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

$\lim_{x\to\infty}\left(\frac{ln\left(x^6-5\right)}{ln\left(x\right)cos\left(\frac{1}{x}\right)}\right)$

$2^{-3}\cdot2^9$

$24a^2-12ab$

$\lim_{x\to\frac{1}{3}}\left(\frac{3x+7}{6x+2}\right)$

$\left(\frac{1}{2}m+\frac{2}{5}\right)\left(\frac{1}{3}m-\frac{1}{2}\right)$

$y=x^2-2x^{-1}$