$\left(a^3+4ab^4\right)^2$
$\frac{0}{x+1}$
$\frac{3x}{4}-9=-6$
$\int\left(\frac{\left(9\right)}{\left(16-25x^2\right)}\right)dx$
$x^2+5x+6=\left(x+2\right)\left(x+3\right)$
$-4x+9<x-1$
$\left(\frac{x-1}{x}\right)^4$
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