$\left(2p^{3}+q\right)^{2}$
$-\:1\:+\:1\:+\:1\:-\:1$
$\int\frac{2x}{\sin\left(x^2\right)}dx$
$x\:+\:6\:\ge\:2x\:-\:8$
$\frac{\left(2m^{-2}p^{-4}q^3\right)^{-3}}{2m^2p^2q^3\cdot2p^0q^0}$
$\left(x+1\right)dy=\left(x+6\right)dx$
$-9-\left(-8\right)$
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