$\frac{1}{sqrt}$
$\left(x^3+8\right)\left(x^4-16\right)\left(x^2-y^2-4x+4\right)$
$\frac{x^{-6}.x^7}{x^6}$
$2x-1>13$
$\int_{\sqrt{3}}^2\left(\frac{\sqrt{x^2-4}}{x}\right)dx$
$\lim_{x\to2}\left(4x^2-8x-5\right)$
$8x-1>6x+4$
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