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

Latest revision as of 00:45, 16 August 2023

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)
Preceded by Problem 1	Followed by Problem 3
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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

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 ==See Also==
-{{AHSME 50p box|year=1967|num-b=1|num-a=3}}
+{{AHSME 40p box|year=1967|num-b=1|num-a=3}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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

Latest revision as of 00:45, 16 August 2023

Problem

Solution

See Also