Lời giải:
\(P=\frac{x+2}{(\sqrt{x})^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(\frac{x+2}{\sqrt{x^3}-1}+\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\sqrt{x^3}-1}+\frac{x-1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}=\frac{2x+1}{\sqrt{x^3}-1}-\frac{x+\sqrt{x}+1}{\sqrt{x^3}-1}\)
\(=\frac{2x+1-(x+\sqrt{x})}{\sqrt{x^3}-1}=\frac{x-\sqrt{x}}{\sqrt{x^3}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(P-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{2\sqrt{x}-(x+1)}{3(x+\sqrt{x}+1)}\)
\(=\frac{-(\sqrt{x}-1)^2}{3(x+\sqrt{x}+1)}\)
Với \(x\neq 1, x\geq 0\Rightarrow -(\sqrt{x}-1)^2< 0; x+\sqrt{x}+1>0\)
Do đó: \(P-\frac{1}{3}< 0\Rightarrow P< \frac{1}{3}\)
