$x^2+3x-10>0$
$x-\frac{1}{8}<-\frac{3}{4}$
$\int\frac{-\left(4x+4\right)}{x^3-4x}dx$
$\lim_{x\to0}\left(\frac{4x-1}{x}\right)$
$2b^3+b^2+b^2+b+b^3+b+b$
$\frac{x^4+2}{x^2+4x+1}$
$\left(2-5\right)+\left(-3\right)-\left(-11\right)$
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