$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 13:\\ a.\ \frac{\sqrt{6}}{3}\\ b.\ \frac{x}{\sqrt{5}} =\frac{x\sqrt{5}}{5}\\ c.\ \frac{a\sqrt{5a}}{7\sqrt{b}} =\frac{a\sqrt{5ab}}{7b}\\ d.\ Đề\ \ sai\ vì\ x< 0;\ y >0\ thì\ xy< 0\\ \Rightarrow \frac{3}{xy} < 0\Rightarrow \sqrt{\frac{3}{xy}} \ không\ có\ nghĩa\\ Bài\ 14:\\ \\ a.=\frac{\left( 10+2\sqrt{10}\right)\left(\sqrt{5} -\sqrt{2}\right)}{\left(\sqrt{5} -\sqrt{2}\right)\left(\sqrt{5} +\sqrt{2}\right)} =\frac{6\sqrt{5}}{5-2} =2\sqrt{5}\\ b.=\frac{\left( 2\sqrt{8} -\sqrt{12}\right)\left(\sqrt{18} -\sqrt{48}\right)}{\left(\sqrt{18} -\sqrt{48}\right)\left(\sqrt{18} +\sqrt{48}\right)} =\frac{48-22\sqrt{6}}{18-48} =\frac{48-22\sqrt{6}}{-30}\\ c.\ =\frac{\sqrt{2}\left(\sqrt{5} +\sqrt{3}\right)}{\left(\sqrt{5} +\sqrt{3}\right)\left(\sqrt{5} -\sqrt{3}\right)} =\frac{\sqrt{2}\left(\sqrt{5} +\sqrt{3}\right)}{2}\\ d.\ \sqrt{\frac{\left( 2-\sqrt{3}\right)^{2}}{4-3}} =\ \sqrt{\frac{\left( 2-\sqrt{3}\right)^{2}}{1}} =2-\sqrt{3}\\ Bài\ 15:\\ a.\ \left[\frac{15\left(\sqrt{6} -1\right)}{\left(\sqrt{6} -1\right)\left(\sqrt{6} +1\right)} +\frac{4\left(\sqrt{6} +2\right)}{\left(\sqrt{6} +2\right)\left(\sqrt{6} -2\right)} -\frac{12\left(\sqrt{6} +3\right)}{\left( 3-\sqrt{6}\right)\left( 3+\sqrt{6}\right)}\right]\left(\sqrt{6} +11\right)\\ =\left[\frac{15\sqrt{6} -15}{6-1} +\frac{4\sqrt{6} +8}{6-4} -\frac{12\sqrt{6} +36}{9-6}\right]\left(\sqrt{6} +11\right)\\ =\left[\frac{15\sqrt{6} -15}{5} +\frac{4\sqrt{6} +8}{2} -\frac{12\sqrt{6} +36}{3}\right]\left(\sqrt{6} +11\right)\\ =\left(\sqrt{6} -11\right)\left(\sqrt{6} +11\right)\\ =6-121=-115\\ b.\ \frac{3+\sqrt{5}}{9-5} -\frac{\sqrt{5} +1}{5-1} =\ \frac{3+\sqrt{5}}{4} -\frac{\sqrt{5} +1}{4} =\frac{2}{4} =\frac{1}{2}\\ c.\ =\frac{\sqrt{5} -1}{\left(\sqrt{5} +1\right)\left(\sqrt{5} -1\right)} +\frac{\sqrt{5} +2}{\left(\sqrt{5} +2\right)\left(\sqrt{5} -2\right)} -\frac{\sqrt{5} +3}{\left( 3-\sqrt{5}\right)\left(\sqrt{5} +3\right)} -\sqrt{5}\\ =\frac{\sqrt{5} -1}{5-1} +\frac{\sqrt{5} +2}{5-4} -\frac{\sqrt{5} +3}{9-5} -\sqrt{5}\\ =\frac{\sqrt{5} -1}{4} +\frac{\sqrt{5} +2}{1} -\frac{\sqrt{5} +3}{4} -\sqrt{5}\\ =\frac{-4}{4} +2=-1+2=1\\ d.