$1+9+4x^2$
$\lim_{x\to3}\left(\frac{x-3}{x^{2}-3}\right)$
$x-4y+1=0$
$\:4\:\:.\:\:3\:\:-\:\:8\:\:.\:\:2\:\:.\:\:30\:\:.\:\:2$
$\frac{x^2+4x+4}{\left(x-3\right)}$
$\frac{8m^3-12m^2+6m-1}{4m^2-1}\cdot\frac{2m^2+m}{4m^2-2}$
$2x+\frac{1}{2}^2$
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