\(a,=\left(\sqrt{5}-3\right)\sqrt{\left(3-\sqrt{5}\right)^2}\\ =\left(5-\sqrt{3}\right)\left(3-\sqrt{5}\right)\\ =-\left(3-\sqrt{5}\right)^2\\ =6\sqrt{5}-14\\ b,=\left(\sqrt{7}+2\right)\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)-\left(3-\sqrt{7}\right)\\ =3-3+\sqrt{7}=\sqrt{7}\)
\(c,=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}-3\sqrt{5}}{\sqrt{\left(3+\sqrt{5}\right)^2}}\\ =\dfrac{\sqrt{5}-1+\sqrt{5}-2-3\sqrt{5}}{3+\sqrt{5}}\\ =\dfrac{-3-\sqrt{5}}{3+\sqrt{5}}=\dfrac{-\left(3+\sqrt{5}\right)}{3+\sqrt{5}}=-1\)
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