Cho E=92-1/9-2/10-3/11-...-92/100
F=1/45+1/50+1/55+...+1/500
Tìm E/F
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\(A=\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5.111}{2.111}+\frac{4.1111}{11.1111}+\frac{3.11111}{22.11111}+\frac{11}{11.30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left[\left(\frac{5}{2 }+\frac{1}{30}+\frac{13}{60}\right)+\left(\frac{4}{11}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left[\left(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}\right)+\left(\frac{8}{22}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left(\frac{11}{4}+\frac{1}{2}\right)\)
\(A=\frac{1}{7}.\frac{13}{4}\)
\(A=\frac{13}{21}\)
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:
\(E=92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...\left(1-\dfrac{92}{100}\right)=\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{100}=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)\(F=\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)
\(\dfrac{E}{F}=\dfrac{8\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}=\dfrac{8}{\dfrac{1}{5}}=40\)