rationalize √a+b+√a-b/√a+b-√a-b
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√a+b+√a-b/√a+b-√a-b
=2√a/0.... wrong question
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Solution:
Given (√a+b+√a-b)/(√a+b-√a-b)
Rationalising the denominator, we get,
= [(√a+b+√a-b)(√a+b+√a-b)]/[(√a+b-√a-b)(√a+b+√a-b)]
= (√a+b+√a-b)²/[(√a+b)²-(√a-b)²]
=[(√a+b)²+(√a-b)²-2*√a+b*√a-b]/[a+b-(a-b)]
= (a+b+a-b-2√a²-b²)/(a+b-a+b)
= (2a-2√a²-b²)/(2b)
= [2(a-√a²-b²)]/(2b)
= (a-√a²-b²)/b
••••
mysticd:
:)
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