$\lim_{x\to\infty}\left(\frac{x+2^n}{2n+3^n}\right)$

$\left(\frac{-4u^7}{u^{\frac{1}{4}}}\right)^2$

$2x=y^2$

$\left[\left(a+2b\right)^2-\left(a-2b\right)^2+a^2+16b^2\:\right]-\left(4b-a\right)^2$

$\frac{3x}{\left(2x-3\right)}+\frac{1}{\left(4x^2-9\right)}-\frac{6x^2}{3+2x}$

$\frac{3x^2y}{6x^2y^2}\frac{+12x^3y^2}{+24x^3y^3}$