$x+6=3x+22$
$-5+8-3+2-10-12-17-16+5+4$
$-20-4-365+55-12-8$
$\lim_{x\to0}\left(\frac{x^2+3}{\left(2x-1\right)\left(x+1\right)}\right)$
$0=\frac{x+3}{7x-28}-\frac{6}{x-4}-\frac{2}{7}$
$\sqrt[12]{81^3}$
$\left(x+3\right)^2+3x=\left(x-3\right)^2+6\left(14-x\right)$
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