Question

Tìm n \(\in\) Z: n²+4n-8 chia hết cho n+3

Answer 1

n²+4n-8 chia hết cho n+3

=>n²+3n+n+3-11 chia hết cho n+3

=>n(n+3)+(n+3)-11 chia hết cho n+3

=>(n+1)(n+3)-11 chia hết cho n+3

Mà (n+1)(n+3) chia hết cho n+3

=>11 chia hết cho n+3

=>n+3\(\inƯ\left(11\right)\)

=>n+3\(\in\left\{-11;-1;1;11\right\}\)

=>n\(\in\left\{-14;-4;-2;8\right\}\)

Answer 2

n²+4n-8 chia hết cho n+3

=>n²+3n+n+3-11 chia hết cho n+3

=>n(n+3)+(n+3)-11 chia hết cho n+3

=>(n+1)(n+3)-11 chia hết cho n+3

Mà (n+1)(n+3) chia hết cho n+3

=>11 chia hết cho n+3

=>n+3$\inƯ\left(11\right)$∈Ư(11)

=>n+3$\in\left\{-11;-1;1;11\right\}$∈{−11;−1;1;11}

=>n$\in\left\{-14;-4;-2;8\right\}$