$\displaystyle (x^2)/(1+x^2) = (1- (1/(1+x^2))$ Why is this so?
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$\displaystyle 1- \frac{1}{1+x^2} = \frac{1+x^2}{1+x^2}- \frac{1}{1+x^2} = \frac{1+x^2-1}{1+x^2} = \frac{x^2}{1+x^2}$
Thanks. I never thought about adding or subtracting one.
Originally Posted by KyleC $\displaystyle (x^2)/(1+x^2) = (1- (1/(1+x^2))$ Why is this so? To go from the left-hand side to the right-hand side, turn the divisor around to get $\displaystyle x^2\, +\, 1$, and then do the long division on $\displaystyle x^2\, +\, 0x\, +\, 0$.
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