$7x\:-\:6\:+\:3x\:\le\:6x$
$\lim_{x\to0}\left(\frac{1+x-e^x}{x^3}\right)$
$-6\:\cdot7$
$4-\left(5-2\right)+\left(3-4\right)-\left(6-3\right)$
$\frac{4x^3+7x+9}{2x-1}$
$\left(\frac{x^2}{2}-\frac{1}{x}\right)^2$
$6+x\le-1$
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