$f\left(x\right)=\frac{\left(1\right)}{\left(8+x\right)^2}$
$2x+3x+12x$
$\left(\frac{1}{4}-3x^2\right)\left(\frac{1}{4}+3x^2\right)$
$\frac{1}{1+\tan\left(x\right)}=1$
$t^2+4t+3$
$\frac{dy}{dx}=\frac{3}{2}\left(y-1\right)^3$
$2x+x+2x+x$
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