$\left(4c+2\right)\left(4c-3\right)$
$24x^3\:-\:36\:ax^2$
$\frac{1}{16}\int xe^{\frac{-x}{4}}dx$
$\left(-7n-1\right)-\left(n+2\right)$
$y'\:=\:\frac{3y^2-x^2}{xy}$
$8x-2\le14$
$\int\frac{1}{\left(x^2+16\right)}dx$
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