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\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{37}-\frac{2}{39}\)
\(=\frac{2}{3}-\frac{2}{39}\)
\(=\frac{8}{13}\)
Ta có:
\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)
\(\frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)
\(\frac{2}{7.9}=\frac{1}{7}-\frac{1}{9}\)
\(......................................\)
\(\frac{2}{37.39}=\frac{1}{37}-\frac{1}{39}\)
nên \(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(C=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
Ta có: \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{97.99}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
\(M=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
Ta có:
A = \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\) + ... + \(\frac{2}{2001.2003}\) + \(\frac{2}{2003.2005}\)
= \(\frac{1}{1}\) - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{9}\) + ... + \(\frac{1}{2001}\)- \(\frac{1}{2003}\)+ \(\frac{1}{2003}\)- \(\frac{1}{2005}\)
= 1 - \(\frac{1}{2005}\)
= \(\frac{2004}{2005}\)
Chúc bạn học tốt nha ^^!!
M=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
M=\(\frac{1}{3}-\frac{1}{99}\)
M=\(\frac{32}{99}\)
TICK ỦNG HỘ NHA
\(Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\)\(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
\(Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(Q=\frac{1}{3}+0+0+0-\frac{1}{11}\)
\(Q=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)
\(A=\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.100}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right).\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}x\frac{32}{99}=\frac{32}{198}\)
bn tự rút gọn nha mk mới làm tắt đó
A= 1/3-1/5+1/5-1/7+....+1/25-1/27
=1/3-1/27
=8/27
vậy A=8/27