$\lim_{x\to\infty}\left(0\right)^{-2x}$
$\left(3-4\right)^3-\left(2^0-6^1\right)^2-5^0$
$500b-6b+14b-6b-\left(7b-2b\right)+\left(4b-b\right)$
$\frac{\left(x-6\right)^2}{n\left(-7\right)^n}$
$32x-4+5x-10+5x^3-2x^2+5x+8$
$3x^2-18x-81=-6$
$\int\:\frac{3}{x^2-4x+3}dx$
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