$\displaystyle \begin{array}{{>{\displaystyle}l}} g) \ \frac{1}{\sqrt{2} -1} =\frac{\left(\sqrt{2} +1\right)}{\left(\sqrt{2} -1\right)\left(\sqrt{2} +1\right)} =\frac{\sqrt{2} +1}{2-1} =\sqrt{2} +1\\ h) \ \frac{\sqrt{2} +1}{\sqrt{2} -1} =\frac{\left(\sqrt{2} +1\right)^{2}}{\left(\sqrt{2} -1\right)\left(\sqrt{2} +1\right)} =\frac{\left(\sqrt{2} +1\right)^{2}}{2-1} =3+2\sqrt{2} \ \\ i) \ \frac{2-\sqrt{2}}{\sqrt{2} -1} =\frac{\sqrt{2}\left(\sqrt{2} -1\right)}{\sqrt{2} -1} =\sqrt{2}\\ j) \ \frac{1}{\sqrt{3} +\sqrt{2}} =\frac{\sqrt{3} -\sqrt{2}}{\left(\sqrt{3} +\sqrt{2}\right)\left(\sqrt{3} -\sqrt{2}\right)} =\frac{\sqrt{3} -\sqrt{2}}{3-2} =\sqrt{3} -\sqrt{2}\\ k) \ \frac{2+\sqrt{3}}{2-\sqrt{3}} =\frac{\left( 2+\sqrt{3}\right)^{2}}{\left( 2-\sqrt{3}\right)\left( 2+\sqrt{3}\right)} =\frac{7+4\sqrt{3}}{2-3} =-7-4\sqrt{3} \ \\ j)\frac{\sqrt{15} -\sqrt{6}}{\sqrt{2} -\sqrt{5}} =\frac{\sqrt{3}\left(\sqrt{5} -\sqrt{2}\right)}{-\left(\sqrt{5} -\sqrt{2}\right)} =-\sqrt{3}\\ \end{array}$
