$3x\:-\:y\:+\:4\:+\:2y\:-\:9\:$
$\frac{\left(3x+4\right)}{5}+2\le\frac{\left(x+3\right)}{4}$
$\lim_{x\to\infty}\left(\frac{x+1}{x^2+4x+3}\right)$
$\frac{2x+x^2-24}{x-4}$
$\lim_{x\to\infty}\left(\frac{5xe^{-2x}}{x^2+3x+1}\right)$
$-2\cos\beta+14=12$
$3x+5\ge7$
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