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2000 AMC 10 Problems/Problem 15

Revision as of 17:56, 7 January 2009 by BOGTRO (talk | contribs) (New page: <math>ab=a-b</math> <math>\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2}{ab}-ab=\frac{-a^2b^2+a^2+b^2}{ab}</math> <math>\frac{-a^2+2ab-b^2+a^2+b^2}{ab}=2</math>. E.)
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$ab=a-b$

$\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2}{ab}-ab=\frac{-a^2b^2+a^2+b^2}{ab}$

$\frac{-a^2+2ab-b^2+a^2+b^2}{ab}=2$.

E.

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