$\left(3p^3+2m\right)^2\left(3m^3+2m\right)^2$
$2x^2+2y^2-4x-20y+40=0$
$\lim_{x\to\infty}\left(\frac{\ln\left(x\right)}{1+2\ln\sin\left(x\right)}\right)$
$x^2-12x^3$
$\sin^2\left(x\right)\cot\left(x\right)=\cos^2\left(x\right)$
$\frac{x^2-8}{x+3}$
$58.3\sqrt{24.3}$
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