$4\:5^2-4$
$4\left(x+1\right)-6x>4$
$8xy'-9y=2x^5,\:y\left(1\right)=8$
$\frac{2}{x+3}=\frac{8}{x+6}$
$\frac{x^2-8x+3}{x-5}$
$\left(14x+9\right)^2$
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