$\frac{13}{7} × \frac{47}{5}$
$= ( \frac{1}{7} × 13 ) × ( \frac{1}{5} × 47 )$
$= ( \frac{1}{7} × \_\_\_\_\_\_ ) × (13 × 47)$
$= \_\_\_\_\_\_ × 611$
$= \_\_\_\_\_\_$
$\frac{611}{42} ÷ \frac{47}{42}$
$= ( \frac{1}{42} × 611 ) ÷ ( \frac{1}{42} × 47 )$
$= ( \_\_\_\_\_\_ ÷ \_\_\_\_\_\_ ) × (611 ÷ 47)$
$= \_\_\_\_\_\_ × 13$
$= \_\_\_\_\_\_$

$\frac{13}{7} × \frac{47}{5}$
$= ( \frac{1}{7} × 13 ) × ( \frac{1}{5} × 47 )$
$= ( \frac{1}{7} × \underline{\frac{1}{5}} ) × (13 × 47)$
$= \underline{\frac{1}{35}} × 611$
$= \underline{\frac{611}{35}}$
$\frac{611}{42} ÷ \frac{47}{42}$
$= ( \frac{1}{42} × 611 ) ÷ ( \frac{1}{42} × 47 )$
$= ( \underline{\frac{1}{42}} ÷ \underline{\frac{1}{42}} ) × (611 ÷ 47)$
$= \underline{1} × 13$
$= \underline{13}$

例2 计算 $(9\dfrac{2}{7} + 7\dfrac{2}{9}) ÷ (\dfrac{5}{7} + \dfrac{5}{9})$。

$9\frac{2}{7}=\frac{65}{7}$，$7\frac{2}{9}=\frac{65}{9}$
$\begin{aligned}原式&=( \frac{65}{7} + \frac{65}{9} ) ÷ ( \frac{5}{7} + \frac{5}{9} )\\&=[ 65×( \frac{1}{7} + \frac{1}{9} ) ] ÷ [ 5×( \frac{1}{7} + \frac{1}{9} ) ]\\&=65÷5\\&=13\end{aligned}$

分析与解答
这道题有什么简便的方法吗？

$(9 \frac{2}{7} +7 \frac{2}{9}) ÷ (\frac{5}{7} + \frac{5}{9})$
$=(\frac{65}{7} + \frac{65}{9}) ÷ (\frac{5}{7} + \frac{5}{9})$
$=[(\quad) ×65] ÷ [(\quad) ×5]$
$=\_\_\_\_\_\_ ÷ \_\_\_\_\_\_$
$=\_\_\_\_\_\_$
练习与实践
计算 $(9 \frac{5}{8} +8 \frac{5}{9}) ÷ (\frac{7}{8} + \frac{7}{9})$。

$(9 \frac{2}{7} +7 \frac{2}{9}) ÷ (\frac{5}{7} + \frac{5}{9})$
$=(\frac{65}{7} + \frac{65}{9}) ÷ (\frac{5}{7} + \frac{5}{9})$
$=[(\frac{1}{7}+\frac{1}{9}) × 65] ÷ [(\frac{1}{7}+\frac{1}{9}) × 5]$
$=65×(\frac{1}{7}+\frac{1}{9}) ÷ 5×(\frac{1}{7}+\frac{1}{9})$
$=13$
$(9 \frac{5}{8} +8 \frac{5}{9}) ÷ (\frac{7}{8} + \frac{7}{9})$
$=(\frac{77}{8}+\frac{77}{9})÷(\frac{7}{8}+\frac{7}{9})$
$=[(\frac{1}{8}+\frac{1}{9})×77]÷[(\frac{1}{8}+\frac{1}{9})×7]$
$=77÷7$
$=11$
答：计算结果为11。