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anonymous
 4 years ago
sqrtcsqrtd/sqrtc+sqrtd
anonymous
 4 years ago
sqrtcsqrtd/sqrtc+sqrtd

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Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1u probably want to rationalize the denominator?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt{c}\sqrt{d}}{\sqrt{c}+\sqrt{d}}\]

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1\[(\sqrt c\sqrt d)/(\sqrt c+\sqrt d)=(\sqrt c\sqrt d)(\sqrt c\sqrt d)/(\sqrt c\sqrt d)(\sqrt c+\sqrt d)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt{c}\sqrt{d}}{\sqrt{c}+\sqrt{d}}\times \frac{\sqrt{c}\sqrt{d}}{\sqrt{c}\sqrt{d}}\] \[\frac{\sqrt{c}\sqrt{d})^2}{cd}\]

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1therefore you have\[(\sqrt c\sqrt d)^2 \over cd\]

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1u missed a bracket satellite :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0all right wise gut \[\frac{(\sqrt{c}\sqrt{d})^2}{cd}\]

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.1therefore ur final answer is \[c  2 \sqrt c \sqrt d + d \over cd\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328147495943:dw
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