Answer
Đặt `a/b = c/d = k \ (k \in RR ; k \ne 0)`
`=>` $\begin{cases} a = bk\\c = dk \end{cases}$
Ta có:
`{1111c - 99d}/{9999c - 11d}`
`= {1111dk - 99d}/{9999dk - 11d}`
`= {d . (1111k - 99)}/{d . (9999k - 11)}`
`= {1111k - 99}/{9999k - 11}` `(1)`
Ta lại có:
`{1111a - 99b}/{9999a - 11b}`
`= {1111bk - 99b}/{9999bk - 11b}`
`= {b . (1111k - 99)}/{b . (9999k - 11)}`
`= {1111k - 99}/{9999k - 11}` `(2)`
Từ `(1)` và `(2)`
`=> {1111k - 99}/{9999k - 11} = {1111k - 99}/{9999k - 11}`
`=> {1111c - 99d}/{9999c - 11d} = {1111a - 99b}/{9999a - 11b}`
Vậy `{1111c - 99d}/{9999c - 11d} = {1111a - 99b}/{9999a - 11b}.`
