$\frac{\sin6x}{1-\cos6x}$
$4\cdot x^4+4\cdot x^3-3\cdot x^2-2\cdot x+1$
$\lim_{n\to\infty}\left(\frac{1}{\sqrt{n}}\right)^2$
$-3\left(3z^2+z-1\right)$
$5.5+4.75$
$\lim_{x\to0}\left(\frac{tan\left(2x\right)}{ln\left(1-x\right)}\right)$
$\left(z^6+9\right)^2$
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