a. A = \(\dfrac{1}{\sqrt{3} -2}\) - \(\dfrac{\sqrt{12}-\sqrt{15}}{\sqrt{5}-2}\) = \(\dfrac{\sqrt{3} +2}{(\sqrt{3} -2)(\sqrt{3}+2)}\)- \(\dfrac{(\sqrt{12}-\sqrt{15})(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}\)= \(\dfrac{\sqrt{3} +2}{3-4}\) - \(\dfrac{-\sqrt{3}}{5-4}\)= \(\dfrac{\sqrt{3} +2}{-1}\) - \(\dfrac{-\sqrt{3}}{1 }\) = \(-\sqrt{3} - 2 + \sqrt{3}\) = -2
b. B = \(\dfrac{1}{\sqrt{3}-2}\) - \(\dfrac{1}{\sqrt{3}+2}\)= \(\dfrac{(\sqrt{3}+2)-(\sqrt{3}-2)}{(\sqrt{3}-2)(\sqrt{3}+2)}\)= \(\dfrac{4}{3-4 }\)= -4
c. C = \(2\sqrt{4 + \sqrt{6 -2\sqrt{5}}}(\sqrt{10}-\sqrt{2})\) = \(2\sqrt{4 + \sqrt{(\sqrt{5} - 1 )^2}}(\sqrt{10}-\sqrt{2})\)= \(2\sqrt{4 + \sqrt{5} - 1}(\sqrt{10}-\sqrt{2})\) = \(2\sqrt{3 + \sqrt{5} }(\sqrt{10}-\sqrt{2})\) = 8
