Question

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Answer 1

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Answer 2

$\dfrac{3}{ \sqrt{5}- \sqrt{2} }+ \dfrac{5}{ \sqrt{7}+ \sqrt{2} }- \dfrac{2}{ \sqrt{7}- \sqrt{5} }= \\ \\ = \dfrac{3( \sqrt{5}+ \sqrt{2}) }{5-2}+ \dfrac{5( \sqrt{7}- \sqrt{2}) }{7-2}- \dfrac{2( \sqrt{7}+\sqrt{5}) }{7-5}= \\ \\ = \sqrt{5}+ \sqrt{2}+ \sqrt{7}- \sqrt{2}- \sqrt{7}- \sqrt{5}= \\ \\ =0$

Вариант с 7-√5:
$\dfrac{3}{ \sqrt{5}- \sqrt{2} }+ \dfrac{5}{ \sqrt{7}+ \sqrt{2} }- \dfrac{2}{ 7- \sqrt{5} }= \\ \\ = \dfrac{3( \sqrt{5}+ \sqrt{2}) }{5-2}+ \dfrac{5( \sqrt{7}- \sqrt{2}) }{7-2}- \dfrac{2(7+ \sqrt{5}) }{49-5} = \\ \\ =\sqrt{5}+ \sqrt{2}+ \sqrt{7}- \sqrt{2}- \dfrac{7+ \sqrt{5} }{22}= \\ \\ = \dfrac{22 \sqrt{5}-7- \sqrt{5} }{22}+ \sqrt{7} = \\ \\ = \dfrac{21 \sqrt{5}-7 }{22}+ \sqrt{7}$