{1,2,4,8,16,32\right\}\left(1\right)\)

Mà \(n\in N\Rightarrow3n+1\equiv1\left(mod3\right)\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow3n+1\in\left\{1,4,16\right\}\)

\(\Leftrightarrow n\in\left\{0,1,5\right\}\)

Vậy.........................