7. Let $f(x)=\frac{1010 x+1009}{1009 x+1010}$, define $f^{(1)}(x)=f(x), f^{(i)}(x)=f\left(f^{(i-1)}(x)\right), i=2,3, \cdots$, then $f^{(n)}(x)=$
Solve the fixed point equation $x=\frac{1010 x+1009}{1009 x+1010} \Rightarrow 1009 x^{2}+1010 x=1010 x+1009 \Rightarrow x= \pm 1$.
Thus, $\frac{f(x)+1}{f(x)-1}=\frac{2019(x+1)}{x-1}$, let $\varphi(x)=\frac{x+1}{x-1}, g(x)=2019 x$,
then $\varphi[f(x)]=g[\varphi(x)] \Rightarrow \varphi \cdot f=g \cdot \varphi \Rightarrow f=\varphi^{-1} \cdot g \cdot \varphi \Rightarrow f^{(n)}=\varphi^{-1} \cdot g^{(n)} \cdot \varphi$.
Since $\varphi^{-1}(x)=\frac{x+1}{x-1}, g^{(n)}(x)=2019^{n} x$, substituting yields
$$
f^{(n)}(x)=\frac{2019^{n} \cdot \frac{x+1}{x-1}+1}{2019^{n} \cdot \frac{x+1}{x-1}-1}=\frac{2019^{n}(x+1)+(x-1)}{2019^{n}(x+1)-(x-1)}=\frac{\left(2019^{n}+1\right) x+2019^{n}-1}{\left(2019^{n}-1\right) x+2019^{n}+1} .
$$