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.)
 
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<math>ab=a-b</math>
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#REDIRECT [[2000 AMC 12 Problems/Problem 11]]
 
 
<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.
 

Latest revision as of 23:54, 26 November 2011