Question

Bài 1, Tìm số nguyên n, sao cho:

a)\(\left(n+5\right)⋮\left(n-4\right)\)

b) \(2n⋮\left(n-1\right)\)

c) \(\left(3n-8\right)⋮\left(n-4\right)\)

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

Answer 1

b)

Để \(2n⋮\left(n-1\right)\)

\(\Rightarrow2.\left(n-1\right)+2⋮\left(n-1\right)\)

\(\Rightarrow2⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)\inƯ\left(2\right)=\left\{1;2\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\end{matrix}\right.\)

Vậy n=2;n=3 thì \(2n⋮\left(n-1\right)\)

Answer 2

c)

Để \(\left(3n-8\right)⋮\left(n-4\right)\)

\(\Rightarrow3.\left(n-4\right)+4⋮\left(n-4\right)\)

\(\Rightarrow4⋮\left(n-4\right)\)

\(\Rightarrow\left(n-4\right)\inƯ\left(4\right)=\left\{1;2;4\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n-4=1\Rightarrow n=5\\n-4=2\Rightarrow n=6\\n-4=4\Rightarrow n=8\end{matrix}\right.\)

Vậy với .....................

Answer 3

d)

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

\(\Rightarrow2.\left(n-5\right)+11⋮\left(n-5\right)\)

\(\Rightarrow11⋮\left(n-5\right)\)

\(\Rightarrow\left(n-5\right)\inƯ\left(11\right)=\left\{1;11\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n-5=1\Rightarrow n=6\\n-5=11\Rightarrow n=16\end{matrix}\right.\)

Vậy với ........................................

Answer 4

a)

Để \(\left(n+5\right)⋮\left(n-4\right)\)

\(\Rightarrow\left(n-4\right)+9⋮\left(n-4\right)\)

\(\Rightarrow9⋮\left(n-4\right)\)

\(\Rightarrow\left(n-4\right)\inƯ\left(9\right)=\left\{1;3;9\right\}\)

Ta có bảng :

n-4	1	3	9
n	5	7	13

Vậy n=5;n=7;n=13 thì \(\left(n+5\right)⋮\left(n-4\right)\)

Answer 5

Toshiro Kiyoshi , Trần Đăng Nhất , Hồng Phúc Nguyễn giúp mik với các bạn !