$\frac{1-z^8}{1-z^4}$
$-4xy-9xy$
$\frac{3x-3y}{2x-4y}+\frac{2y}{x-y}$
$\lim_{x\to\infty}\left(\frac{e^x+1}{e^x-1}\right)^{\ln\left(x\right)}$
$\left(-13r\:-15\right)-\left(-9r\:+\:16\right)$
$3e^xtany+y'\left(2-e^x\right)sec^2y=0$
$x^2+6x\le0$
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