Question

tìm n thuộc Z sao cho:n+3 chia hết n^2-7

Answer 1

Answer 2

Ta có: \(\left(n+3\right)⋮n^2-7\)

\(\Leftrightarrow\left(n+3\right)\left(n-3\right)⋮n^2-7\)

\(\Leftrightarrow n^2-9⋮n^2-7\)

\(\Leftrightarrow n^2-7-2⋮n^2-7\)

mà \(n^2-7⋮n^2-7\)

nên \(-2⋮n^2-7\)

\(\Leftrightarrow n^2-7\inƯ\left(-2\right)\)

\(\Leftrightarrow n^2-7\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow n^2\in\left\{8;6;9;5\right\}\)

\(\Leftrightarrow n\in\left\{2\sqrt{2};-2\sqrt{2};\sqrt{6};-\sqrt{6};3;-3;\sqrt{5};-\sqrt{5}\right\}\)

mà \(n\in Z\)

nên \(n\in\left\{3;-3\right\}\)

Vậy: Để \(\left(n+3\right)⋮n^2-7\) thì \(n\in\left\{3;-3\right\}\)