$4\cdot49$

$\lim_{x\to0}\left(\frac{1-\ln\left(1+x\right)}{x^2}\right)$

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

$\left(x-120\right)\left(x+120\right)$

$\left(5^6\right)^3$

$\left(2n+5\right)\left(2n-5\right)$

$\left(-3+\frac{6}{x}\right)$