Difference between revisions of "2000 AMC 10 Problems/Problem 15"

(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.)
(No difference)

Revision as of 17:56, 7 January 2009

$ab=a-b$

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

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

E.