$\frac{3x+2}{5}>3x+4$

$\sqrt[3]{3x^2-y}\cdot\sqrt[3]{36x}$

$\lim_{x\to0}\left(\frac{\sqrt[4]{x}}{\ln\left(2x+1\right)}\right)$

$-4\left(x+6\right)+2\left(x-2\right)$

$\lim_{x\to a}\left(\frac{6^x-2^x}{x}\right)$

$\frac{-3}{\sqrt{1-9x^2}}$

$\frac{m}{3}=\frac{m+4}{7}$