$\frac{4a^2b^5}{-2ab}$
$z^2\:+\:z^2\:-\:18z\:+\:18z\:+\:18\:+18$
$3x+1=1000$
$\frac{a^1}{a^{-1}}$
$\lim_{x\to\infty}\left(\left(6x+1\right)\cdot\frac{1}{n+1}\right)$
$\:x^3\:-\:3x^2\:+\:2x-\:2\:entre\:x\:-\:1$
$\left(\frac{3}{5}k^2-2y^7\right)^5$
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