Ta có:
\(\overline{0,a}\times\overline{0,b}\times\overline{a,b}=\overline{0,bbb}\)
\(=\left(a\times0,1\right)\times\left(b\times0,1\right)\times\left(ab\times0,1\right)=\overline{bbb}\div1000\)
\(=a\times b\times\overline{ab}\times0,1\times0,1\times0,1=\overline{bbb}\div1000\)
\(=a\times b\times\overline{ab}\times0,001=\overline{bbb}\div1000\)
\(=a\times b\times\overline{ab}\div1000=\overline{bbb}\div1000\)
\(=a\times b\times\overline{ab}=\overline{bbb}\)
\(=a\times b\times\overline{ab}=b\times111\)
\(\Rightarrow a\times\overline{ab}=111\)
\(\Rightarrow a=3\)
\(\Rightarrow3\times\overline{3b}=111\)
\(\overline{3b}=111\div3\)
\(\overline{3b}=37\)
\(\Rightarrow b=7\)
Vậy \(b=7\)