$5=\frac{g}{8}$
$3^3-3$
$x^2+4x\:=\:20$
$\arccos\left(x+y\right)=\arcsin\left(xy\right)$
$\left(6x-2\right)\left(4x-3\right)\:$
$\int\left(\frac{1-x^2}{x^4}\right)dx$
$\frac{2}{x+1}-\frac{x+2}{x^2-1}$
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