$\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{1}{2}\sqrt{48} -2\sqrt{75} -\frac{\sqrt{33}}{\sqrt{11}} +5\sqrt{1\frac{1}{3}} \ \\ =\frac{1}{2} .4\sqrt{3} -2.5\sqrt{3} -\sqrt{\frac{33}{11}} +5\sqrt{\frac{4}{3}} \ \\ =2\sqrt{3} -10\sqrt{3} -\sqrt{3} +\frac{10}{\sqrt{3}} \ \\ =-11\sqrt{3} +\frac{10}{\sqrt{3}} =\frac{-33+10}{\sqrt{3}} =\frac{-23}{\sqrt{3}} =\frac{-23\sqrt{3}}{3} \ \\ d)\sqrt{9-4\sqrt{5}} -\sqrt{9+\sqrt{80}}\\ =\sqrt{4-2.2\sqrt{5} +5} -\sqrt{9+4\sqrt{5}}\\ =\sqrt{\left( 2-\sqrt{5}\right)^{2}} -\sqrt{4+2.2\sqrt{5} +5} \ \\ =|2-\sqrt{5} |-|2+\sqrt{5} |\\ =\sqrt{5} -2-2-\sqrt{5} =-4\ \\ e) \ 2\sqrt{3}\left(\sqrt{27} +2\sqrt{48} -\sqrt{75}\right) \ \\ =2\sqrt{3}\left( 3\sqrt{3} +2.4\sqrt{3} -5\sqrt{3}\right)\\ =2\sqrt{3}\left( 6\sqrt{3}\right) =12.3=36\ \\ f)\sqrt{3-2\sqrt{2}} -\sqrt{6-4\sqrt{2}}\\ =\sqrt{2-2\sqrt{2} +1} -\sqrt{4-2.2\sqrt{2} +2} \ \\ =\sqrt{\left(\sqrt{2} -1\right)^{2}} -\sqrt{\left( 2-\sqrt{2}\right)^{2}}\\ =|\sqrt{2} -1|-|2-\sqrt{2} |\\ =\sqrt{2} -1-2+\sqrt{2} =-\left( 3-2\sqrt{2}\right) =-\left( 1-\sqrt{2}\right)^{2} \ \\ Bài\ 2:\ \\ a)\frac{1}{\sqrt{2} +1} +\frac{1}{\sqrt{3} +\sqrt{2}} +\frac{1}{\sqrt{3} +\sqrt{4}} \ \\ =\frac{\sqrt{2} -1}{2-1} +\frac{\sqrt{3} -\sqrt{2}}{3-2} +\frac{\sqrt{4} -\sqrt{3}}{4-3} \ \\ =\sqrt{2} -1+\sqrt{3} -\sqrt{2} +\sqrt{4} -\sqrt{3} \ \\ =2-1=1\ \\ b)\frac{8+2\sqrt{2}}{3-\sqrt{2}} -\frac{2+3\sqrt{2}}{\sqrt{2}} +\frac{\sqrt{2}}{1-\sqrt{2}} \ \\ =\frac{2\sqrt{2} +2\sqrt{2}}{3-\sqrt{2}} -\frac{\sqrt{2}\left(\sqrt{2} +3\right)}{\sqrt{2}} +\frac{\sqrt{2}}{1-\sqrt{2}}\\ =\frac{4\sqrt{2}}{3-\sqrt{2}} -\left(\sqrt{2} +3\right) -\frac{\sqrt{2}\left(\sqrt{2} +1\right)}{2-1} \ \\ =\frac{4\sqrt{2}\left(\sqrt{2} +3\right)}{3-1} -\sqrt{2} -3-2-\sqrt{2}\\ =8+12\sqrt{2} -2\sqrt{2} -5=3-11\sqrt{2} \ \\ \end{array}$