$\frac{-3x+1}{6}<\frac{1}{3}$
$\frac{d}{dx}\left(10e^y+\tan\left(x\right)\right)=y$
$\frac{-1+b}{2}\le\:6$
$17+13+32-2x$
$\left(14x+9\right)-\left(8x+2\right)$
$0.9995\:-\:0.9877$
$\left(3x^2-2y^2+1\right)^2$
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