$\lim_{x\to\infty}\left(3x+9\right)^{\frac{1}{x}}$
$9x^2+9$
$\frac{d^2}{dx^2}\left(x^{-7}\right)$
$\left(e^{\frac{y}{x}}\right)^2$
$\frac{x^2-10x+21}{3x-12}=\frac{x-5}{x-4}$
$-4+5-6\left(3+1\right)$
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