The answer is $v$ .
We define real functions $U$ and $V$ as follows: $$\begin{align*} U(x) &= (x+x^2 + \dotsb + x^8) + 10x^9 = \frac{x^{10}-x}{x-1} + 9x^9 \\ V(x) &= (x+x^2 + \dotsb + x^{10}) + 10x^{11} = \frac{x^{12}-x}{x-1} + 9x^{11} . \end{align*}$$ We wish to show that if $U(u)=V(v)=8$ , then $u <v$ .
We first note that when $x \le 0$ , $x^{12}-x \ge 0$ , $x-1 < 0$ , and $9x^9 \le 0$ , so \ Similarly, $V(x) \le 0 < 8$ .
We also note that if $x \ge 9/10$ , then $$\begin{align*} U(x) &= \frac{x-x^{10}}{1-x} + 9x^9 \ge \frac{9/10 - 9^9/10^9}{1/10} + 9 \cdot \frac{9^{9}}{10^9} \\ &= 9 - 10 \cdot \frac{9^9}{10^9} + 9 \cdot \frac{9^9}{10^9} = 9 - \frac{9^9}{10^9} > 8. \end{align*}$$ Similarly $V(x) > 8$ . It then follows that $u, v \in (0,9/10)$ .
Now, for all $x \in (0,9/10)$ , $$\begin{align*} V(x) &= U(x) + V(x)-U(x) = U(x) + 10x^{11}+x^{10} -9x^9 \\ &= U(x) + x^9 (10x -9) (x+1) < U(x) . \end{align*}$$ Since $V$ and $U$ are both strictly increasing functions over the nonnegative reals, it then follows that $$V(u) < U(u) = 8 = V(v),$$ so $u<v$ , as desired. $\blacksquare$