Các câu hỏi tương tự
Tìm x biết: \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Tìm x biết (1/1.101+2/1.102+...+10/10.110).X=1/1.11+1/2.12+...+1/100.110
tìm x biết ( \(\frac{1}{1.101}\)+ \(\frac{1}{2.102}\)+\(\frac{1}{3.103}\)+......+ \(\frac{1}{10.110}\)) .x =\(\frac{1}{1.11}\)+\(\frac{1}{2.12}\)+.......+\(\frac{1}{100.110}\)
Tìm x
:\(\left(\frac{2011}{1.11}+\frac{2011}{2.12}+...+\frac{2011}{100.110}\right)=\frac{2012}{1.101}+\frac{2012}{2.102}+...+\frac{2012}{10.110}\)
(\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.100}\)).x=\(\frac{1}{1.12}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
Tìm x bt :
\(\left(\frac{1}{1\cdot101}+\frac{1}{2.102}+....+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1.11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
Cho Afrac{frac{1}{1.101}+frac{1}{2.102}+frac{1}{3.103}+...+frac{1}{25.125}}{frac{1}{1.26}+frac{1}{2.27}+frac{1}{3.28}+...+frac{1}{100.125}}. . Bfrac{frac{16}{9}-frac{16}{127}+frac{16}{2017}}{frac{5}{2017}+frac{5}{9}-frac{5}{127}}-frac{frac{6000}{43}-frac{6000}{257}-frac{125}{42}}{frac{2000}{43}-frac{250}{252}-frac{2000}{257}}.Chứng minh rằng Afrac{1}{2007^2}+frac{1}{2006^2}+frac{1}{2005^2}+...+frac{1}{7^2}+frac{1}{6^2}+frac{1}{5^2}B.
Đọc tiếp
Cho \(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}.\) .
\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}.\)
Chứng minh rằng \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B.\)
Cho: B = \(\frac{1}{101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
C =\(\frac{1}{26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
Tìm thương B : C
cho : \(A=\frac{1}{1.51}+\frac{1}{2.52}+\frac{1}{3.53}+...+\frac{1}{10.60}\)và \(B=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{50.60}\)
tính \(\frac{B}{A}\)