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algebra intermediate

Problem

An equivalent of the expression

(\frac{x ^{2} + 1}{x}) (\frac{y ^{2} + 1}{y}) + (\frac{x ^{2} - 1}{y}) (\frac{y ^{2} - 1}{x})

,

x y \neq = 0

, is:

(A)

1

(B)

2 x y

(C)

2 x^{2} y^{2} + 2

(D)

2 x y + \frac{2}{x y}

Solution

(\frac{x ^{2} + 1}{x}) (\frac{y ^{2} + 1}{y}) + (\frac{x ^{2} - 1}{y}) (\frac{y ^{2} - 1}{x})

\frac{( x ^{2} + 1 ) ( y ^{2} + 1 )}{x y} + \frac{( x ^{2} - 1 ) ( y ^{2} - 1 )}{x y}

\frac{( x ^{2} y ^{2} + x ^{2} + y ^{2} + 1 ) + ( x ^{2} y ^{2} - x ^{2} - y ^{2} + 1 )}{x y}

\frac{x ^{2} y ^{2} + x ^{2} + y ^{2} + 1 + x ^{2} y ^{2} - x ^{2} - y ^{2} + 1}{x y}

\frac{2 x ^{2} y ^{2} + 2}{x y}

\frac{2 x ^{2} y ^{2}}{x y} + \frac{2}{x y}

\frac{2 x y * x y}{x y} + \frac{2}{x y}

2 x y + \frac{2}{x y}

The answer is

2 x y + \frac{2}{x y}

Final answer

D