$f^2-6f+8$
$4x+10\le-2x-8$
$\left(x-6\right)\left(4x+3\right)$
$\lim\:_{x\to\:0}\left(\frac{3x^3-6x}{5x^3+7}\right)$
$2x^2-3\le5$
$3x+5<x+13$
$4^{10}$
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