$\left(3m^2-4n^4\right)^3\left(5m^3+\right)$
$\frac{-5}{4}x\le-1$
$0=-0$
$\int\frac{1}{x\left(2x+5\right)}dx$
$\lim_{x\to-8}3x^2-22\left(x+25\right)$
$-3\left\{-2+6\left[-3+2-\left(-3+0\right)+1\right]+4\right\}+0$
$\frac{x^3+0x^2-2x+4}{3x^2-1}$
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