$x^4-49c^8$
$x+\frac{9}{x}=-10$
$\left(0.3\right)^2$
$\left(2x^{2}y^{3}z^{4}-4x^{3}y^{4}z^{5}\right)^{3}$
$\frac{dy}{dx}=\:\frac{x\left(e^{x^2}+\:2\right)}{6y^2}$
$-17+1+80+9$
$\frac{x^3-1}{x^2\left(x-2\right)^3}$
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