$\displaystyle \begin{array}{{>{\displaystyle}l}} a) \ \left(\sqrt{3} -\sqrt{2}\right)\left(\sqrt{3} +\sqrt{2}\right) :\left(\frac{\sqrt{3}}{\sqrt{3} +\sqrt{2}} +\frac{\sqrt{2}}{\sqrt{3} -\sqrt{2}}\right)\\ =3-2:(\frac{\sqrt{3}\left(\sqrt{3} -\sqrt{2}\right) +\sqrt{2}\left(\sqrt{3} +\sqrt{2}\right)}{3-2}\\ =\frac{1}{3-\sqrt{6} +\sqrt{6} +2} =\frac{1}{5} \ \\ b) \ \frac{3+2\sqrt{3}}{\sqrt{3}} +\frac{2+\sqrt{2}}{\sqrt{2} +1} -\left(\sqrt{3} +2\right)\\ =\frac{\sqrt{3}\left(\sqrt{3} +2\right)}{\sqrt{3}} +\frac{\sqrt{2}\left(\sqrt{2} +1\right)}{\sqrt{2} +1} -\sqrt{3} -2\\ =\sqrt{3} +2+\sqrt{2} -\sqrt{3} -2\\ =\sqrt{2}\\ Bài\ 5:\ \\ a) \ \sqrt{9-x^{2}} \geqslant 0\ với\ mọi\ x\ \\ Dấu\ =\ xảy\ ra\ khi\ ( 3-x)( 3+x) =0\ \\ \rightarrow x=\pm 3\ \\ Vậy\ giá\ trị\ nhỏ\ nhất\ là\ 0\ tại\ x=\pm 3\\ b) \ 1+\sqrt{x-2}\\ Ta\ có\ :\ \sqrt{x-2} \geqslant 0\ với\ mọi\ x\ \\ \rightarrow \sqrt{x-2} +1\geqslant 1\ \\ Dấu\ bằng\ xảy\ ra\ khi\ x-2=0\ \rightarrow x=2\ \\ Vậy\ gtnn\ là\ 1\ tại\ x=2\ \\ c) \ 1-2\sqrt{1-3x} \ \\ Ta\ có\ :\ 2\sqrt{1-3x} \ \geqslant 0\ \\ \rightarrow -2\sqrt{1-3x} \leqslant 0\ \\ \rightarrow 1-2\sqrt{1-3x} \leqslant 1\ \\ Dấu\ bằng\ xảy\ ra\ khi\ 1-3x=0\ \rightarrow x=\frac{1}{3} \ \\ Vậy\ gtln\ là\ 1\ tại\ x=\frac{1}{3} \ \\ d) \ \sqrt{x} -x\ ( x >0) \ \\ \rightarrow -\left( x-\frac{1}{2} .2\sqrt{x} +\frac{1}{4}\right) +\frac{1}{4}\\ \rightarrow -\left(\sqrt{x} -\frac{1}{2}\right)^{2} +\frac{1}{4} \ \\ Ta\ có\ :\ \left(\sqrt{x} -\frac{1}{2}\right)^{2} \geqslant 0\ \\ \rightarrow -\left(\sqrt{x} -\frac{1}{2}\right)^{2} \leqslant 0\ \\ \rightarrow -\left(\sqrt{x} -\frac{1}{2}\right)^{2} +\frac{1}{4} \leqslant \frac{1}{4} \ \\ Dấu\ bằng\ xảy\ ra\ khi\ \sqrt{x} -\frac{1}{2} =0\ \rightarrow x=\frac{1}{4} \ \\ Vậy\ Giá\ trị\ lớn\ nhất\ là\ \frac{1}{4} \ tại\ x=\frac{1}{4} \ \\ e) \ \sqrt{2x^{2} -2x+5}\\ =\sqrt{2\left( x^{2} -2.\frac{1}{2} x+\frac{1}{4}\right) +\frac{9}{2}}\\ =\sqrt{2\left( x-\frac{1}{2}\right)^{2} +\frac{9}{2}} \ \\ Ta\ có:\ \sqrt{2\left( x-\frac{1}{2}\right)^{2}} \geqslant 0\ \\ \rightarrow \sqrt{2\left( x-\frac{1}{2}\right)^{2} +\frac{9}{2}} \geqslant \sqrt{\frac{9}{2} \ }\\ Dấu\ bằng\ xảy\ ra\ tại\ x-\frac{1}{2} =0\ \rightarrow x=\frac{1}{2} \ \\ Vậy\ min=\frac{3}{\sqrt{2} \ } \ tại\ x=\frac{1}{2} \end{array}$