$\frac{a^{-5}}{a^3}$
$6x+11>18-x$
$\int\frac{x+3}{x^2\cdot\left(x+1\right)}$
$\left(3\:-2\:+\:7\right)\:+\:\left(-5\:+\:4\:-\:6\:+\:2\right)+\:8$
$4\left(x-6\right)+6=6x-2$
$16\cdot64\cdot256$
$-933-158$
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