$5t^2\left(-t^{-2}\right)$
$\frac{d}{dx}2x\sqrt{y}+4y=6$
$\frac{2\left(a^2b^{-1}\right)^2}{4\left(3a\right)^{-5}}$
$\int-3x\:\cos\left(4x\right)\:dx$
$6^8\cdot6^3$
$2x+1+3x-1$
$4x^2-16x+4y^2-32y=0$
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