$6\left(3\right)^2\left(-2^2+3\right)\left(-2^2\right)\left(3\right)$
$-9+-5+1$
$\lim_{x\to\infty}\left(\frac{lnx}{ln\left(4x\right)}\right)$
$\frac{x^2-13x+39}{x-6}$
$3x+8=-2x-17$
$ln\left(\sqrt[3]{1+x^2}\right)$
$\sin^2x\left(\frac{senx}{cosx}+\frac{cosx}{senx}\right)$
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