$\frac{x^2}{3x}$
$2x-3=-x-9$
$\frac { 4 } { 2 } | _ { x }$
$x^2-9>0$
$\frac{5\left(x+2\right)}{x+8}-\frac{3\left(x-2\right)}{x+8}$
$\left(a^2+b^3-c^2+3xy^3\right)^0$
$\left(7+a^2b\right)^3$
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