Question

Rationalize to denominators 1/9+√7 this is a question​

Answer 1

Solution :-

$\frac{1}{9 + \sqrt{7} } = \frac{1}{9 + \sqrt{7} } \times \frac{9 - \sqrt{7} }{9 - \sqrt{7} } \\ \\ = \frac{9 - \sqrt{7} }{ {9}^{2} - {( \sqrt{7)} }^{2} } = \frac{9 - \sqrt{7} }{81 - 7} \\ \\ = \frac{9 - \sqrt{7} }{74}$

Answer 2

Answer:

$\dfrac{1}{9+\sqrt{7} }$

Multiplying both numerator and denominator by conjugate

$\dfrac{1}{9+\sqrt{7} }\times \dfrac{9-\sqrt{7}}{9-\sqrt{7} }$

$\dfrac{9-\sqrt{7}}{(9-\sqrt{7})(9+\sqrt{7} ) }$

$\dfrac{9-\sqrt{7}}{9^2-(\sqrt{7})^2}$

$\dfrac{9-\sqrt{7}}{81-7}$

$\dfrac{9-\sqrt{7}}{74}$