$\frac{5x-3}{6}+\frac{x-5}{18}>\frac{x+1}{3}$
$5x+7=2\left(x+8\right)$
$\int\left(cotan\left(ax\right)\right)dx$
$2x+1=x-8$
$\left(2t^5-3\right)^2$
$18x-5\:4x+1\:18x-5\:4x+1$
$\left(-3x+7\right)\left(4x-9\right)$
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