$$8-2\left(x-8\right)<4+x$$

$$y-2xy'\:=0$$

$$\left(x-\frac{2}{3}\right)\left(x-\frac{1}{6}\right)$$

$$\int\left(\frac{4}{\sqrt[3]{3+6x}}\right)dx$$

$$y'\:-\:y^2\:-\:y\:=\:0$$

$$1+\tan^{2}\beta=\frac{1}{\cos^{2}\beta}$$

$$\lim_{x\to2}\frac{\sqrt[3]{\left(x^3-5x^2+8x-4\right)}}{x-2}$$