$\left(\frac{3x^2}{2y^2}\right)^5$
$4\pi3$
$\left(x^2+3\right)y''$
$2x^3-4$
$0\:-\:8\:+\:4\cdot0^2\:$
$\lim\:_{x\to\:\:-3}\left(-3x\right)$
$\cos\left(\frac{4}{5}\right)\cos\left(\frac{3\pi}{2}\right)+\sin\left(\frac{4}{5}\right)\sin\left(\frac{3\pi}{2}\right)$
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