\(c,\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
\(d,\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}}{\sqrt{2}}\\ =\dfrac{\left(\sqrt{5}+\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
c: Ta có: \(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(=\dfrac{\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1+\sqrt{7}+1}{\sqrt{2}}\)
\(=\sqrt{14}\)
