$\frac{x-6}{3}\le x+4$
$\frac{2\cos^{2}x+5\cos x+3}{\sen^{2}x}=\frac{2\cos x+3}{1-\cos x}$
$274.32\:x\:2\:\frac{1}{4}$
$3125\:x\:3$
$\left(3n^{-3}\right)^2$
$\frac{27x^3+8}{x^2+16x+28}$
$\lim_{x\to\infty}\left(\frac{x^3}{\ln\left(x^5\right)}\right)$
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