$\lim_{x\to+\infty}\left(\frac{\ln\left(x^2e^x\right)}{1+e^{3x}}\right)$

$\frac{2\cdot\:3^{n+4}-3^4\cdot\:3^n}{3\cdot\:3^{3+n}}$

$\frac{dy}{dx}=\left(\frac{2xz}{z^2-x^2}\right)$

$\frac{dy}{dx}=\cos^2\left(2x\right)\cos^2\left(2y\right)$

$9x^2y^4-121z^2$

$\frac{3x}{3x+2}\cdot\frac{3x-2}{2x+3}$

$4x+10>4-2x$