$\frac{2x^3-4x^2-5}{x^2+3}$
$\sqrt[4]{y^3}$
$\left(\frac{-1}{2}\right)^{-3}$
$2x^2<18$
$\left(\sqrt[2]{3}-2\right)\left(\sqrt[2]{3}+2\right)$
$6x^2-3x+8<0$
$\frac{d}{dx}\left(y\sqrt{1-\sin\left(x\right)}=\sqrt{1+\cos\left(x\right)}\right)$
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