$1=ae^{-x+1}$
$\lim_{x\to0}\left(\frac{x}{\arctan\left(x\right)^2}\right)$
$\left(xy+4z^2\right)\left(xy-4z^2\right)$
$\frac{-2x^2-5x+12}{x+4}$
$-\left(x+3\right)\left(x-5\right)$
$24m^3-4y^2+9$
$5\csc x-8=0$
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