Question

Answer 1

\(a,=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}-\sqrt{7}-\sqrt{5}=-2\sqrt{5}\\ b,=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}=2\sqrt{5}\\ c,=\sqrt{\left(3+2\sqrt{2}\right)^2}-\sqrt{\left(3-2\sqrt{2}\right)^2}=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}\\ d,=\dfrac{\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{3}+\sqrt{5}\right)^2}}{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}=\dfrac{2\sqrt{5}-\sqrt{3}-2\sqrt{3}-\sqrt{5}}{\sqrt{5}-\sqrt{3}}\\ =\dfrac{\sqrt{5}-3\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\dfrac{\left(\sqrt{5}-3\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{-4-2\sqrt{15}}{2}=-2-\sqrt{15}\)