How do you rationalize the denominator #2/(sqrt[3] +sqrt[2])#?

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How do you rationalize the denominator #2/(sqrt[3] +sqrt[2])#?

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Answer 1 · 2015-04-12T19:41:13

You can multiply and divide by #sqrt(3)-sqrt(2)# to get:
#2/(sqrt(3)+sqrt(2))(sqrt(3)-sqrt(2))/(sqrt(3)-sqrt(2))=#
#=(2*(sqrt(3)-sqrt(2)))/(3-2)=2*(sqrt(3)-sqrt(2))#

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How do you rationalize the denominator #2/(sqrt[3] +sqrt[2])#?

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