$\frac{dy}{dx}=\sqrt{x}y^3$
$10y\:+12+7y\:+8+3y$
$2.8+\frac{2.8^2}{2}+2.8$
$\int\left(\left(x^3\left(1+x^4\right)^7\right)\right)dx$
$3j^{-3}x-6j^{4\:}$
$5ab+7b-12ab-5b$
$\frac{d}{dx}\left(y=\frac{x^2\left(x-1\right)^3}{x+3}\right)$
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