$3a^59$
$\:\left(-2x^2+3\right)^2$
$-2.5\left(8-b\right)$
$\lim_{x\to\infty}\left(\frac{2x^3+4}{x^5}\right)$
$\frac{\left(2x^3z^2\right)^3}{x^3y^4z^2\cdot x^4z^3}$
$\left(3x^3-2\right)\cdot\left(x^2-1\right)$
$\frac{d}{dx}\log\left(x^2+3\right)$
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