$\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{5}{2\sqrt{7}} \ =\frac{5\sqrt{7}}{2.\left(\sqrt{7}\right)^{2}} =\frac{5\sqrt{7}}{2.7} =\frac{5\sqrt{7}}{14} \ \\ b) \ \frac{1}{2+\sqrt{3}} =\frac{2-\sqrt{3}}{\left( 2+\sqrt{3}\right)\left( 2-\sqrt{3}\right)} =\frac{2-\sqrt{3}}{4-3} =2-\sqrt{3}\\ c)\frac{35}{2\sqrt{2} -1} =\frac{35\left( 2\sqrt{2} +1\right)}{\left( 2\sqrt{2} -1\right)\left( 2\sqrt{2} +1\right)}\\ =\frac{35\left( 2\sqrt{2} +1\right)}{8-1} =5\left( 2\sqrt{2} +1\right) \ \\ d) \ \frac{\sqrt{3} -\sqrt{5}}{\sqrt{3} +\sqrt{5}} =\frac{\left(\sqrt{3} -\sqrt{5}\right)^{2}}{\left(\sqrt{3} -\sqrt{5}\right)\left(\sqrt{3} +\sqrt{5}\right)}\\ =\frac{3-2\sqrt{15} +5}{3-5} =\frac{8-2\sqrt{15}}{-2}\\ =\frac{-2\left( 4-\sqrt{15}\right)}{2} =\sqrt{15} -4\\ VD3:\ \\ a) \ \frac{3+\sqrt{3}}{5\sqrt{3}} =\frac{\sqrt{3}\left(\sqrt{3} +1\right)}{5\sqrt{3}} =\frac{\sqrt{3} +1}{5} \ \\ b) \ \frac{2+\sqrt{2}}{\sqrt{2} +1} =\frac{\sqrt{2}\left(\sqrt{2} +1\right)}{\sqrt{2} +1} =\sqrt{2} \ \\ c) \ \frac{\sqrt{15} -\sqrt{5}}{\sqrt{3} -1} =\frac{\sqrt{5}\left(\sqrt{3} -1\right)}{\sqrt{3} -1} =\sqrt{5} \ \\ d) \ \frac{2-\sqrt{10}}{\sqrt{5} -\sqrt{2}} =\frac{-\sqrt{2}\left(\sqrt{5} -\sqrt{2}\right)}{\sqrt{5} -\sqrt{2}} =-\sqrt{2} \ \end{array}$
