$\frac{-3x^3-18x^2+21x+180}{x+4}$
$\frac{e^2\left(1-3x^4\right)}{\left(x^4+1\right)^2}$
$-33=3\left(x-4\right)-6x$
$\frac{3x+6}{3-3x}$
$2,5\:+\:18,75$
$36a^4-49y^8$
$\frac{4}{x}+\frac{x}{x+3}=\frac{9}{x^2+3x}$
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