a) $(\frac{x+2}{x\sqrt{x}-1}+$ $\frac{\sqrt{x}}{x+\sqrt{x}+1}+$ $\frac{1}{1-\sqrt{x}}):$ $\frac{\sqrt{x}-1}{2}$
=$(\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+$ $\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+$ $\frac{x+\sqrt{x}+1}{(1-\sqrt{x})(x+\sqrt{x}+1)}).$ $\frac{2}{\sqrt{x}-1}$
= $\frac{x+2+\sqrt{x}(\sqrt{x}-1)-(x+\sqrt{x}+1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{2}{\sqrt{x}-1}$
=$\frac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{2}{\sqrt{x}-1}$
=$\frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{2}{\sqrt{x}-1}$
=$\frac{2}{(x+\sqrt{x}+1)}$
b)
Với x≥0, $x\neq1$ thì P xác định
Ta có: $x+\sqrt{x}+1=(\sqrt{x}+\frac{1}{2} )^2+\frac{3}{4}≥0+\frac{3}{4}>0$
Vậy P=$\frac{2}{(x+\sqrt{x}+1)}>0$
