Question

tìm n để 2-7n/n+1 có giá trị nguyên

Answer 1

Để -7n+1/n+1 là số nguyên thì

-7n-7+8 chia hết cho n+1

\(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

hay \(n\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\)

Answer 2

Ta có:\(\dfrac{2-7n}{n+1}\)

=\(\dfrac{-7n+2}{n+1}\)

=\(\dfrac{-7n+\left(-7\right)+7+2}{n+1}\)

=\(\dfrac{-7\left(n+1\right)+9}{n+1}\)

=\(-7+\dfrac{9}{n+1}\)

Để \(\dfrac{2-7n}{n+1}\)\(\) \(\in\) Z suy ra \(-7+\dfrac{9}{n+1}\)\(\in\) Z suy ra\(\dfrac{9}{n+1}thu\text{ộc}\) Z

suy ra n+1 thuộc Ư(9)={ \(\pm1\);\(\pm3\);\(\pm9\)}

suy ra n\(\in\){0;-2;2;-4;8;-10}