$f\left(x\right)=\left(\frac{1+2x^2}{7-8x}\right)^4$

$\left(\tan+\cot\right)^2$

$\tan\left(x\right)=\sin\left(x\right)\cdot\cos\left(x\right)$

$\lim_{x\to1}\left(\frac{\sqrt{x+3}-2}{\ln3x}\right)$

$\left(-x\right)\cdot\left(4x+5\right)$

$-\frac{x^2}{x^2+1}+\frac{x^4+1}{x^4-1}$

$10-15d-2+17d$