Question 110

The simplest form of the expression $$\frac{p^2-p}{2p^3+6p^2}+\frac{p^2-1}{p^2+3p}+\frac{p^2}{p+1}$$

Solution

$$\frac{p^2-p}{2p^3+6p^2}$$ = $$\frac{p(p-1)}{2p^2(p+3)}$$ = $$\frac{(p-1)}{2p(p+3)}$$

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

$$\frac{p^2}{p+1}$$ = $$\frac{p^2}{p+1}$$

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

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