Question

√7-3/√7+3 by rationalize denominator​

Answer 1

Answer:

$\sqrt{7}$ - 3 / $\sqrt{7$ +3 = ($\sqrt{7}$ - 3)($\sqrt{7}$ - 3 ) /( $\sqrt{7$ +3)($\sqrt{7}$ - 3 )

                       =$\frac{ \sqrt{7}^2 - 2 (\sqrt{7} )(3) + 3^{2} }{\sqrt{7}^2 - 3^{2} }$

                      = $\frac{7 - 6\sqrt{7} + 9 }{7 - 9}$ = $\frac{16 - 6\sqrt{7} }{2}$

                                      = $\frac{2( 8 - 3\sqrt{7}) }{2}$                        

( now we cancel the 2 from both the denominators and numerators)

                                      =$8-3\sqrt{7}$

       ∴ $8 - 3\sqrt{7}$ is the final answer

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