Question

tìm n nguyên để b  có giá trị là stn b= 2n+2/n+2 +5n+17/n+2 -3n/n-2

Answer 1

\(b=\dfrac{2n+2}{n+2}+\dfrac{5n+17}{n+2}-\dfrac{3n}{n-2}\)

\(b=\dfrac{7n+19}{n+2}-\dfrac{3n}{n-2}\)

\(b=\dfrac{7\left(n+2\right)+5}{n+2}-\dfrac{3\left(n-2\right)+6}{n-2}\)

\(b=7+\dfrac{5}{n+2}-3-\dfrac{6}{n-2}\)

để b là STN thì \(\left\{{}\begin{matrix}n+2\inƯ\left(5\right)\\n-2\inƯ\left(6\right)\end{matrix}\right.\)      \(\Rightarrow\left\{{}\begin{matrix}n+2\in\left\{1;5\right\}\\n-2\in\left\{1;2;3;6\right\}\end{matrix}\right.\)

           \(\Rightarrow\left\{{}\begin{matrix}n\in\left\{-1;3\right\}\\n\in\left\{3;4;5;8\right\}\end{matrix}\right.\)     => n = 3 thỏa mãn

vậy n=3