\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right).\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x+2}{\sqrt{x}+1}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x+\sqrt{x}-x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
Để P âm \(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Mà \(\sqrt{x}+2>0\forall x\Rightarrow\sqrt{x}-1< 0\Rightarrow x< 1\)
Để \(P\in Z\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}\in Z\)
\(\Rightarrow1-\frac{3}{\sqrt{x}+2}\in Z\Rightarrow\frac{3}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ_3\)
Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=3\Rightarrow x=1\)
Mà để \(P\in Z^-\Rightarrow\hept{\begin{cases}x< 1\\x=1\end{cases}}\)\(\Rightarrow x\in\varnothing\)
Vậy không có giá trị nào của x để P nguyên âm
