\(a,=0\\ b,=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\dfrac{2-\sqrt{3}}{4-3}+\dfrac{1}{\sqrt{3}}-\dfrac{2\left(3-\sqrt{3}\right)}{6}\\ =2-\sqrt{3}+\dfrac{\sqrt{3}}{3}-\dfrac{3-\sqrt{3}}{3}\\ =2-\sqrt{3}+\dfrac{2\sqrt{3}-3}{3}=\dfrac{6-3\sqrt{3}+2\sqrt{3}-3}{3}=\dfrac{3-\sqrt{3}}{3}\\ d,=\dfrac{5\left(4+\sqrt{11}\right)}{5}+\dfrac{3-\sqrt{7}}{2}-\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{\sqrt{7}-5}{2}\\ =4+\sqrt{11}+\dfrac{3-\sqrt{7}-\sqrt{7}+5}{2}-2\sqrt{7}-4\\ =\sqrt{11}-2\sqrt{7}+\dfrac{8-2\sqrt{7}}{2}=\sqrt{11}-2\sqrt{7}+4-\sqrt{7}=\sqrt{11}-3\sqrt{7}+4\)
