$8\left(2j+3\right)+4j$
$\lim_{x\to0}\left(\frac{2x^2+2x}{x^2-5}\right)$
$\frac{x^6}{x^3}$
$14ab-28a^2b^2^2$
$3+2\cdot10+\left(1-\left(12+1\right)\right)-8-8\cdot\left(-2\right)$
$\frac{u}{6}\cdot6$
$3m-6=24$
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