Difference between revisions of "1951 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
| − | {{ | + | Note that <math>(x + y)^{-1}(x^{-1} + y^{-1}) = \dfrac{1}{x + y}\cdot\left(\dfrac{1}{x} + \dfrac{1}{y}\right) = \dfrac{1}{x + y}\cdot\dfrac{x + y}{xy} = \dfrac{1}{xy} = x^{-1}y^{-1}</math>. The answer is <math>\textbf{(C)}</math>. |
== See Also == | == See Also == | ||
Revision as of 23:55, 24 April 2013
Problem
When simplified and expressed with negative exponents, the expression $(x \plus{} y)^{ \minus{} 1}(x^{ \minus{} 1} \plus{} y^{ \minus{} 1})$ (Error compiling LaTeX. Unknown error_msg) is equal to:
$\textbf{(A)}\ x^{ \minus{} 2} \plus{} 2x^{ \minus{} 1}y^{ \minus{} 1} \plus{} y^{ \minus{} 2} \qquad\textbf{(B)}\ x^{ \minus{} 2} \plus{} 2^{ \minus{} 1}x^{ \minus{} 1}y^{ \minus{} 1} \plus{} y^{ \minus{} 2} \qquad\textbf{(C)}\ x^{ \minus{} 1}y^{ \minus{} 1}$ (Error compiling LaTeX. Unknown error_msg) $\textbf{(D)}\ x^{ \minus{} 2} \plus{} y^{ \minus{} 2} \qquad\textbf{(E)}\ \frac {1}{x^{ \minus{} 1}y^{ \minus{} 1}}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Note that 
. The answer is 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
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| All AHSME Problems and Solutions | ||
