$\left(x+5\right)\left(x+4\right)^2$
$\int t^{-u}dt$
$x\cdot2-6=x-6$
$2x+5>17$
$-7y\cdot\left(\frac{-5y}{y^2}\right)$
$\left(\frac{x}{2}+2\right)^2$
$\left(-10^8\right)$
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