$\frac{2}{4+4x}$
$3x+2<x-8$
$\lim_{x\to1}\left(\frac{ln\left(x\right)-x-1}{e^x-e}\right)$
$-1+\frac{a^2}{25}$
$\int\frac{x-3}{x^2-6x+8}dx$
$1+\:\left(-77\right)\:+\:10$
$ln\left(\frac{\left(x+\sqrt[2]{x^2-4}\right)}{2}\right)$
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