2. 102 Real numbers $u, v$ satisfy the relation
$$\begin{aligned}
\left(u+u^{2}+u^{3}+\cdots+u^{8}\right)+10 u^{9} & =\left(v+v^{2}+v^{3}+\cdots+v^{10}\right)+ \\
& 10 v^{11} \\
& =8 .
\end{aligned}$$

Determine which of $u$ and $v$ is larger, and prove your conclusion.
[Solution] From the given relation, we know that
$$u0, v>0$. Therefore,
$$00 \\
g(0)=-80
\end{array}$$

Thus, $u$ and $v$ are the unique roots of the equations $f(x)=0$ and $g(x)=0$ in the interval $(0,1)$, respectively.

From equation (1), we know that $u$ is the root of
$$f_{1}(x)=(10 x-9)\left(x^{9}-1\right)+x-1=0$$

in the interval $(0,1)$. Since $f_{1}(0)=8>0$ and $f_{1}\left(\frac{9}{10}\right)=-\frac{1}{10}<0$, there exists a unique root $u$ in $(0,1)$. 

Similarly, $v$ is the root of
$$g(x)=10 x^{10}-9 x+1=0$$

in the interval $(0,1)$. Since $g(0)=-80<0$ and $g(1)=2>0$, there exists a unique root $v$ in $(0,1)$. 

Since $f_{1}(u)=0$ and $g(v)=0$, and $f_{1}(x)$ is continuous and strictly decreasing in $(0,1)$, while $g(x)$ is continuous and strictly increasing in $(0,1)$, it follows that $u < v$.

Therefore, $v$ is in the interval $(u, 1)$, hence $u < v$.