$1x^2-6x+6$
$\lim_{x\to\infty}\left(3x^3-6\right)^{\frac{1}{\ln\left(x\right)}}$
$-4\ge2x+2$
$\frac{2\left(4m+1\right)}{m-1}=\frac{5m}{m-3}$
$x^2-3x-108$
$-\left(-1^4\right)$
$x=\frac{6-x}{x}$
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