\(10,\\ a,=\dfrac{\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(\sqrt{2}-\sqrt{3}\right)^2}=\dfrac{\sqrt{2}-2+\sqrt{6}}{1-5+2\sqrt{6}}=\dfrac{\sqrt{2}-2+\sqrt{6}}{2\sqrt{6}-4}\\ =\dfrac{1-\sqrt{2}+\sqrt{3}}{2\sqrt{3}-2\sqrt{2}}=\dfrac{\left(1-\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{3}+2\sqrt{2}\right)}{4}\\ =\dfrac{2\sqrt{3}+2\sqrt{2}-2\sqrt{6}-4+6+2\sqrt{6}}{4}\\ =\dfrac{2\sqrt{3}+2\sqrt{2}+2}{4}=\dfrac{\sqrt{3}+\sqrt{2}+1}{4}\)
\(b,=\dfrac{1}{3\sqrt{2}+2\sqrt{2}-2\sqrt{2}}=\dfrac{1}{4\sqrt{2}}=\dfrac{\sqrt{2}}{8}\\ c,=\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{5}}{3-\left(\sqrt{2}-\sqrt{5}\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{5}}{3-7+2\sqrt{10}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{2}}{2\sqrt{10}-4}=\dfrac{\left(\sqrt{5}+\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{10}+4\right)}{24}\\ =\dfrac{10\sqrt{2}+4\sqrt{5}+2\sqrt{30}+4\sqrt{3}-4\sqrt{5}-4\sqrt{2}}{24}\\ =\dfrac{6\sqrt{2}+2\sqrt{30}+4\sqrt{3}}{24}=\dfrac{3\sqrt{2}+\sqrt{30}+2\sqrt{3}}{12}\)
