Six The terms of the sequence $x_{1}, x_{2}, x_{3}, \cdots$ are non-zero real numbers, and satisfy $x_{\mathrm{a}+2}=\frac{x_{\square} \cdot x_{\mathrm{n}+1}}{2 x_{\mathrm{a}}-x_{\mathrm{a}+1}} \quad(n=1,2, \cdots)$,
(1) Find the necessary and sufficient conditions for $x_{1}$ and $x_{2}$ such that all $x_{0}$ are integers;
(2) If $x_{1}=\lg 5, x_{2}=\lg 2$, find the smallest integer $n$ such that $x_{\mathrm{n}}<\frac{\lg 5 \lg 2}{10+\lg 2}$ (1 \mathrm{~g} 2=0.3010).
Six, Solution (1) Hint: By mathematical induction, it is easy to prove
$$
x_{\mathrm{n}}=\frac{x_{1} x_{2}}{(n-1) x_{1}-(n-2) x_{2}},
$$

i.e., $x_{\mathrm{n}}=\frac{x_{1} x_{2}}{n\left(x_{1}-x_{2}\right)+\left(2 x_{2}-x_{1}\right)}$.
When $x_{1} \neq x_{2}$ and $n$ is sufficiently large, there must be $x_{0}<1$. When $x_{1}=x_{2}$, $x_{0}=\frac{x_{1} x_{2}}{2 x_{2}-x_{1}}=x_{1}$ (constant).

Therefore, the necessary and sufficient condition for all $x_{0}$ to be integers is $x_{1}=x_{2}$ (integer).
(2) From $\frac{\lg 2 \cdot \lg 5}{n(\lg 5-\lg 2)+2 \lg 2-\lg 5}$ $<\frac{\lg 5 \lg 2}{10+\lg 2}$, the smallest integer solution is $n=27$.