$\lim_{x\to6}\left(\frac{3+\sqrt{x}}{\sqrt{3+x}}\right)$

$\frac{\left(3\left(-\frac{1}{2}\right)^3+\left(-\frac{1}{2}\right)^2+\left(\frac{3}{4}\right)-\left(\frac{1}{2}\right)+\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)}$

$\frac{x^2-y^2}{x^3+y^3}$

$\left(r-7s\right)^2$

$m^9-8$

$12x^2-5x-3$

$x+14=16$