$\left(\frac{3a^4b^5}{3ab^2}\right)^3$
$\lim_{x\to1}\left(\frac{\sin\left(2x-2\right)}{x-1}\right)$
$3\sec^2x=4$
$24x+18y$
$\lim_{x\to0}\left(\frac{x}{\sqrt{\left(\left(4x^2+25\right)\right)}}\right)$
$2+e^{infinito}$
$\left(1\right)=\log_a\left(1\right)$
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