$\left(3x-4\right)\left(x+3\right)-\left(x-2\right)^2=3$
$\frac{d}{dx}\left(\frac{x^4\cdot\left(x+1\right)^2}{\left(x-3\right)^3}\right)$
$\frac{dy}{dx}=\frac{x}{9y}$
$4a^2+4a+1$
$3x-2y+z=8$
$3x^2+5=18x$
$\frac{2}{5}\left(5x^2-3x^{\frac{1}{3}}\right)$
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