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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| + | ==Problem== | ||
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| + | ==Solution== | ||
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<math>ab=a-b</math> | <math>ab=a-b</math> | ||
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E. | E. | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2000|num-b=14|num-a=16}} | ||
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