\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=1-\dfrac{1}{11}\)

\(=\dfrac{11}{11}-\dfrac{1}{11}\)

\(=\dfrac{10}{11}\)

\((\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11})\cdot y=\frac{2}{3}\)

\(\Rightarrow(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11})\cdot y=\frac{2}{3}\)

\(\Rightarrow1-\frac{1}{11}\cdot y=\frac{2}{3}\)

\(\Rightarrow\frac{10}{11}\cdot y=\frac{2}{3}\)

\(\Rightarrow y=\frac{2}{3}:\frac{10}{11}=\frac{11}{15}\)

Vậy :\(y=\frac{11}{15}\)

Bạn có muốn mình thử lại không?

không cảm ơn bạn

SOS

hép☹

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)

=> \(\frac{5}{11}y=\frac{2}{3}\)

=>y = \(\frac{2}{3}:\frac{5}{11}\)

=> y = \(\frac{22}{15}\)

cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm

bạn phạm khánh hà ơi dấu chấm ở giữa các phân số có nghĩa là dấu nhân đó

Giải:

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)y=-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}.\dfrac{10}{11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{5}{11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow15y=-22\)

\(\Leftrightarrow y=-\dfrac{22}{15}\)

Vậy ...

ta có : ( 1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 ).y = -2/3

⇒ ( 1- 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 ) . y = -2/3

⇒ ( 1 - 1/11 ) . y = -2/3

⇒ 10/11 . y = -2/3

⇒ y = -2/3 : 10/11

⇒ y = -11/15 .

Vậy y = -11/15

\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{9.11}\right)\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{11}\right)\)

\(=2.\left(1-\frac{1}{11}\right)\)

\(=2.\left(\frac{11}{11}-\frac{1}{11}\right)\)

\(=2.\frac{10}{11}\)

\(=\frac{20}{11}\)

\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)

\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(2.\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(2.\frac{10}{11}.y=\frac{2}{3}\)

\(\frac{20}{11}.y=\frac{2}{3}\)

\(\Rightarrow y=\frac{11}{30}\)

Study well 

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

\(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}\right)\)

\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=1-\dfrac{1}{11}\)

\(=\dfrac{11}{11}-\dfrac{1}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{2}{11}\)

\(=1-\dfrac{2}{11}\)

\(=\dfrac{9}{11}\)