$\frac{d}{dx}\left(sqrt\left(\frac{x-4}{x^8+1}\right)\right)$
$\left(y+3\right)^2=-8\left(x+5\right)$
$\frac{c+10}{c+4}\:=\:\frac{c-4}{c+2}$
$x^2-10x+21=0$
$81x^4+198x^2+121$
$-18x^3-9x^3-6$
$\frac{3}{x}+\frac{5}{x-2}=1$
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