\ =\frac{5-3\sqrt{2}}{\left( 5+3\sqrt{2}\right)\left( 5-3\sqrt{2}\right)} +\frac{5+3\sqrt{2}}{\left( 5+3\sqrt{2}\right)\left( 5-3\sqrt{2}\right)}\\ =\frac{5-3\sqrt{2} +5+3\sqrt{2}}{25-18} =\frac{10}{7}\\ Bài\ 16:\\ a.\ \frac{\sqrt{2}\sqrt{7}}{7}\\ b.\ \frac{\sqrt{31x^{2}}}{31}\\ c.\frac{b\sqrt{5b}}{7\sqrt{a}} =\frac{b\sqrt{5ab}}{7a}\\ d.\ -7xy\frac{4}{\sqrt{xy}} =-28\sqrt{xy}\\ Bài\ 17:\\ a.=\frac{\left( 5+2\sqrt{5}\right)\left(\sqrt{5} -\sqrt{2}\right)}{\left(\sqrt{5} -\sqrt{2}\right)\left(\sqrt{5} +\sqrt{2}\right)} =\frac{5\sqrt{5} +10-5\sqrt{2} -4\sqrt{10}}{5-2}\\ =\frac{5\sqrt{5} +10-5\sqrt{2} -4\sqrt{10}}{3}\\ b.=\frac{\left( 2\sqrt{6} -\sqrt{10}\right)\left( 4\sqrt{3} +2\sqrt{5}\right)}{\left( 4\sqrt{3} +2\sqrt{5}\right)\left( 4\sqrt{3} -2\sqrt{5}\right)} =\frac{14\sqrt{2}}{48-20} =\frac{\sqrt{2}}{2}\\ c.\ =\frac{\left( 2\sqrt{2} +3\sqrt{3}\right)}{\left( 2\sqrt{2} +3\sqrt{3}\right)\left( 2\sqrt{2} -3\sqrt{3}\right)} =\frac{\left( 2\sqrt{2} +3\sqrt{3}\right)}{8-27} =\frac{\left( 2\sqrt{2} +3\sqrt{3}\right)}{-19}\\ d.\ \sqrt{\frac{\left( 3-\sqrt{5}\right)^{2}}{9-5}} =\ \ \sqrt{\frac{\left( 3-\sqrt{5}\right)^{2}}{4}} =\frac{3-\sqrt{5}}{2}\\ Bài\ 18:\\ a.\frac{\left( 3+2\sqrt{3}\right)\sqrt{3}}{3} +\frac{\left( 2+\sqrt{2}\right)\left(\sqrt{2} -1\right)}{2-1} -\left( 2+\sqrt{3}\right)\\ =\frac{3\sqrt{3} +6}{3} +\sqrt{2} -\left( 2+\sqrt{3}\right)\\ =\sqrt{3} +2+\sqrt{2} -\left( 2+\sqrt{3}\right)\\ =\sqrt{2}\\ b.\ \frac{1+\sqrt{5} -5-\sqrt{5}}{1+\sqrt{5}} .\frac{5-\sqrt{5} -1+\sqrt{5}}{1-\sqrt{5}} =\frac{( -4) .6}{1-5} =6\\ c.\ \frac{5-2\sqrt{5} -4+2\sqrt{5}}{2-\sqrt{5}} .\frac{5+3\sqrt{5} -6-3\sqrt{5}}{3+\sqrt{5}} =\frac{1.( -1)}{1-\sqrt{5}} =\frac{\sqrt{5} +1}{4}\\ d.\frac{3\left(\sqrt{5} +\sqrt{2}\right)}{\left(\sqrt{5} +\sqrt{2}\right)\left(\sqrt{5} -\sqrt{2}\right)} -\frac{2\left( 2+\sqrt{2}\right)}{\left( 2-\sqrt{2}\right)\left( 2+\sqrt{2}\right)} +\frac{\sqrt{3} -\sqrt{2}}{\left(\sqrt{3} -\sqrt{2}\right)\left(\sqrt{3} +\sqrt{2}\right)}\\ =\frac{3\left(\sqrt{5} +\sqrt{2}\right)}{3} -\frac{2\left( 2+\sqrt{2}\right)}{2} +\frac{\sqrt{3} -\sqrt{2}}{1}\\ =\sqrt{5} +\sqrt{2} -2-\sqrt{2} +\sqrt{3} -\sqrt{2}\\ =\sqrt{5} -2 +\sqrt{3} -\sqrt{2}\\ \end{array}$