$-9x^2y\left(-7x^3y^4-8x^{-2}y^2\right)$
$\left(2x^2-x-3\:\cdot\:x^3-1\right)\left(3\right)$
$\frac{6z}{2\left(z^{2}\right)^{5}}$
$-3+2a^3-3a^2+4a$
$\lim_{x\to4}\left(\sqrt{4-x}\right)$
$\frac{5}{3}z-3=7$
$\lim_{x\to1}\left(\frac{x^2-x}{\sqrt{2x+2}-2}\right)$
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