Solution to Sum of Fractions Equation

Note that the RHS of the equation can be rewritten as $\begin{aligned}&\,\,\,\,\,\,\dfrac1{x(x+1)}+\dfrac1{(x+1)(x+2)}+\dfrac1{(x+2)(x+3)}\\&=\left(\dfrac1x-\dfrac1{x+1}\right)+\left(\dfrac1{x+1}-\dfrac1{x+2}\right)+\left(\dfrac1{x+2}-\dfrac1{x+3}\right)\\&=\dfrac1x-\dfrac1{x+3}=\dfrac3{x(x+3)}.\end{aligned}$ Hence $\dfrac1{10x}=\dfrac3{x(x+3)}\implies x+3=30\implies x=\boxed{27}.$ (We can multiply out the $x$ from both denominators since it is clear that $x=0$ makes both sides undefined.)