$\frac{-3a^5+11a^3-16a^2+32}{-3a^2-6a+8}$
$\lim_{x\to4}\frac{\tan\left(5x\right)\left(\pi\right)}{4}$
$-8m>16$
$5.43\cdot10^7$
$\frac{9\left(a^2+1\right)}{n^2}+\frac{6\left(a-1\right)}{n}+1$
$\left(c^4+c^2-4\right)\left(c^4-c^{2\:}-4\right)$
$x^2\le3x+40$
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