$\frac{3}{6}+\frac{1}{6}+\frac{4}{12}$
$\lim_{x\to\infty}\left(\sqrt[3]{n^2}-n^3\right)$
$\left(\sqrt[4]{2}\:-\frac{3}{2}x\right)^2$
$-\sqrt[4]{625}$
$-\left(\frac{1}{2}\right)^{-2}$
$\frac{dy}{dx}=\frac{2+\sin\left(x\right)}{\frac{1}{y}}$
$\frac{x}{-8}+9>13$
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