$\frac{p}{q\:}q'=\:\frac{3p^2}{p^2+4}$
$1+64m^3$
$dr-2dx=0$
$2x^2-24z-4+0$
$\frac{a^4-1^4}{a+1}$
$1+\ln\left(1\right)$
$\cos\left(d\right)\cdot\tan\left(d\right)=\sin\left(d\right)$
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