Question

$\sqrt{5 + \sqrt[2]{6} } + \sqrt{8 - \sqrt[2]{15 } }$
simplify​

Answer 1

$\sqrt{5 + \sqrt{6} } + \sqrt{8 - \sqrt{15} }$

Use, this Relation:

$\sqrt{x \pm \sqrt{y} } = \sqrt{ \frac{x + \sqrt{ {x }^{2} - y} }{2} } \pm \sqrt{ \frac{x - \sqrt{ {x}^{2} - y} }{2} }$

You can easily prove this with some simple algebra

$\sqrt{5 + \sqrt{6} } = \sqrt{ \frac{5 + \sqrt{25 - 6} }{2} } + \sqrt{ \frac{5 - \sqrt{25 - 6} }{2} }$

$\sqrt{5 + \sqrt{6} } = \sqrt{ \frac{5 + \sqrt{19} }{2} } + \sqrt{ \frac{5 - \sqrt{19} }{2} }$

therefore, its not possible to Denest this square root.

$\sqrt{8 - \sqrt{15} } = \sqrt{ \frac{8 + \sqrt{64 - 15} }{2} } \small{ - }\sqrt{ \frac{8 - \sqrt{64 - 15} }{2} }$

$\sqrt{8 - \sqrt{15} } = \sqrt{ \frac{8 + 7}{2} } - \sqrt{ \frac{8 - 7}{2} }$

$\sqrt{8 - \sqrt{15} } = \sqrt{ \frac{15}{2} } - \sqrt{ \frac{1}{2} } = \frac{ \sqrt{30} - \sqrt{2} }{2}$

Therefore, Your answer is

$\boxed{ \sqrt{5 + \sqrt{6} } + \sqrt{8 - \sqrt{15} } = \sqrt{5 + \sqrt{6} } + \frac{ \sqrt{30} - \sqrt{2} }{2} }$