Question

Cho P = \(\dfrac{4\sqrt{x}}{x+2}\)

Tìm \(x\in R\) để \(P\in Z\)

Answer 1

ĐKXĐ: \(x\ge0\)

\(\left\{{}\begin{matrix}4\sqrt{x}\ge0\\x+2>0\end{matrix}\right.\) \(\Rightarrow P=\dfrac{4\sqrt{x}}{x+2}\ge0\)

\(P-2=\dfrac{4\sqrt{x}}{x+2}-2=\dfrac{4\sqrt{x}-2x-4}{x+2}=\dfrac{-x-\left(\sqrt{x}-2\right)^2}{x+2}< 0\)

\(\Rightarrow P< 2\Rightarrow0\le P< 2\)

Mà \(P\in Z\Rightarrow\left[{}\begin{matrix}P=0\\P=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{4\sqrt{x}}{x+2}=0\\\dfrac{4\sqrt{x}}{x+2}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\x-4\sqrt{x}+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=2+\sqrt{2}\\\sqrt{x}=2-\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6+4\sqrt{2}\\x=6-4\sqrt{2}\end{matrix}\right.\)