fractional equation

http://www.wolframalpha.com/input/?i=1%2F(x(x-1))+%2B+1%2F(x(x%2B1))%2B1%2F((x%2B1)(x%2B2))%2B+...%2B+1%2F((9%2Bx)(10%2Bx))%3D11%2F12

1/(x(x-1)) + 1/(x(x+1))+1/((x+1)(x+2))+  1/((2+x)(3+x))+ 1/((3+x)(4+x))+ 1/((4+x)(5+x))+ 1/((5+x)(6+x))+ 1/((6+x)(7+x))+ 1/((7+x)(8+x))+ 1/((8+x)(9+x))+ 1/((9+x)(10+x))=11/12

$\begin{array}{ll} \left.\begin{array}{rcl} && \frac{1} {(9+x)(10+x)} \\ &+& \frac{1} {(8+x)(9+x)} \\ \end{array}\right\} &= \frac{1}{9+x}\cdot (\frac{1}{8+x} + \frac{1}{10+x} )=\frac{2}{(8+x)(10+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(7+x)(8+x)} \\ &+& \frac{1} {(6+x)(7+x)} \\ \end{array}\right\} &= \frac{1}{7+x}\cdot (\frac{1}{6+x} + \frac{1}{8+x} )=\frac{2}{(6+x)(8+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(5+x)(6+x)} \\ &+& \frac{1} {(4+x)(5+x)} \\ \end{array}\right\} &= \frac{1}{5+x}\cdot (\frac{1}{4+x} + \frac{1}{6+x} ) =\frac{2}{(4+x)(6+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(3+x)(4+x)} \\ &+& \frac{1} {(2+x)(3+x)} \\ \end{array}\right\} &= \frac{1}{3+x}\cdot (\frac{1}{2+x} + \frac{1}{4+x} )=\frac{2}{(2+x)(4+x)}\\ \left.\begin{array}{rcl} &+& \frac{1} {(1+x)(2+x)} \\ &+& \frac{1} {(0+x)(1+x)} \\ \end{array}\right\} &= \frac{1}{1+x}\cdot (\frac{1}{0+x} + \frac{1}{2+x} )=\frac{2}{x(2+x)}\\ \begin{array}{rcl} &+& \frac{1} {x(x-1)} & \\ \end{array} \\ \hline \begin{array}{rcl} &=& \frac{11}{12} & \end{array} \\ \end{array}$

$\begin{array}{ll} \left.\begin{array}{rcl} && \frac{2}{(8+x)(10+x)} \\ &+& \frac{2}{(6+x)(8+x)} \\ \end{array}\right\} &= \frac{2}{8+x}\cdot (\frac{1}{6+x} + \frac{1}{10+x} )=\frac{4}{(6+x)(10+x)}\\ \left.\begin{array}{rcl} &+& \frac{2}{(4+x)(6+x)} \\ &+& \frac{2}{(2+x)(4+x)} \\ \end{array}\right\} &= \frac{2}{4+x}\cdot (\frac{1}{2+x} + \frac{1}{6+x} )=\frac{4}{(2+x)(6+x)}\\ \left.\begin{array}{rcl} &+& \frac{2}{x(2+x)} \\ &+& \frac{1} {x(x-1)} \\ \end{array}\right\} &= \frac{1}{x}\cdot (\frac{2}{2+x} + \frac{1}{x-1} ) =\frac{3}{(2+x)(x-1)}\\ \hline \begin{array}{rcl} &=& \frac{11}{12} & \end{array} \\ \end{array}$

$\begin{array}{ll} \left.\begin{array}{rcl} && \frac{4}{(6+x)(10+x)} \\ &+& \frac{4}{(2+x)(6+x)} \\ \end{array}\right\} &= \frac{4}{6+x}\cdot (\frac{1}{2+x} + \frac{1}{10+x} )=\frac{8}{(2+x)(10+x)}\\ \begin{array}{rcl} &+& \frac{3}{(2+x)(x-1)} \\ \end{array}\\ \hline \begin{array}{rcl} &=& \frac{11}{12} & \end{array} \\ \end{array}$

$\begin{array}{rcll} \frac{3}{(2+x)(x-1)} + \frac{8}{(2+x)(10+x)} &=& \frac{11}{12}\\ \frac{1}{2+x}\cdot (\frac{3}{x-1} + \frac{8}{10+x} ) &=& \frac{11}{12}\\ \frac{1}{2+x}\cdot \left(\frac{11x+22}{(x-1)(10+x)} \right) &=& \frac{11}{12}\\ \frac{1}{(2+x)}\cdot \frac{(11x+22)}{(x-1)(10+x)} &=& \frac{11}{12} \qquad & | \qquad : 11\\ \frac{1}{(2+x)}\cdot \frac{\frac{11x+22}{11}}{(x-1)(10+x)} &=& \frac{1}{12}\\ \frac{1}{(2+x)}\cdot \frac{(x+2)}{(x-1)(10+x)} &=& \frac{1}{12}\\ \frac{1}{(x-1)(10+x)} &=& \frac{1}{12}\\ (x-1)(10+x) &=& 12 \\ 10x+x^2-10-x &=& 12 \\ x^2 +9x -22 &=& 0 \\\\ x_{1,2} &=& \frac{-9\pm\sqrt{81-4(-22)}}{2} \\ x_{1,2} &=& \frac{-9\pm\sqrt{81+88}}{2} \\ x_{1,2} &=& \frac{-9\pm\sqrt{196}}{2} \\ x_{1,2} &=& \frac{-9\pm 13}{2} \\\\ x_1 &=& \frac{-9+ 13}{2} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2} \\\\ x_2 &=& \frac{-9- 13}{2} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{-11} \end{array}$

Very impressive Heureka