Bài toán: Cho $P=\dfrac{4\sqrt {x}+1}{x+3},x\ge 0$Tìm $x$ để $P\in \mathbb{Z}$.Giải$\begin{aligned} P & =\dfrac{4 \sqrt{x}+1}{x+3}>0 \\ P & =\dfrac{4 \sqrt{x}+1}{x+3}=\dfrac{(-2 x+4 \sqrt{x} *-2)+2 x+3}{x+3} \\ & =\dfrac{-2(\sqrt{x}-1)^2+2 x+3}{x+3}<\dfrac{2 x+6}{x+3}=2\end{aligned}$$\Rightarrow 0<p<2$ mà $ p\in \mathbb{Z}\Rightarrow p=1$$P=1\Rightarrow 4\sqrt {x}+1=x+3\Leftrightarrow x-4\sqrt {x}+4=2\Leftrightarrow {(\sqrt {x}-2)^2}=2\Leftrightarrow x={(2\pm \sqrt {2})^2}$