$\frac{2x+6}{2}-2x=3$
$\left(\frac{2+9x}{3x}\right)\left(9-x\right)$
$5x-8\le6x-12$
$2+6c+5c+9$
$\frac{cos\:x}{1+sin\:x}+\tan x=cos\:x$
$\lim_{x\to\infty}\left(\frac{1+e^{-3x}}{9+5x^2}\right)$
$\:\left(x+6\right)^2-\left(x+4\right)^2-2\left(2x+5\right)$
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