$\frac{3\left(3x+1\right)+2\left(x+2\right)}{6}<-x$
$x^2-12x-56=0$
$\frac{x^3-6x^2+3x+8}{x+1}$
$\left(x^2+9x\right)\left(-1x-9x\right)$
$2x+y=8$
$\frac{4^2-121}{2n+11}$
$\lim_{x\to\infty}\left(\frac{ln\left(4x+5\right)}{ln\left(2x+8\right)+1}\right)$
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