$8^2-3^2\cdot\sqrt{16}$
$\frac{d}{dx}\:x^5-6y^3+y^4=1$
$\frac{d}{dx}\left(x^{\frac{3}{5}}+y^{\frac{3}{5}}=3\right)$
$y=6y+10\sin\left(2x\right)$
$17+1$
$\frac{1}{3x^2+x-2}$
$x^2+\:x\:\ge\:12$
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