$\frac{du}{dx}=2u$
$\left(3y+4z\right)\left(3y-4z\right)$
$18a^5b-9a^3b^2-6a^2b^3$
$\lim_{x\to\infty}\left(\frac{7x^4-x^3+3}{4x^5+x^2+4}\right)$
$4x^2+7x=0$
$7a\cdot\left(-4\right)$
$\left|\left(+45\right)-\left(-4\right)\right|-\left(+17\right)$
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