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Simplify $\dfrac{{x}^{2}-1}{3}\times\dfrac{1}{x-1}-\dfrac{1}{2}$
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factorize difference of 2 squares and carry cross multiplication

$\frac{(x+1)(x-1)}{3} \times \frac{1}{(x-1)} - \frac{1}{2}$

$= \frac{(x+1)}{3} - \frac{1}{2}$

$= \frac{(2x+2-3)}{6}$

$= \frac{(2x-1)}{6}$
by Diamond (39,577 points)