1967 AHSME Problems/Problem 2

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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 $\boxed{D}$

See Also

1967 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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