Difference between revisions of "1981 AHSME Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | + | Converting all negative exponents to fractions yields: | |
− | <cmath>(\frac{1}{x+y+z})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(\frac{1}{xy+yz+xz})(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz})</cmath> | + | <cmath>\left(\frac{1}{x+y+z}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{xy+yz+xz}\right)\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)</cmath> |
+ | Adding fractions within each factor yields: | ||
+ | <cmath>\left(\frac{1}{x+y+z}\right)\left(\frac{xy+yz+xz}{xyz}\right)\left(\frac{1}{xy+yz+xz}\right)\left(\frac{x+y+z}{xyz}\right)</cmath> | ||
+ | Multiplying and cancelling common factors yields: | ||
+ | <cmath>\frac{1}{x^2 y^2 z^2} = \boxed{\textbf{(A)}\ x^{-2}y^{-2}z^{-2}}</cmath> | ||
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-edited by coolmath34 | -edited by coolmath34 |
Latest revision as of 16:09, 18 August 2025
Problem 8
For all positive numbers ,
,
, the product
equals
Solution
Converting all negative exponents to fractions yields:
Adding fractions within each factor yields:
Multiplying and cancelling common factors yields:
-edited by coolmath34
See also
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.