\(\dfrac{x}{2}+\dfrac{x}{6}+\dfrac{x}{12}+...+\dfrac{x}{90}=\dfrac{p^2+10}{p^3+2}\\ \Rightarrow x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)=\dfrac{p^2+10}{p^3+2}\\ \Rightarrow x.\left(1-\dfrac{1}{10}\right)=\dfrac{p^2+10}{p^3+2}\\ \Rightarrow x.\dfrac{9}{10}=\dfrac{p^2+10}{p^3+2}\\ \Rightarrow x=\dfrac{p^2+10}{p^3+2}.\dfrac{10}{9}=\dfrac{10p^2+100}{9p^3+18}\)
\(\begin{align*} \frac{x}{2}+\frac{x}{6}+\frac{x}{12}+...+\frac{x}{90}&=\frac{p^{2}+10}{p^{3}+2} \\ \Rightarrow x\cdot\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)&=\frac{p^{2}+10}{p^{3}+2} \\ x\cdot\left(1-\frac{1}{10}\right)&=\frac{p^{2}+10}{p^{3}+2} \\ \frac{9}{10}&=\frac{p^{2}+10}{p^{3}+2} \\ \Rightarrow x&=\frac{p^{2}+10}{p^{3}+2}\cdot\frac{10}{9} \\ &=\frac{10p^{2}+100}{9p^{3}+18} \end{align*}\)
