$\frac{1}{x+3}=\frac{1}{x-3}$
$\ln\left(x^3-4\right)-\ln\left(x^2-2x\right)$
$\frac{z^3}{\left(1+3z^2\right)^2}$
$\frac{d}{dx}\left(2xy+3x^2y^2=1\right)$
$6x-7<5x$
$2x-10\:en\:x=2$
$519-\left(-1294\right)$
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