Notice that
[tex]\dfrac1{(x-1)x} = \dfrac1{x-1} - \dfrac1x \\\\ \dfrac1{x(x+1)} = \dfrac1x - \dfrac1{x+1} \\\\ \vdots \\\\ \dfrac1{(x+9)(x+10)} = \dfrac1{x+9} - \dfrac1{x+10}[/tex]
That is, the n-th term (where n = -1, 0, 1, …, 9) in the sum on the left side has a partial fraction decomposition of
[tex]\dfrac{1}{(x+n)(x+n+1)} = \dfrac1{x+n} - \dfrac1{x+n+1}[/tex]
and in the sum, some adjacent terms will cancel and leave you with
[tex]\dfrac1{(x-1)x} + \dfrac1{x(x+1)} + \cdots + \dfrac1{(x+9)(x+10)} = \dfrac1{x-1} - \dfrac1{x+10} = \dfrac{11}{12}[/tex]
Now solve for x, bearing in mind that we cannot have x = 0, -1, -2, …, -10 :
[tex]\dfrac1{x-1} - \dfrac1{x+10} = \dfrac{11}{12}[/tex]
Combine the fractions on the left side:
[tex]\dfrac{(x+10)-(x-1)}{(x-1)(x+10)} = \dfrac{11}{(x-1)(x+10)} = \dfrac{11}{12}[/tex]
Then we must have
[tex](x-1)(x+10) = x^2 + 9x - 10 = 12 \implies x^2+9x-22 = (x+11)(x-2) = 0[/tex]
so that either x = 2 or x = -11.
