$-6-x\ge0$
$\lim_{x\to-1}\left(\frac{x^3-1}{x+\left|1\right|}\right)$
$x+n=6$
$\frac{1}{w^{-5}}$
$\left(x+\frac{q}{s}\right)^2$
$-\frac{1}{3}.3\:mn\:.3\:m\:^2$
$f\left(x\right)=\frac{5}{x^{10}}-\frac{5x^{-4}}{6}$
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