Question

7+3√5/3+√5 - 7-3√5/3-√5

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } \times \frac{3 - \sqrt{5} }{3 - \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } \times \frac{3 + \sqrt{5} }{3 + \sqrt{5} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ (\frac{7 \times 3 - 7 \times \sqrt{5} + 3 \sqrt{5} \times 3 - 3 \sqrt{5} \times \sqrt{5} }{ {(3)}^{2} - {( \sqrt{5} )}^{2} } ) \\ - ( \frac{7 \times 3 + 7 \times \sqrt{5} - 3 \sqrt{5} \times 3 - 3 \sqrt{5} \times \sqrt{5} }{ {(3)}^{2} - {( \sqrt{5}) }^{2} } )\\ \\ =( \frac{21 - 7 \sqrt{5} + 9\sqrt{5} - 15}{9 - 5} ) \\ - ( \frac{21 + 7 \sqrt{5} - 9 \sqrt{5} - 15}{9 - 5} ) \\ \\ = ( \frac{6 + 2 \sqrt{5} }{4} ) - ( \frac{6 - 2 \sqrt{5} }{4} ) \\ \\ = \frac{6 + 2 \sqrt{5} - 6 + 2 \sqrt{5} }{4} \\ \\ = \frac{2 \sqrt{5} + 2 \sqrt{5} }{4} \\ \\ = \frac{4 \sqrt{5} }{4} \\ \\ = \sqrt{5}$

Hope this helps!!!!

@Mahak24

Thanks...
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