\(P=\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}+3\left(1-\sqrt{x}\right).\)
\(=\frac{\sqrt{x^3}-2^3}{x+2\sqrt{x}+4}+3-3\sqrt{x}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}+3-3\sqrt{x}\)
\(=\sqrt{x}-2+3-3\sqrt{x}=-2\sqrt{x}+1\)
\(Q=\frac{2P}{1-P}=\frac{2\left(-2\sqrt{x}+1\right)}{1-\left(-2\sqrt{x}+1\right)}\)
\(=\frac{-4\sqrt{x}+2}{1+2\sqrt{x}-1}=\frac{-2\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{-2\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=-2+\frac{1}{\sqrt{x}}\)
\(Q\in Z\Leftrightarrow-2+\frac{1}{\sqrt{x}}\in Z\Rightarrow\frac{1}{\sqrt{x}}\in Z\)
\(\Rightarrow1\)\(⋮\)\(\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ_1\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=1\\\sqrt{x}=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x\in\varnothing\end{cases}}}\)
Vậy \(Q\in Z\Leftrightarrow x=1\)
