\(\overline{ab3}=\dfrac{3}{4}\overline{3ab}\)
\(\Rightarrow4.\overline{ab3}=3.\overline{3ab}\)
\(\Rightarrow4\left(10.\overline{ab}+3\right)=3\left(300+\overline{ab}\right)\)
\(\Rightarrow40.\overline{ab}+12=900+3.\overline{ab}\)
\(\Rightarrow900-12=40\overline{ab}-3\overline{ab}\)
\(\Rightarrow888=37\overline{ab}\)
\(\Rightarrow\overline{ab}=888:37=24\)
Ta có: \(\overline{ab3}=\dfrac{3}{4}\overline{3ab}\)
\(\Leftrightarrow10\overline{ab}+3=\dfrac{3}{4}\left(300+\overline{ab}\right)\)
\(\Leftrightarrow10\overline{ab}+3=\dfrac{3}{4}.300+\dfrac{3}{4}\overline{ab}\)
\(\Leftrightarrow10\overline{ab}-\dfrac{3}{4}\overline{ab}=225-3\)
\(\Leftrightarrow\dfrac{37}{4}\overline{ab}=222\)
\(\Leftrightarrow\overline{ab}=222:\dfrac{37}{4}=222.\dfrac{4}{37}=24\)
Vậy \(\overline{ab}\) = 24.
Câu hỏi của Đặng Trọng Hoàng - Toán lớp 6
Ta cos : \(ab3=\dfrac{3}{4}\overline{3ab}\)
\(\Rightarrow10\overline{ab}+3=\dfrac{3}{4}\left(300+\overline{ab}\right)\)
\(\Rightarrow10\overline{ab}+3=\dfrac{3}{4}.300+\dfrac{3}{4}\overline{ab}\)
\(\Rightarrow10\overline{ab}-\dfrac{3}{4}\overline{ab}=225-3\)
\(\Rightarrow\dfrac{37}{4}\overline{ab}=222\)
\(\Rightarrow\overline{ab}=222:\dfrac{37}{4}\)
\(\Rightarrow\overline{ab}=222.\dfrac{4}{37}=24\)
\(\)Vậy \(\overline{ab}=24\)
