I'm still not sure exactly what you're trying to do, but as far as finding products that equal 1, 2, etc.

Rewriting the rows:

$\displaystyle \frac{1}{2x^2} \ \ \ \ \frac{1}{2\sqrt{x}} \ \ \ \ \frac{2}{x} \ \ \ \ \frac{2}{\sqrt{x}} \ \ \ \ \frac{1}{2x}$

$\displaystyle 2\sqrt{x} \ \ \ \ \frac{2}{x^2} \ \ \ \ \frac{1}{2\sqrt{x}} \ \ \ \ \frac{1}{2x^2} \ \ \ \ 2x^2$

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$\displaystyle \frac{1}{2x^2}\cdot 2x^2 = 1$

$\displaystyle 2\sqrt{x} \cdot \frac{1}{2\sqrt{x}}=1$

$\displaystyle \frac{2}{x^2}\cdot 2x^2=4$

Is this what you're talking about? If so, try some products and see what happens.