Question

Tìm x nguyên để P=\(\dfrac{x-3}{\sqrt{x}+3}\) nguyên

Answer 1

\(P=\dfrac{\left(x-9\right)+6}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}\\ P=\sqrt{x}-3+\dfrac{6}{\sqrt{x}+3}\in Z\\ \Leftrightarrow6⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;3\right\}\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x\in\left\{0;9\right\}\)