$\left(3y-2\right)\left(5\right)$
$\frac{2x^2+3x+4}{x-1}$
$\frac{\left(6y+9y^3-5y^4-8+6y^5\right)}{\left(2y-1\right)}$
$\:\left(\frac{x\left(x+n\right)}{n}\right)^n$
$x^2\:+\:5x\:+\:2$
$x+x+1+1+1+1$
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