$\lim_{x\to-6}\left(\frac{\left(x^3-6x-130\right)}{x^2-x-42}\right)$
$\left(4z^m\right)\left(nz^2\right)$
$\left(2x^2+4y\right)^2$
$\left(-4\right)^2\left(4\right)$
$-8=4.x$
$\frac{4^{x+3}}{2^{x-1}}=64$
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