Câu 3:
a, Chứng minh rằng nếu:
(\(\overline{ab}\)+\(\overline{cd}\)+\(\overline{eg}\)) ⋮ 11 thì \(\overline{abcdeg}\) ⋮ 11
b, Cho E = 92-\(\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}\); F= \(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\)Tính \(\dfrac{E}{F}\)
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Câu 3:
a, Chứng minh rằng nếu:
(\(\overline{ab}\)+\(\overline{cd}\)+\(\overline{eg}\)) ⋮ 11 thì \(\overline{abcdeg}\) ⋮ 11
b, Cho E = 92-\(\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}\); F= \(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\)Tính \(\dfrac{E}{F}\)
a) Ta có:
\(\overline{abcdeg}=10000.\overline{ab}+100.\overline{cd}+eg=9999.\overline{ab}+99.\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)\(9999.\overline{ab}⋮11\)
\(99.\overline{cd}⋮11\)
\(\overline{ab}+\overline{cd}+\overline{eg}⋮11\)
\(\Rightarrow9999.\overline{ab}+99.\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)hay \(\overline{abcdeg}⋮11\)(đpcm)
b) Ta có:
\(E=92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...\left(1-\dfrac{92}{100}\right)=\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{100}=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)\(F=\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)
\(\dfrac{E}{F}=\dfrac{8\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}=\dfrac{8}{\dfrac{1}{5}}=40\)