Question

Math question. plz answer it.

$a) \frac{1}{3 + \sqrt{7} } \\ \\ \\ \\ \\ b) \frac{5 + \sqrt{5} }{7 + \sqrt{3} }$
? ? ? ? ? ?

Answer 1

Hola Mate!!

Your answer :-

$a) = > \frac{1}{3 + \sqrt{7} } \\ \\ = > \frac{1}{3 + \sqrt{7} } \times \frac{3 - \sqrt{7} }{3 - \sqrt{7} } \\ \\ = > \frac{3 - \sqrt{7} }{ {(3)}^{2} - {( \sqrt{7}) }^{2} } \\ \\ = > \frac{3 - \sqrt{7} }{9 - 7 } \\ \\ = > \frac{3 - \sqrt{7} }{2} \\ \\ b) = > \frac{5 + \sqrt{5} }{7 + \sqrt{3} } \\ \\ = > \frac{5 + \sqrt{5} }{7 + \sqrt{3} } \times \frac{7 - \sqrt{3} }{7 - \sqrt{3} } \\ \\ = > \frac{(5 + \sqrt{5})(7 - \sqrt{3}) }{ {(7)}^{2} - {( \sqrt{3}) }^{2} } \\ \\ = > \frac{5(7 - \sqrt{3}) + \sqrt{5} (7 - \sqrt{3} )}{49 - 3} \\ \\ = > \frac{35 - 5 \sqrt{3 } + 7 \sqrt{5} - \sqrt{15} }{46}$

☆ Hope it helps ☆

Answer 2

$\frac{1}{3 + \sqrt{7} } \\ \frac{1}{3 + \sqrt{7} } \times \frac{3 - \sqrt{7} }{3 - \sqrt{7} } \\ \frac{3 - \sqrt{7} }{3 {}^{2} - \sqrt{7 {}^{2} } } \\ \frac{3 - \sqrt{7} }{2}$
2.
$\frac{5 + \sqrt{5} }{7 + \sqrt{3} } \times \frac{7 - \sqrt{3} }{7 - \sqrt{3} } \\ \frac{5(7 - \sqrt{3}) + \sqrt{5}(7 - \sqrt{3}) }{7 {}^{2} - 3 } \\ \frac{35 - 5 \sqrt{3} + 7 \sqrt{5} - \sqrt{15} }{49 - 3} \\ \frac{35 - 5 \sqrt{3} + 7 \sqrt{5} - \sqrt{15} }{46}$
hope this helps you
don't forget to do it brainleist.