6. For all non-negative values of the real variable $x$, the function $f(x)$ satisfies the condition $f(x+1)+1=f(x)+\frac{20}{(x+1)(x+2)}$. Calculate $\frac{2019}{f(2019)}$, if $\quad f(0)=2019$.
# Solution.

Notice that

$f(x+2019)-f(x)=(f(x+2019)-f(x+2018))+(f(x+2018)-f(x+2017))+\cdots+(f(x+$ 1) $-f(x))=\frac{20}{(x+2019)(x+2020)}-1+\frac{20}{(x+2018)(x+2019)}-1+\cdots+\frac{20}{(x+1)(x+2)}-1$. Therefore, $f(2019)-f(0)=20\left(\frac{1}{2019}-\frac{1}{2020}+\ldots+1-\frac{1}{2}\right)-2019=20\left(1-\frac{1}{2020}\right)-2019$. Thus, $\frac{2019}{f(2019)}=\frac{2020}{20}=101$

Answer: 101.