Question

tìm số nguyên a : \(\dfrac{2a+9}{a+3}-\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}\) là số nguyên

Answer 1

Có điều kiện a ko b?

Ta có: \(\dfrac{2a+9}{a+3}-\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}=\dfrac{2a+9-5a+17-3a}{a+3}=\dfrac{-6a-8}{a+3}\)

\(=\dfrac{-6a-18+10}{a+3}=\dfrac{-6\left(a+3\right)+10}{a+3}=\dfrac{-6\left(a+3\right)}{a+3}+\dfrac{10}{a+3}=-6+\dfrac{10}{a+3}\)

Để \(\dfrac{2a+9}{a+3}-\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}\) là số nguyên

\(\Leftrightarrow\left(a+3\right)\in U\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Leftrightarrow a\in\left\{-13;-8;-5;-4;-2;-1;2;7\right\}\)