Find $x$

$\begin{aligned} \frac {22}{45} & = \left(\frac 1{1\cdot 2\cdot 3} + \frac 1{2\cdot 3\cdot 4} + \frac 1{3\cdot 4\cdot 5} +\cdots + \frac 1{8\cdot 9\cdot 10} \right)x \\ & = x \sum_{n=1}^8 \frac 1{n(n+1)(n+2)} & \small \color{#3D99F6} \text{By partial fractions} \\ & = \frac x2 \sum_{n=1}^8 \left(\frac 1n -\frac 2{n+1}+\frac 1{n+2} \right) \\ & = \frac x2 \left(\frac 11 -\frac 19 - \frac 12 +\frac 1{10} \right) \\ & = \frac x2 \left(\frac {90-10-45+9}{90} \right) \\ & = \frac {11}{45}x \\ \implies x & = \boxed{2} \end{aligned}$

1/[k(k+1)(k+2)] = (1/2)[1/(k(k+1)) - 1/((k+1)(k+2)). So (1/2)[1/2 - 1/(9(10))]x = 22/45, so x=2