Question

Answer the question attached.

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Answer 1

$\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } \\ \\ = \frac{(7 + 3 \sqrt{5} )(3 - \sqrt{5} ) - (7 - 3 \sqrt{5})(3 + \sqrt{5} )}{(3 + \sqrt{5} )(3 - \sqrt{5} )} \\ \\ = \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - (21 + 7 \sqrt{5} - 9 \sqrt{5} - 15)}{ {3}^{2} - { \sqrt{5} }^{2} } \\ \\ = \frac{18 \sqrt{5} - 14 \sqrt{5} }{9 - 5} \\ \\ = \frac{4 \sqrt{5} }{4} \\ \\ = \sqrt{5} \\ 0 + 1\sqrt{5}= a + b \sqrt{5}$

Hence, by comparing,

$a = 0 \\ b = 1$

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Answer 2

Answer:

refer to the attachment..

I hope it helpful to you..