Question

Tìm n nguyên biết:

1/ 2n + 7 chia hết n - 3

2/ n+2 chia hết n - 3

3/ 3n+ 2 chia hết 2n -1

4/ n.n - 7 chia hết n + 3

 

Answer 1

1/

Ta có 2n+7=2n-6+13=2(n-3)+13

Vì \(2\left(n-3\right)⋮\left(n-3\right)\)

Để \(\left[2\left(n-3\right)+13\right]⋮\left(n-3\right)\Leftrightarrow13⋮\left(n-3\right)\Leftrightarrow\left(n-3\right)\inƯ_{ }_{_{ }\left(13_{ }\right)_{ }}=\left\{\pm1;\pm13\right\}\)Ta có bảng:

n-3	-13	-1	1	13
n	-10	2	4	16

Vậy...

Answer 2

Câu 2 tt

3/3n+2 chia hếy 2n-1

Ta có \(\hept{\begin{cases}\left(3n+2\right)⋮\left(2n-1\right)\\\left(2n-1\right)⋮\left(2n-1\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2\left(3n+2\right)⋮\left(2n-1\right)\\3\left(2n-1\right)⋮\left(2n-1\right)\end{cases}}\)

\(\Rightarrow2\left(3n+2\right)-3\left(2n-1\right)⋮\left(2n-1\right)\)

\(\Rightarrow1⋮\left(2n-1\right)\)

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

Kẻ bảng như trên nhá bn

T.i.c.k cho mik

#TM