$\frac{9x^3-12x^4+21x^2+3}{3x}$
$10x+9=4x-9$
$x+10+x+12$
$\left(\left(-4\right)x\left(-4\right)x\left(-4\right)x\left(-4\right)\right)$
$-\left(-6x-6\right)^2$
$\lim_{x\to\infty}\left(6sqrt\left(x^2+7x\right)-6x\right)$
$\lim_{x\to\infty}\left(\frac{x^2}{lnx}\right)$
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