$\lim_{x\to0}\left(\frac{x+5}{x}\right)^{-4x}$
$y'=\frac{x^2}{1+y^2}$
$2.6r\:+\:1.3\:=\:-16.9$
$\frac{\left(x+2\right)}{2x-3}>4$
$\lim_{x\to0}\left(\frac{-e^x+1}{e^x}\right)$
$3x+4y=5x-4y$
$\left|b^2cz^5+\frac{2}{3}b^2cz^5-\frac{5}{3}b^2cz^5\right|\cdot4$
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