Question

Tìm số nguyên n, biết:

(2n+3) chia hết cho(n-3)

Answer 1

Ta có:

2n + 3 = 2n - 6 + 9 = 2(n - 6) + 9 \(⋮\) n - 3

=> 9 \(⋮\) n - 3 => n - 3 \(\in\) Ư (9) = {\(\pm\)1; \(\pm\)9}

=> n \(\in\) {-2; -4; 6; -12}

Answer 2

Lộn:

Ta có:

\(2n+3=2n-6+9=2\left(n-3\right)+9⋮n-3\)

\(\Rightarrow9⋮n-3\Rightarrow n-3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

\(\Rightarrow n\in\left\{4;2;6;0;12;-6\right\}\)

Answer 3

Vì : \(2n+3⋮n-3\) ; \(n-3⋮n-3\)

\(\Rightarrow2\left(n-3\right)⋮n-3\Rightarrow2n-6⋮n-3\)

\(\Rightarrow\left(2n+3\right)-\left(2n-6\right)⋮n-3\)

\(\Rightarrow2n+3-2n+6⋮n-3\Rightarrow9⋮n-3\)

\(\Rightarrow n-3\in\left\{\pm1;\pm3;\pm9\right\}\)

+) Nếu : n - 3 = 1 => n = 1 + 3 = 4

n - 3 = -1 => n = -1 + 3 = 2

n - 3 = 3 => n = 3 + 3 = 6

n - 3 = -3 => n = -3 + 3 = 0

n - 3 = 9 => n = 9 + 3 = 12

n - 3 = -9 => n = -9 + 3 = -6

Vậy \(n\in\left\{4;2;6;0;12;-6\right\}\)