$x^2+9x+0$
$\:8a\left(3a\:-\:5y\:-\:2z\right)\:-\:6y\left(4a\:-\:6y\:+\:3z\right)$
$\left(x+1\right)\left(y+2\right)=1$
$y'+y=xy^3$
$\frac{9x^6-y^2}{3x^3+y}$
$x-10\leq5$
$\frac{x^3}{\left(x^4+3\right)^2}$
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