$\displaystyle \begin{array}{{>{\displaystyle}l}} P=\frac{\sqrt{x} +2}{\sqrt{x} -1}\\ a) \ Với\ x=16\ \rightarrow \sqrt{x} =4\ \\ do\ đó\ :\ P=\frac{4+2}{4-1} =\frac{6}{3} =2\\ b) \ x=3+2\sqrt{2} \ \\ \rightarrow x=1+2\sqrt{2} +2\\ \rightarrow x=\left( 1+\sqrt{2}\right)^{2}\\ \rightarrow \sqrt{x} =1+\sqrt{2} \ \\ P=\frac{1+\sqrt{2} +2}{1+\sqrt{2} -1} =\frac{3+\sqrt{2}}{\sqrt{2}} =\frac{3\sqrt{2} +2}{2}\\ P=\frac{\sqrt{x} +2}{\sqrt{x} -1} \ ( \ DK\ :x\#1;x\geqslant 0)\\ a) \ P=2\ \rightarrow \frac{\sqrt{x} +2}{\sqrt{x} -1} =2\ \\ \rightarrow \sqrt{x} +2=2\left(\sqrt{x} -1\right)\\ \rightarrow \sqrt{x} +2=2\sqrt{x} -2\\ \rightarrow \sqrt{x} =4\\ \rightarrow x=16\ ( tm)\\ b) \ Ta\ có\ :\ \sqrt{x} \geqslant 0\ \rightarrow \sqrt{x} +2 >0\ với\ mọi\ x\ \\ Để\ P >0\ thì\ \sqrt{x} -1 >0\ \\ \rightarrow x >1\ \\ c) \ P< 1\ \rightarrow P-1< 0\ \\ do\ đó\ :\ \frac{\sqrt{x} +2}{\sqrt{x} -1} -1< 0\ \\ \rightarrow \frac{\sqrt{x} +2-\sqrt{x} +1}{\sqrt{x} -1} < 0\ \\ \rightarrow \frac{3}{\sqrt{x} -1} < 0\ \\ Để\ P-1< 0\ thì\ \sqrt{x} -1< 0\ \\ \rightarrow x< 1\ \\ Vậy......................\ \\ \\ \\ \ \end{array}$
