Question

Tìm n\(\varepsilon\)Z để n\(^2\)+13n chia hết n+3

Answer 1

\(n^2+13n=n^2+6n+7n+9-9=\left(n^2+6n+9\right)+\left(7n-9\right)\)

\(=\left(n^2+3n+3n+9\right)+\left(7n-9\right)=\left[n\left(n+3\right)+3\left(n+3\right)\right]+\left(7n-9\right)=\left(n+3\right)^2+\left(7n-9\right)\)

Mà (n+3)² chia hết cho n+3

=>7n-9 chia hết cho n+3

=>7(n+3)-30 chia hết cho n+3

=>-30 chia hết cho n+3 (vì 7(n+3) chia hết cho n+3))

=>n+3 \(\in\) Ư(-30)={-30;-15;-10;-6;-5;-3;-2;-1;;1;2;3;5;6;10;15;30}

=>n \(\in\) {-33;-18;-13;-9;.......27}

Vậy..............

Answer 2

n²+13n chia hết cho n+3

=>n²+3n+10n+30-30 chia hết cho n+3

=>n.(n+3)+10.(n+3)-30 chia hết cho n+3

=>(n+10).(n+3)-30 chia hết cho n+3

Mà (n+10).(n+3) chia hết cho n+3

=>30 chia hết cho n+3

=>n+3\(\in\){-30;-15;-10;-6;-5;-3;-2;-1;1;2;3;5;6;10;15;30}

=>n\(\in\){-33;-18;-13;-9;-8;-6;-5;-4;-2;-1;0;2;3;7;12;27}

Answer 3

n²+13n chia hết cho n+3

=>n²+3n+10n+30-30 chia hết cho n+3

=>n.(n+3)+10.(n+3)-30 chia hết cho n+3

=>(n+10).(n+3)-30 chia hết cho n+3

Mà (n+10).(n+3) chia hết cho n+3

=>30 chia hết cho n+3

=>n+3$\in$

{-30;-15;-10;-6;-5;-3;-2;-1;1;2;3;5;6;10;15;30}

=>n$\in$

{-33;-18;-13;-9;-8;-6;-5;-4;-2;-1;0;2;3;7;12;27}