$\left(2x+3\right)^2=\left(2x+3\right)\:^2$
$8n=120$
$\frac{\left(3x^2-1\right)}{\left(x^3-x\right)^2}$
$\lim_{x\to-2}\left(\frac{x^2-2x-8}{x+2}\right)$
$7y^2+5y-38$
$2x-14=0$
$\sqrt{2^2}-2\cdot2$
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