$\frac{3\left(x+4\right)}{6}-\frac{x-2}{5}=\frac{3}{10}\left(x+8\right)$
$\int_1^2\left(3x^2+1\right)dx$
$\frac{d}{dx}\left(5-x^2-4x\right)$
$-18+2$
$ydx-2xdy=0$
$\left(8x^2+y\right)^2$
$\frac{x^3+x^2-12}{x-2}$
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