$\frac{d^3}{dx^3}\:\frac{3}{\left(x-1\right)^2}$
$\left(x^2+4x+8\right)-\left(x^2-4x+8\right)$
$y'=\frac{-1}{4y-16x}$
$x^2+2+15$
$\frac{4}{x}-\frac{x}{2}=\frac{12}{x}$
$\left(-\sqrt{9}\right)+\sqrt{-4}-\left(2\sqrt{576}+\sqrt{-64}\right)$
$\left(-2x-2y+3z\right)^2$
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