$x^3+2^2-4x-8$
$a^2-a+4a-3a^2+1$
$133\:,\:y\:=\:19$
$\int\frac{4x+12}{\left(x-2\right)\left(x^2+4x+\right)}dx$
$0+4cos\left(\pi\right)\sin\left(\frac{\pi}{8}\right)$
$\left(-20xy^2+6x\right)dx+\left(3y^2-20x^2y\right)dy=0$
$4\left(a+10\right)-8a$
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