$4+3x_{-1+x-5y}$
$x+12=8$
$\frac{m^{-2}}{4m^2\:}\:$
$8\frac{1}{4}+3+3$
$\frac{dy}{dx}=\frac{x\sqrt{\left(1+y^2\right)}}{y\sqrt{\left(1+x^2\right)}}$
$cos^2\left(x\right)tan\left(x\right)=1-cos^2\left(x\right)$
$\lim_{x\to-f}\left(\frac{y+f}{f^2-y^2}\right)$
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