$2^3.3^3.5^3$
$\left(5\right)+\left(7\right)^2$
$\:3x-14>x-8$
$\left(2x^{-5}-7\right)\left(2x^{-5}+7\right)$
$x-5+x+3$
$\frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{\left(x-1\right)\left(x-2\right)}$
$5\sqrt{x^3y}$
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