$\frac{x^2}{x^4}$

$\lim_{x\to\infty}\ln\left(x\right)^{\frac{16}{x}}$

$\frac{\cot\left(x\right)-1}{\csc\left(x\right)}=\cos\left(x\right)+\sin\left(x\right)$

$5ab-4ac$

$\frac{d}{dx}\left(x^2-2xy+y^2-6x+2y=0\right)$

$5b^3-2\left(b^3+2\right)-4\cdot\left(b^2+2\right)+5-4ab^2$

$x^2+4x=1$