$\lim_{x\to3}\left(\frac{x^2-6x+24}{x^2-4x+3}\right)$
$\frac{1}{3a+3}=\frac{1}{6}-\frac{2}{a+1}$
$\sqrt{a^2b\sqrt{a^3b^5}}$
$\left(5\sqrt{2}\right)^2$
$5w\left(w-6\right)$
$15\left(x-2\right)$
$\frac{3x^3+13x^2-13x+2}{x-1}$
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