$\int_0^4x\left(3x-x^2\right)dx$
$\lim_{x\to\infty}\left(\frac{x\:^3-8}{x^2-2x}\right)$
$\tan\left(x\right)=0.024$
$\frac{1}{2x}+\frac{3}{4}-\frac{5}{12x}=\frac{5}{6}$
$\left(5x^2-4x\right)^2$
$x^6-1=0$
$\left(3-8\right)\cdot\left(-2.5\right)$
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