7. (7 points) Given a sequence of numbers with a certain pattern: $1, \frac{2}{3}, \frac{5}{8}, \frac{13}{21}, \frac{34}{55} \cdots$. Then, in this sequence, the 10th number from left to right is . $\qquad$
【Answer】Solution: From the original problem, we derive the rule that starting from the third number, the numerator is the sum of the numerator and denominator of the previous fraction, and the denominator is the sum of its own numerator and the denominator of the previous fraction.
Therefore, the subsequent fractions are: $\frac{8923361015974181}{14437798725846765}$, and the 10th number is $\frac{4181}{6765}$.
The 10th number is $\frac{4181}{6765}$.
Therefore, the answer is $\frac{4181}{6765}$.