$\left(3x^{\frac{1}{4}}+4\right)^2$
$\lim_{x\to+\infty}\left(\frac{1\cos\left(x\right)}{x^2}\right)$
$cosh\left(jx\right)=jcosh\left(x\right)$
$\frac{20x^5-15x^6}{-5x^5}$
$14m^2+22m-15\:entre\:7m-3$
$8-x>4$
$2x+5>8x-3$
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