\(b,=\sqrt{\dfrac{8-2\sqrt{15}}{2}}-\sqrt{\dfrac{8+2\sqrt{15}}{2}}=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\\ =\dfrac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\\ c,=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}-\sqrt{\left(3+\sqrt{6}\right)^2}\\ =\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}-3-\sqrt{6}\\ =\dfrac{2\sqrt{3}}{\sqrt{2}}-3-\sqrt{6}=\sqrt{6}-3-\sqrt{6}=-3\)
\(d,=\sqrt{\dfrac{12-6\sqrt{3}}{2}}-\sqrt{\dfrac{12+6\sqrt{3}}{2}}+\dfrac{\sqrt{6}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\\ =\dfrac{3-\sqrt{3}}{\sqrt{2}}-\dfrac{3+\sqrt{3}}{\sqrt{2}}+\sqrt{6}\\ =\dfrac{-2\sqrt{3}}{\sqrt{2}}+\sqrt{6}=-\sqrt{6}+\sqrt{6}=0\)
