$30+20+50+100+12$
$3\left(-2\left(1\left(7\right)\right)\right)$
$6x-6=48$
$\int\frac{\left(x^2+4x+3\right)}{\left(x+5\right)^3}dx$
$-4\left(u+5\right)+2u$
$-a^2+3a+28$
$\left(-\:1\:.\:3\right)^3$
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