$12\sqrt[3]{a^2b}$
$\lim_{x\to\infty}\left(\frac{x^2+3}{x+1}\right)$
$x^2+2x=6$
$-\left(-2\right)^2-5\left(-2\right)+7$
$\left(-234\right)+\left(-63\right)+\left(-91\right)+\left(-21\right)$
$5x+12\:con\:x=5$
$10^{x-3}=1$
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