$4x\:3\:+4x^3+4x\:3$
$\frac{dy}{dx}=\frac{y^2+1}{8x^2-6x+1},\:y\left(0\right)=\sqrt{3}$
$16+8x^2+x^4$
$\left(4c\right)^3$
$\frac{3xy^2-y^2}{y}$
$\lim_{x\to2}\left(2x^2-2x^3+5x-9\right)$
$2\cdot10^8$
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