Question

[Nguyen Thi Ha Vi]

tìm n thuộc N

a )3n +1 chia hết cho 2n + 3

b)  n^2  +5 chia hết cho n+ 1

Answer 1

[Nguyễn Thùy Trang ( team ASL)]

a) \(3n+1⋮2n+3\)

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

\(\Leftrightarrow n-2⋮2n+3\)

\(\Rightarrow2\left(n-2\right)⋮2n+3\)

\(\Leftrightarrow2n-4⋮2n+3\)

\(\Leftrightarrow\left(2n+3\right)-7⋮2n+3\)

\(\Leftrightarrow7⋮2n+3\)

\(\Leftrightarrow2n+3\inƯ\left(7\right)\) mà \(n\in N\Rightarrow2n+3\in N,2n+3\ge3\)

\(\Leftrightarrow2n+3=7\Leftrightarrow n=2\)

b) \(n^2+5⋮n+1\)

\(\Leftrightarrow\left(n^2+n\right)+5-n⋮n+1\)

\(\Leftrightarrow n\left(n+1\right)+5-n⋮n+1\)

\(\Leftrightarrow5-n⋮n+1\)

\(\Leftrightarrow6-\left(n+1\right)⋮n+1\)

\(\Leftrightarrow6⋮n+1\) mà \(n\in N\Rightarrow n+1\in N,n+1\ge1\)

\(\Leftrightarrow n+1\in\left\{1;2;3;6\right\}\Leftrightarrow n\in\left\{0;1;2;5\right\}\)

Answer 2

[Edogawa Conan]

a) Ta có: 3n + 1 \(⋮\)2n + 3

=>2(3n + 1) \(⋮\)2n + 3

=> 6n + 2 \(⋮\)2n + 3

=> 3(2n + 3) - 7 \(⋮\)2n + 3

=> 7 \(⋮\)2n + 3

=> 2n + 3 \(\in\)Ư(7) = {1; -1; 7; -7}

=> 2n \(\in\){-2; -4; 4; -10}

=> n \(\in\){-1; -2; 2; -5}

b) Ta có: n² + 5 = n(n + 1) - (n + 1) + 6 = (n - 1)(n + 1) + 6

Do (n - 1)(n + 1) \(⋮\)n + 1 => 6 \(⋮\)n + 1

                          => n + 1 \(\in\)Ư(6) = {1; -1; 2; -2; 3; -3; 6; -6}

                        => n \(\in\){0; -2; 1; -3; 2; -4; 5; -7}