$37,279+5,4$
$\int\left(3x\:-\:6\right)^5dx$
$\lim_{x\to0}\left(\frac{x-3}{x^2-4x+3}\right)$
$\left(x+1\right)\left(x-1\right)+\left(3+x\left(3-x\right)\right)$
$6x-2\left(x+3\right)\le x\left(5-x\right)-7$
$2-x^{2}=4x$
$\int_{0}^{2}\left(\frac{1}{x^{2}}\right)dx$
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