Question

tìm n thuộc Z để :

a> n^2-7 chia hết cho n+3

b> n+3 chia het cho n^2-7

c> n+4 chia hết cho n+1

d> 4n+3 chia hết cho n-2

e> 5n+2 chia het cho n-3

2. timf x , y thuôc X

a> (x-3)(2y+1)=6

b> xy+3x-7y=21

c> xy+2y+3x-2y=11

d> 4x-xy +2y=10

e>3y-xy-x=-2

Answer 1

a. Ta có: n²-7 \(⋮\) n+3

<=> n²-9+2 \(⋮\) n+3

<=> (n-3)(n+3)+2\(⋮\) n+3

<=> 2 \(⋮\) n+3

=> n+3\(\in\)Ư₍₂₎=\(\left\{\pm1;\pm2\right\}\)

Ta có bảng sau:

n+3	1	-1	2	-2
n	-2	-4	-1	-5

Vậy n\(\in\left\{-2;-4;-1;-5\right\}\)