$f\left(x\right)=\frac{\left(x+3\right)^8}{\left(3x-3\right)^9}$
$\lim_{x\to5}\left(4x-20\right)\left(x^2-3x-10\right)$
$\frac{3x^2-x+1}{x+1}$
$-27+476$
$16y^2-x^2$
$6-\left|3+\left(4-8\right)-2\right|$
$\frac{x}{\left(x^2-3x-4\right)}$
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