$\frac{x-3}{2}=\frac{2x+4}{5}$
$\int\sec^2\left(x\right)\cos\left(\tan\left(x\right)+1\right)dx$
$\frac{x^2+3x+11}{x+1}$
$4r^{-3}2r^2$
$9m^2+24mn+16n^2$
$\frac{\left(cosx\right)}{1-sinx\:}=secx+tanx$
$\lim_{x\to6}\left(\frac{x^2-x-30}{x-6}\right)$
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