Difference between revisions of "1967 AHSME Problems/Problem 2"

Revision as of 11:40, 5 July 2013

Problem

An equivalent of the expression

$\left(\frac{x^2+1}{x}\right)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)$ , $xy \not= 0$ ,

is:

$\text{(A)}\ 1\qquad\text{(B)}\ 2xy\qquad\text{(C)}\ 2x^2y^2+2\qquad\text{(D)}\ 2xy+\frac{2}{xy}\qquad\text{(E)}\ \frac{2x}{y}+\frac{2y}{x}$

Solution

$\left(\frac{x^2+1}{x}\right)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)$ $\frac{(x^2+1)(y^2+1)}{xy} + \frac{(x^2-1)(y^2-1)}{xy}$ $\frac{(x^2y^2+x^2+y^2+1) + (x^2y^2-x^2-y^2+1)}{xy}$ $\frac{x^2y^2+x^2+y^2+1 + x^2y^2-x^2-y^2+1}{xy}$ $\frac{2x^2y^2+2}{xy}$ $\frac{2x^2y^2}{xy} + \frac{2}{xy}$ $\frac{2xy * xy}{xy} + \frac{2}{xy}$ $2xy + \frac{2}{xy}$ The answer is $\boxed{D}$

1967 AHSC (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 21:12, 6 June 2012 (view source) Silvernight (talk \| contribs) (Created page with "==Problem== An equivalent of the expression <math>\left(\frac{x^2+1}{x}\right)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)</math>, <mat...")		Revision as of 11:40, 5 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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Difference between revisions of "1967 AHSME Problems/Problem 2"

Revision as of 11:40, 5 July 2013

Problem

Solution

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