$\frac{\left(1-3\right)^2}{-1}+\frac{\left(-3-1\right)^2}{1}$
$\lim_{x\to infinito}\left(\frac{2x+1}{2x-4}\right)^{6x}$
$5x-7=4x+3$
$4^2+2\left(0\right)$
$8x^2+2x-8xy+2$
$\int_{-2}^1\left(\frac{x}{\sqrt{6x^2+1}}\right)dx$
$\:21\left(6x\right)^3$
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