Question : What is the simplified value of:
$\frac{1}{8}\left\{\left(x+\frac{1}{y}\right)^2-\left(x-\frac{1}{y}\right)^2\right\}$
Option 1: $\frac{x}{y}$
Option 2: $\frac{2x}{y}$
Option 3: $\frac{x}{2y}$
Option 4: $\frac{4x}{y}$
Correct Answer: $\frac{x}{2y}$
Solution :
$\frac{1}{8}\left\{\left(x+\frac{1}{y}\right)^2-\left(x-\frac{1}{y}\right)^2\right\}$
$=\frac{1}{8}(x^2 +\frac{1}{y^2} + 2×x×\frac{1}{y} - x^2 - \frac{1}{y^2} + 2×x×\frac{1}{y})$
$=\frac{1}{8}(\frac{4x}{y})$
$=\frac{x}{2y}$
Hence, the correct answer is $\frac{x}{2y}$.
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