Question

((√7-1)÷(√7+1))-((√7+1)÷(√7-1))
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Answer 1

$\frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1} - \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1} \\ \\ \bf \: on \: rationaizing \: the \\ \bf \: denominator \: we \: get \\ \\ = \frac{ \sqrt{7} - 1}{ \sqrt{7} + 1 } \times \frac{ \sqrt{7} - 1}{ \sqrt{7} - 1} - \frac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1} \\ \\ = \frac{ {( \sqrt{7} )}^{2} + {(1)}^{2} - 2( \sqrt{7} )(1)}{ {( \sqrt{7}) }^{2} - {(1)}^{2} } - \frac{ {( \sqrt{7} )}^{2} + {(1)}^{2} + 2( \sqrt{7} )(1) }{ {( \sqrt{7} )}^{2} - {(1)}^{2} } \\ \\ = \frac{7 + 1 - 2 \sqrt{7} }{7 - 1} - \frac{7 + 1 + 2 \sqrt{7} }{7 - 1} \\ \\ = \frac{8 - 2 \sqrt{7} }{6} - \frac{8 + 2 \sqrt{7} }{6} \\ \\ = \frac{8 - 2 \sqrt{7} - 8 - 2 \sqrt{7} }{6} \\ \\ = - \frac{4 \sqrt{7} }{6} \\ \\ = - \frac{2 \sqrt{7} }{3}$