$\frac{16}{-8}$
$4x^2\:+\:8x\:+\:16$
$-39\:\cdot\:7$
$\int\frac{1}{\left(2+\sqrt{x}\right)^4}dx$
$x^2-40x$
$x^2+3x-2>0$
$\frac{5}{y-1}-\frac{5}{y+1}=\frac{2}{y-2}-\frac{2}{y+3}$
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