$\frac{n-3}{n+9}=\frac{n-7}{n+3}$
$\frac{d}{dx}\left(3x+\frac{12}{2-x}\right)$
$\lim_{x\to\infty}\left(\frac{4x+4}{3}\right)$
$9-4x^2=0$
$-9\:+\:-19\:+\:-6\:+\:1\:+\:50$
$\left(+10\right)-\:\left(-9\right)$
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