Using whole numbers, consider that you look for the zero of function
$$f(x)=(1-40x)^{-\frac {100}{5133}}-x-1$$ It is clear that $x=0$ it a trivial solution. Discarding this one, we can approximate the other building the $[1,n]$ Padé approximant of $\frac{f(x)}x$ around $x=0$. This will write
$$\frac{f(x)}x \sim \frac{ a^{(n)} x -\frac{1133}{5133}} {1+\sum _{i=1}^n b_i^{(n)} x^i }$$ where the coefficients $ a^{(n)}$ and $b_i^{(n)}$ are compued from the function and derivative values at $x=0$.
Then the approximate solution is just given by
$$x_{(n)}=\frac{1133}{5133\,a^{(n)}}$$
This will generate a sequence of rational numbers, the very first ones being
$$\left\{\frac{5815689}{418640000},\frac{17447067}{1725707120},\frac{24440327087}{243
9163381610},\frac{21844121555788638898671}{2184051507726409040779072}\right\}$$
Let us look at their decimal representation
$$\left(
\begin{array}{cc}
n & x_{(n)} \\
1 & 0.0101100974 \\
2 & 0.0100199631 \\
3 & 0.0100016513 \\
4 & 0.0099970864 \\
5 & 0.0099958134 \\
6 & 0.0099954326 \\
7 & 0.0099953131 \\
8 & 0.0099952742 \\
9 & 0.0099952613 \\
10 & 0.0099952569 \\
11 & 0.0099952553 \\
12 & 0.0099952548 \\
13 & 0.0099952546 \\
14 & 0.0099952546 \\
15 & 0.0099952545
\end{array}
\right)$$
Just for the fun
$$x_{(8)}=\frac{2076678199429529816581224448972052669538078027653922391}{20776600532441929227
8132724022546069652413472361607062796}$$