- TeX source:
- \begin{array}{rclll} \dfrac{x}{8-x^2} &=& -\dfrac{1}{x^2-8} &\vert & \cdot \left(8-x^2\right) \cr x &=& -\dfrac{-(-8+x^2)}{x^2-8} \cr x &=& \dfrac{x^2-8}{x^2-8} \cr x &=&1 \;\in\;\mathbb{D} \cr\cr\mathbb{L} &=& \left\{1\right\} \end{array}