Chị em hk lớp 6 đó, mấy ac giúp chị e vs nha
Cho E= 92-1/9-2/10-3/11-...-92/100
F= 1/45=1/50+1/55=...=1/500
Tính E/F.
Những câu hỏi liên quan
Cho E=92-1\9-2\10-3\11-...-92\100; F=1\45+1\50+1\55+...+1\500
Tính E/F
Cho E=92-1/9-2/10-3/11-...-92/100
F=1/45+1/50+1/55+...+1/500
Tìm E/F
E= 92 - \(\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-..........-\frac{92}{100}\) ,F=\(\frac{1}{45}-\frac{1}{50}-\frac{1}{55}-........-\frac{1}{500}\)
Tính \(\frac{E}{F}\)
Cho E= \(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\)
F= \(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\)
Tính \(\frac{E}{F}\)
P/S Giải chi tiết hộ mik vs, lm ơn đó, mk cần gấp !
Cho:E=92-1/9-2/10-3/11.......-92/100
F=1/45+1/50+1/55+...+1/500
Tính:E/F
giúp mik với ai nhanh mik tích cho!Thank you!!!!
câu 1: tính giá trị biểu thức A=\(\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)
câu 2 :Cho E=92-\(\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}\) và F =\(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}\) .Tính\(\frac{E}{F}\)
Xin lỗi, mình chỉ làm được câu 1 thôi
\(A=\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5.111}{2.111}+\frac{4.1111}{11.1111}+\frac{3.11111}{22.11111}+\frac{11}{11.30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left[\left(\frac{5}{2 }+\frac{1}{30}+\frac{13}{60}\right)+\left(\frac{4}{11}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left[\left(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}\right)+\left(\frac{8}{22}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left(\frac{11}{4}+\frac{1}{2}\right)\)
\(A=\frac{1}{7}.\frac{13}{4}\)
\(A=\frac{13}{21}\)
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}
Đọc tiếp
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:
\(E=92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...\left(1-\dfrac{92}{100}\right)=\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{100}=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)\(F=\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)
\(\dfrac{E}{F}=\dfrac{8\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}=\dfrac{8}{\dfrac{1}{5}}=40\)
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(92-1/9-2/10-3/11-...-92/100)/(1/45+1/50+1/55+...+1/500)
[92-1/9-2/10-3/11-....-92/100]:[1/45+1/50+1/55+....1/500]