$0.5\left(5-7x\right)=8-\left(4x+6\right)$
$\lim_{x\to3}\left(\frac{e^{x-4}}{x^2-9}\right)$
$5\left(m-8\right)\left(m+15\right)$
$\lim_{x\to\infty}\left(x^3+7x^2+1\right)$
$\frac{x^2-4x+1}{x+3}$
$\left(e^2+4\right)dy=ydx$
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