$3x+1<4$
$\left(13a^{-2}b^3\right)x\left(8ab^5\right)$
$2x^2+16x-8<=2x^3+10$
$xdx+ydy\:=\:0$
$2b^2-6b+4$
$x^2\frac{dy}{dx}=\left(x+1\right)y$
$\lim_{x\to0}\left(\frac{6\cos\left(x\right)}{x^2}\right)$
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