Lời giải:
Ta có:
\(\underbrace{333....3}_{2017}.\underbrace{666....6}_{2017}=3.\underbrace{111....1}_{2017}.6.\underbrace{111...1}_{2017}.\)
\(=18.\underbrace{111...1}_{2017}.\frac{10^{2017}-1}{9}\)
\(=2.\underbrace{111....1}_{2017}(10^{2017}-1)\)
\(=2[\underbrace{111....1}_{2017}\underbrace{000...0}_{2017}-\underbrace{111...11}_{2017}]\)
\(=2. \underbrace{1111....1}_{2016}0\underbrace{88...8}_{2016}9\)
\(=\underbrace{222....2}_{2016}1\underbrace{77....7}_{2016}8\)
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