$\frac{1-a^5}{1+a}$

$\frac{1}{\sec\left(x\right)^2}=\sin\left(x\right)^2.\cos\left(x\right)^2+\cos\left(x\right)^4$

$-2\tan^2x=2\sec^2x$

$\frac{1}{\left(2x\right)^{\frac{5}{2}}}$

$\frac{x}{8}=8$

$\left(x\frac{1}{x}\right)^3$

$\lim_{x\to\infty}\frac{5^{x-5}}{2^{x+2}}$