$\lim_{x\to\infty}\left(\frac{2^xln2}{2x}\right)$
$\frac{2x+5}{-4x+8}$
$\cos^2\left(\theta\right)=\frac{1}{15}$
$9m^2-36n^2$
$4x^{2\:}+\:12x\:+5$
$800-170<630$
$-\frac{168}{7}$
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