$2^3-2^2$
$e^8=y$
$\sqrt[5]{a^7b^4c^{12}}$
$\frac{1}{yx}=\frac{\left(1+e^{-x}\right)^{2x}}{x}$
$\cos\left(6x\right)^2$
$\frac{1}{cos^2\alpha\:}+\frac{1}{sen^2\alpha\:}=\frac{1}{sen^2\alpha\:\cdot cos^2\alpha\:}$
$\lim_{x\to-0.1}\left(\frac{sen\:2x}{4x}\right)$
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