$\left(y-20y\right)^2-400$
$\sqrt[3]{x}-\frac{1}{\sqrt{x}}$
$\left(\frac{3x}{x-3}\right)+\left(2\right)=\left(3x-1\right)$
$\frac{\left(2x-1\right)}{x^2-12x+35}$
$\int\frac{6x^2}{\sqrt{x^3}-1}dx$
$\lim_{x\to\infty}\left(\frac{3x+4}{4x^3+3}\right)$
$9y^2-16x^2-36y-64x+116=0$
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