\(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{31.33}+\frac{2}{33.35}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{31}-\frac{1}{33}+\frac{1}{33}-\frac{1}{35}\)
\(=\frac{1}{5}-\frac{1}{35}\)
\(=\frac{6}{35}\)
Chúc bạn học giỏi nha!!!
K cho mik vs nhé danggiahuy
đặt A=2/5.7+2/7.9+2/9.11+.....+2/31.33+2/33.35
A=1/5-1/7+1/7-1/9+1/9-1/11+.....+1/31-1/33+1/33-1/35
A=1/5-1/35
A=6/35
2/5.7+2/7.9+2/9.11+.....+2/31.33+2/33.35
\(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}...+\frac{2}{31.33}+\frac{2}{33.35}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....\frac{1}{31}-\frac{1}{33}+\frac{1}{33}-\frac{1}{35}\)
\(\frac{1}{5}-\frac{1}{35}\)
\(=\frac{6}{35}\)
theo mình nghĩ chắc là vậy
=> \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{31}-\frac{1}{33}+\frac{1}{33}-\frac{1}{35}\)
=> \(\frac{1}{5}-\frac{1}{35}\)
=> \(\frac{6}{35}\)
ta có:\(\frac{2}{5\times7}=\frac{1}{5}-\frac{1}{7}\)
Tương tự ....\(\frac{2}{33\times35}=\frac{1}{33}-\frac{1}{35}\)
\(\Rightarrow\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{33.35}=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{33}-\frac{1}{35}\)
\(=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
