$\frac{\frac{\frac{4t^2-16}{8}}{t-2}}{6}$
$+8-7+6-5+6-4-2+15$
$\lim_{x\to\infty}\left(\frac{\left(3n^2-4\right)}{2n^2+n-1}\right)^{4n}$
$2\cos\left(3\theta\:\right)=-1$
$\lim_{x\to\infty}\left(4^{\frac{x+2}{x+1}}\right)$
$1-x-x^2>0$
$12x-4y=20$
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