$\frac{x+3}{3}\:-\frac{x+2}{4}\:>\frac{x}{3}$

$\int\frac{5x^2}{\left(x^2+6\right)^{\frac{3}{2}}}dx$

$\frac{x+3}{4}=\frac{2x-1}{7}$

$\int\left(\left(8x-12\right)\left(4x^2-12x\right)^2\right)dx$

$x+\left(x+2\right)+\left(x+2\right)+x+\left(x+2\right)+\left(x+2\right)+\left(x-1\right)+\left(x-1\right)$

$3x^2+6y+12xy$

$7x<42$