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

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

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

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

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

Vậy :\(y=\frac{11}{15}\)

Bạn có muốn mình thử lại không?

không cảm ơn bạn

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

\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}+\frac{1}{11x13}\)

\(=\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{1}{2}x\frac{10}{39}\)

\(=\frac{5}{39}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)

Đặt A=\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{11x13}\)

Ax2=\(2x\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{11x13}\right)\)

Ax2=\(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{11x13}\)

Ax2=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

Ax2=\(\frac{1}{3}-\frac{1}{13}\)

Ax2=\(\frac{10}{39}\)

=>A=\(\frac{5}{39}\)

~~~hok tốt~~~

1/1 x 3 + 1/3 x 5 + 1/5 x 7 + 1/7 x 9 + 1/9 x 11

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

= 1 - 1/11

= 10/11

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

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

\(=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

                            Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

                              \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

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

                            \(2A=1-\frac{1}{9.11}=1-\frac{1}{99}=\frac{98}{99}\)

                              \(A=\frac{98}{99}:2=\frac{49}{99}\)

                                Ủng hộ mk nha!!!

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

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

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

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

=> \(\frac{5}{11}y=\frac{2}{3}\)

=>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

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

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

A = \(\frac{1}{2}.\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}\)

A = \(\frac{5}{11}\)