.急...(要有过程...)

题目

.急...(要有过程...)
(9又2/7+7又2/9)÷(1/7+91/9)

答案

(9又2/7+7又2/9)÷(1/7+91/9)
=(9+2/7+7+2/9)÷(1/7+10+1/9)
=(16+2/7+2/9)÷(10+1/7+1/9)
=[16+2(1/7+1/9)]÷[10+(1/7+1/9)]
=[16+2(9/7*9+7/7*9)]÷[10+(9/7*9+7/7*9)]
=(16*7*9+2*16)÷(10*7*9+16)
=(20*7*9+2*16-4*7*9)÷(10*7*9+16)
=(20*7*9+2*16)÷(10*7*9+16)-(4*7*9)÷(10*7*9+16)
=2-(4*7*9)÷(10*7*9+16)
=2-252÷646
=2-126÷323
=1+197/323(即:1又323分之197)