$\lim_{x\to2}\left(2x+3\right)\left(x^2+5\right)$
$\frac{2}{x^{\frac{1}{3}}}=2$
$3b^2-9b$
$7\sin^2$
$16a^2+12a$
$\left(y^2+1\right)=\left(1+xy\right)\left(\frac{dy}{dx}\right)$
$-4x+3y-z=16$
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