$\begin{array}{l}b)\, \sqrt{4 - \sqrt7} + \sqrt{4 + \sqrt7}\\ =\sqrt{\dfrac{8 - 2\sqrt7}{2}}+\sqrt{\dfrac{8 + 2\sqrt7}{2}}\\ = \sqrt{\dfrac{(\sqrt7-1)^2}{2}}+\sqrt{\dfrac{(\sqrt7+1)^2}{2}}\\ =\dfrac{\sqrt7 -1 + \sqrt7 + 1}{\sqrt2}\\ = \dfrac{2\sqrt7}{\sqrt2}\\ =\sqrt2.\sqrt7 = \sqrt{14}\\ c)\, \sqrt{2 + \sqrt3} -\sqrt{2 - \sqrt3}\\ = \sqrt{\dfrac{4 + 2\sqrt3}{2}}-\sqrt{\dfrac{4 - 2\sqrt3}{2}}\\ =\sqrt{\dfrac{(\sqrt3+1)^2}{2}}-\sqrt{\dfrac{(\sqrt3-1)^2}{2}}\\ =\dfrac{\sqrt3 + 1 - \sqrt3 + 1}{\sqrt2}\\ =\dfrac{2}{\sqrt2} = \sqrt2\\ d)\,\sqrt{3 + \sqrt5} + \sqrt{3 - \sqrt5}\\ = \sqrt{\dfrac{6 + 2\sqrt5}{2}}+\sqrt{\dfrac{6 - 2\sqrt5}{2}}\\ = \sqrt{\dfrac{(\sqrt5+1)^2}{2}}+\sqrt{\dfrac{(\sqrt5-1)^2}{2}}\\ =\dfrac{\sqrt5 + 1 + \sqrt5 - 1}{\sqrt2}\\ = \dfrac{2\sqrt5}{\sqrt2}\\ =\sqrt2.\sqrt5 = \sqrt{10}\end{array}$
