Question

answer this question fast please

Answer 1

check in attachment.....

Answer 2

$\underline \bold {Solution :-}$

$\frac{ \sqrt{8 + \sqrt{28} } - \sqrt{8 - \sqrt{28} } }{ \sqrt{8 + \sqrt{28} } + \sqrt{8 - \sqrt{28} } } \\ \\ On \: rationalising \: them \\ \\ = \frac{ \sqrt{8 + \sqrt{28} } - \sqrt{8 - \sqrt{28} } }{ \sqrt{8 + \sqrt{28} } + \sqrt{8 - \sqrt{28} } } \times \frac{ \sqrt{8 + \sqrt{28} } - \sqrt{8 - \sqrt{28} } }{ \sqrt{8 + \sqrt{28} } - \sqrt{8 - \sqrt{28} } } \\ \\ = \frac{ ({\sqrt{8 + \sqrt{28} } - \sqrt{8 - \sqrt{28} })}^{2}}{ {( \sqrt{8 + \sqrt{28} }) }^{2} - { (\sqrt{8 - \sqrt{28} } )}^{2} } \\ \\ = \frac{ {( \sqrt{8 + \sqrt{28} } )}^{2} + {( \sqrt{8 - \sqrt{28} }) }^{2} - 2 \sqrt{8 + \sqrt{28} } \sqrt{8 - \sqrt{28} } }{8 + \sqrt{28} - 8 + \sqrt{28} } \\ \\ = \frac{8 + \sqrt{28} + 8 - \sqrt{28} - 2 \sqrt{ {8}^{2} - {( \sqrt{28} )}^{2} } }{2 \sqrt{28} } \\ \\ = \frac{16 - 2 \sqrt{64 - 28} }{2 \sqrt{28} } \\ \\ = \frac{16 - 2 \sqrt{36} }{2 \sqrt{28} } \\ \\ = \frac{16 - 2 \times 6}{2 \sqrt{28} } \\ \\ = \frac{16 - 12}{2 \sqrt{28} } \\ \\ = \frac{4}{2 \sqrt{2 \times 2 \times 7} } \\ \\ = \frac{4}{4 \sqrt{7} } \\ \\ \bold{= \frac{1}{\sqrt{7}}}$