$-2\sin x=\tan x$
$-4x^4+16$
$\lim_{x\to\infty}\:\frac{arctan\left(x^3\right)}{x^8}$
$\frac{2}{5}x-3\ge2x+8$
$x^2+64=-20x$
$\frac{2\infty^3\infty^2+3}{\infty^2}$
$\int\left(x^2-9x+7\right)\cdot e^{-x}dx$
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