$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \frac{\sqrt{a} +a\sqrt{b} -\sqrt{b} -b\sqrt{a}}{ab-1}\\ ĐK\ a >0;b >0;\ ab\neq 1\\ =\frac{\left(\sqrt{a} -\sqrt{b}\right) -\sqrt{ab}\left(\sqrt{a} -\sqrt{b}\right)}{ab-1}\\ =-\frac{\left( 1-\sqrt{ab}\right)\left(\sqrt{a} -\sqrt{b}\right)}{(\left( 1-\sqrt{ab}\right)(\left( 1+\sqrt{ab}\right)}\\ =\frac{\sqrt{b} -\sqrt{a}}{1+\sqrt{ab}}\\ b.\ \frac{2\sqrt{15} -2\sqrt{10} +\sqrt{6} -3}{2\sqrt{5} -2\sqrt{10} -\sqrt{3} +\sqrt{6}}\\ =\ \frac{2\sqrt{5}\left(\sqrt{3} -\sqrt{2}\right) +\sqrt{3}\left(\sqrt{2} -\sqrt{3}\right)}{2\sqrt{5}\left( 1-\sqrt{2}\right) -\sqrt{3}\left( 1-\sqrt{2}\right)}\\ =\frac{\left( 2\sqrt{5} -\sqrt{3}\right)\left(\sqrt{3} -\sqrt{2}\right)}{\left( 2\sqrt{5} -\sqrt{3}\right)\left( 1-\sqrt{2}\right)}\\ =\frac{\sqrt{3} -\sqrt{2}}{1-\sqrt{2}} \end{array}$