Question

3+√7/3-√7-3-√7/3+√7 please answer class 8 question

Answer 1

Hey friend here your answer.....
____________________________________3-3+√7/3-√7/3-√7+√7
=0. (ans)

Answer 2

Hey friend,
Here is the answer you were looking for:
$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } - \frac{3 - \sqrt{7} }{3 + \sqrt{7} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \frac{3 + \sqrt{7} }{3 + \sqrt{7} } - \frac{3 - \sqrt{7} }{3 + \sqrt{7} } \times \frac{3 - \sqrt{7} }{3 - \sqrt{7} } \\ \\ using \: the \: identities \\ \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {(3)}^{2} + {( \sqrt{7}) }^{2} + 2 \times 3 \times \sqrt{7} }{ {(3)}^{2} - {( \sqrt{7} })^{2} } - \frac{ {(3)}^{2} + {( \sqrt{7}) }^{2} - 3 \times 3 \times \sqrt{7} }{ {(3)}^{2} - {( \sqrt{7} )}^{2} } \\ \\ = \frac{9 + 7 + 6 \sqrt{7} }{9 - 7} - \frac{9 + 7 - 6 \sqrt{7} }{9 - 7} \\ \\ = \frac{9 + 7 + 6 \sqrt{7} - 9 - 7 + 6 \sqrt{7} }{2} \\ \\ = \frac{6 \sqrt{7} + 6 \sqrt{7} }{2} \\ \\ = \frac{12 \sqrt{7} }{2} \\ \\ = 6 \sqrt{7}$


Hope this helps!!!

@Mahak24

Thanks...
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