$4a+8$
$\lim_{x\to\infty}\left(\frac{1}{x^3-5x^2-3}\right)$
$-4+5-2+5+2-2+3$
$f\left(4\right)=\frac{4\left(4\right)-10}{x-2}$
$5x\cdot\left(5x+3\right)\cdot\left(5x-3\right)$
$\frac{\cos\left(x\right)\cdot\tan\left(x\right)}{\frac{1}{cos\left(x\right)}}$
$x^3-5x^2+4x-6\:entre\:\left(x-4\right)$
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