Difference between revisions of "1981 AHSME Problems/Problem 8"

(See also)
(Solution)
 
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==Solution==
 
==Solution==
Start simplifying:
+
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>
  
<cmath>(\frac{1}{x+y+z})(\frac{xy+yz+xz}{xyz})(\frac{1}{xy+yz+xz})(\frac{x+y+z}{xyz})</cmath>
 
 
The <math>xy+yz+xz</math> and <math>x+y+z</math> cancel out:
 
 
<cmath>\frac{1}{(xyz)^2}</cmath>
 
 
The answer is <math> \textbf{(A)}\ x^{-2}y^{-2}z^{-2}.</math>
 
  
 
-edited by coolmath34
 
-edited by coolmath34

Latest revision as of 16:09, 18 August 2025

Problem 8

For all positive numbers $x$, $y$, $z$, the product \[(x+y+z)^{-1}(x^{-1}+y^{-1}+z^{-1})(xy+yz+xz)^{-1}[(xy)^{-1}+(yz)^{-1}+(xz)^{-1}]\] equals

$\textbf{(A)}\ x^{-2}y^{-2}z^{-2}\qquad\textbf{(B)}\ x^{-2}+y^{-2}+z^{-2}\qquad\textbf{(C)}\ (x+y+z)^{-1}\qquad \textbf{(D)}\ \dfrac{1}{xyz}\qquad \\ \textbf{(E)}\ \dfrac{1}{xy+yz+xz}$


Solution

Converting all negative exponents to fractions yields: \[\left(\frac{1}{x+y+z}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{xy+yz+xz}\right)\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)\] Adding fractions within each factor yields: \[\left(\frac{1}{x+y+z}\right)\left(\frac{xy+yz+xz}{xyz}\right)\left(\frac{1}{xy+yz+xz}\right)\left(\frac{x+y+z}{xyz}\right)\] Multiplying and cancelling common factors yields: \[\frac{1}{x^2 y^2 z^2} = \boxed{\textbf{(A)}\ x^{-2}y^{-2}z^{-2}}\]


-edited by coolmath34

See also

1981 AHSME (ProblemsAnswer KeyResources)
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

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