$\lim_{h\to0}\left(\frac{\frac{1}{5+h}-\frac{1}{5}}{5}\right)$
$\frac{x^5+x^4-8}{x^3+4x}$
$\frac{151x+1}{5}-\frac{1}{3}$
$y-x^2+y$
$\left(-m^2n\right)\left(3m^2\right)\left(-4mn^3\right)$
$4\sin^2x=1$
$m^2+81+9m^2$
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