Question

tìm số nguyên a sao cho: 3a+11chia hết cho a+2 

Answer 1

3a+11=\(3a+6+5=3\left(a+1\right)+5\)

Để 3a+11 \(⋮\left(a+2\right)\Leftrightarrow3\left(a+2\right)+5⋮\left(a+2\right)\Leftrightarrow5⋮\left(a+2\right)\Rightarrow a+2\in\left\{-5;-1;1;5\right\}\Rightarrow a\in\left\{-7;-3;-1;3\right\}\)

Answer 2

Ta có: \(3a+11⋮a+2\)

<=> \(3\left(a+2\right)+5⋮a+2\)

<=> \(5⋮a+2\)

Vì a nguyên => \(a+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng sau: 

a+2	1	-1	5	-5
a	-1	-3	3	-7

Vậy \(a\in\left\{-1;-3;3;-7\right\}\)