\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
sai zùi các bn ấy còn thiếu nhân với 1/2 because tui làm bài này rất nhiều mà làm như các bn ấy tui toàn bị nghe chửi thôi
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{96}{297}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{2}{3}.\frac{2}{5}+\frac{2}{5}.\frac{2}{7}+\frac{2}{7}.\frac{2}{9}+...+\frac{2}{97}.\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{97}-\frac{2}{99}\)
\(=\frac{2}{3}+\left[\frac{2}{5}-\frac{2}{5}\right]+\left[\frac{2}{7}-\frac{2}{7}\right]\left[\frac{2}{9}-\frac{2}{9}\right]+...+\left[\frac{2}{97}-\frac{2}{97}\right]-\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{99}\)
\(=\frac{64}{99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)\(=\frac{32}{99}\)
\(\frac{2}{3.5}\)\(+\)\(\frac{2}{5.7}\)\(+\)\(\frac{2}{7.9}\)\(+\)\(...\)\(+\)\(\frac{2}{97.99}\)
\(=\)\(\frac{1}{3}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{7}\)\(+\)\(\frac{1}{7}\)\(+\)\(...\)\(\frac{1}{96}\)\(-\)\(\frac{1}{97}\)\(+\)\(\frac{1}{97}\)\(-\)\(\frac{1}{99}\)
\(=\)\(\frac{1}{3}\)\(-\)\(0\)\(-\)\(0\)\(-\)\(...\)\(-\)\(0\)\(-\)\(\frac{1}{99}\)
\(=\)\(\frac{1}{3}\)\(-\)\(\frac{1}{99}\)
\(=\)\(\frac{32}{99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
= 1/3 - 1/5 + 1/5 - 1/7 +...+ 1/97 - 1/99
=1/3 - 1/99
=33/99 - 1/99
=32/99
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
học tốt!!!
Đặt A=2/3*5+2/5*7+2/7*9+...+2/97*99
A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
A=1/3-1/99
A=32/99
Vậy A= 32/99
Nhớ k cho mình nhé
2/3.5 + 2/5.7+ 2/7.9 +... + 2/97.99
= 2/2 . (2/3.5+ 2/5.7+ 2/7.9 +...+2/97.99)
= 1. (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +...+ 1/97 - 1/99)
= 1. ( 1/3 - 1/99)
=1. 32/99 = 32/99
Dấu / là phần nha.
Đặt A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.................+\frac{2}{97.99}\)
A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..........+\frac{1}{97}-\frac{1}{99}\)
A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
Vậy A=\(\frac{32}{99}\)
Chúc bn học tốt
Ta có:
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)\(=\)\(\frac{32}{99}\)
Hok tốt!
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+......+\frac{2}{97\cdot99}\)
\(=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+......+\frac{99-97}{97\cdot99}\)
\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+......+\frac{2}{97}-\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{99}\)
\(=\frac{66}{99}-\frac{2}{99}\)
\(=\frac{66-2}{99}=\frac{64}{99}\)
