$\frac{160x^{-3}y^2}{-40x^2y}$
$3x+2+5$
$\left(-2.3\right)-\left(1.25\right)+\left(34.28\right)$
$\lim_{x\to\infty}\left(\frac{3x-2}{3x+4}\right)^{3x+1}$
$\left(x^2\:z^3+2xy^2\:z^2\:\right)^5$
$\int\frac{2}{x^3\sqrt{x^2-9}}dx$
$3x-2>0$
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