$3x^2=81$
$\frac{24x^6}{12x^{-8}}$
$\lim_{x\to0}\left(\frac{3^{x}-2^{x}}{4^{x}-3^{x}}\right)$
$4\left(n+3\right)-\left(3+n\right)$
$-4x+14+-2x-9$
$-3=6+u$
$36x^2-9x+36=-1$
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