$x^5\sqrt{x^3}+1$
$\frac{2}{3\:}\left(x+\frac{1}{2}\right)=-1$
$b^2+6b+5$
$\frac{w}{6}+13=-2.36$
$y^5+y=0$
$\lim_{x\to infinity}\left(\frac{\sqrt{x+1}}{x^2-1}\right)$
$\left(5x+2\right)\left(x-1\right)$
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