Question

simplify
$simplify \\ \sqrt{7} - 1 \div \sqrt{7 } + 1 + \sqrt{7} + 1 \div \sqrt{7} - 1$

Answer 1

Identities used :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$


$\frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1 } + \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1 }$


On rationalizing the denominator we get,


$= \frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1 } \times \ \frac{ \sqrt{7} - 1 }{ \sqrt{7} - 1 } + \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1 }{ \sqrt{7} + 1 } \\ \\ = \frac{ {( \sqrt{7}) }^{2} + {(1)}^{2} - 2( \sqrt{7} )(1) }{ {( \sqrt{7} )}^{2} - {(1)}^{2} } + \frac{ {( \sqrt{7} )}^{2} + {(1)}^{2} + 2( \sqrt{7} )(1) }{ {( \sqrt{7} )}^{2} - {(1)}^{2} } \\ \\ = \frac{7 + 1 - 2 \sqrt{7} }{7 - 1} + \frac{7 + 1 + 2 \sqrt{7} }{7 - 1} \\ \\ = \frac{8 - 2 \sqrt{7} }{6} + \frac{8 + 2 \sqrt{7} }{6} \\ \\ = \frac{2(4 - \sqrt{7}) }{2(3)} + \frac{2(4 + \sqrt{7} )}{2(3)} \\ \\ = \frac{4 - \sqrt{7} }{3} + \frac{4 + \sqrt{7} }{3} \\ \\ = \frac{4 - \sqrt{7} + 4 + \sqrt{7} }{3} \\ \\ = \frac{4 + 4}{3} \\ \\ = \frac{8}{3} \\ \\ \bf \: = 2.66 \: \: (approx)$

If any doubt, please ask :)