$\frac{\sqrt{6b^2}}{\sqrt{3b}}$
$\lim_{x\to\infty}\left(\frac{\left(3x^2-3x^3\right)}{\left(x^3-x\right)}\right)$
$\left(x+7\right)\left(x-7\right)=-3x$
$8\cdot30$
$-4x+20x+40=8$
$\frac{dy}{dx}=\frac{3y-3}{x}$
$-6\:.\:3$
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