$\left(4\right)^4\left(3\right)^3\left(\frac{1}{2}\right)^3$
$\frac{1}{2}y-2=\frac{1}{3}y$
$\frac{24x^2y-36xy^2+60xy}{12xy}$
$xy'-xy=4y$
$\sin^{-1}\left(-0.75\right)$
$\left(m-12\right)\left(m+12\right)$
$\sqrt{81}\::\:\left(-3\right)^2$
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