$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ A=\left(\frac{x+2}{x\sqrt{x} -1} +\frac{\sqrt{x}}{x+\sqrt{x} +1} +\frac{1}{1-\sqrt{x}}\right) :\frac{\sqrt{x} -1}{2}\\ A=\left(\frac{x+2}{\left(\sqrt{x} -1\right)\left( x+\sqrt{x} +1\right)} +\frac{\sqrt{x}}{x+\sqrt{x} +1} -\frac{1}{\sqrt{x} -1}\right) .\frac{2}{\sqrt{x} -1}\\ A=\frac{x+2+\sqrt{x}\left(\sqrt{x} -1\right) -\left( x+\sqrt{x} +1\right)}{\left(\sqrt{x} -1\right)\left( x+\sqrt{x} +1\right)} .\frac{2}{\sqrt{x} -1}\\ A=\frac{x+2+x-\sqrt{x} -x-\sqrt{x} -1)}{\left(\sqrt{x} -1\right)\left( x+\sqrt{x} +1\right)} .\frac{2}{\sqrt{x} -1}\\ A=\frac{x-2\sqrt{x} +1}{\left(\sqrt{x} -1\right)\left( x+\sqrt{x} +1\right)} .\frac{2}{\sqrt{x} -1}\\ A=\frac{\left(\sqrt{x} -1\right)^{2}}{\left(\sqrt{x} -1\right)\left( x+\sqrt{x} +1\right)} .\frac{2}{\sqrt{x} -1}\\ A=\frac{2}{x+\sqrt{x} +1}\\ Vậy\ A=\frac{2}{x+\sqrt{x} +1} \ với\ x\geqslant 0,\ x\neq 1\\ b.\ Ta\ có\ x+\sqrt{x} +1=\left(\sqrt{x} +\frac{1}{2}\right)^{2} +\frac{3}{4} >0\ \forall x\neq 1\\ \Rightarrow A >0\ \forall x\neq 1\\ \\ \end{array}$
