\(a,A=\sqrt{5}-\sqrt{2}+2\sqrt{5}+\dfrac{\sqrt{2}}{4}=3\sqrt{5}-\dfrac{3\sqrt{2}}{4}\\ b,A'=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}=2-15+8=-5\\ c,B=\sqrt{2\left(6-3\sqrt{3}\right)}\left(3+\sqrt{3}\right)=\sqrt{12-6\sqrt{3}}\left(3+\sqrt{3}\right)\\ =\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)=6\\ d,B'=\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
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