\(\frac{4}{9\times11}+\frac{4}{11\times13}+\frac{4}{13\times15}+...+\frac{4}{95\times97}+\frac{4}{97\times99}\)

\(=2\times\left(\frac{2}{9\times11}+\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{95\times97}+\frac{2}{97\times99}\right)\)

\(=2\times\left(\frac{11-9}{9\times11}+\frac{13-11}{11\times13}+\frac{15-13}{13\times15}+...+\frac{97-95}{95\times97}+\frac{99-97}{97\times99}\right)\)

\(=2\times\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\times\left(\frac{1}{9}-\frac{1}{99}\right)=\frac{20}{99}\)

4/9x11 + 4/11x13 + 4/13X15 + .............+ 4/95X97 + 4/97X99

=2 x (2/9x11 + 2/11x 13 + .........+2/95x97 + 2/97x99)

=2 x ( 1/9 - 1/11 + 1/11- 1/13 +...... + 1/97 - 1/99)

=2 x (1/9 - 1/99)

=2 x10/99

=20/99

Học tốt!

Mọi người giúp mik với mik đang cần gấp

\(=\frac{1}{2}\times\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{9}-\frac{1}{99}\right)\)

\(=\frac{1}{2}\times\frac{10}{99}\)

\(=\frac{5}{99}\)

\(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+....+\frac{1}{97}-\frac{1}{99}=\frac{1}{9}_{ }-\frac{1}{99}=\)\(\frac{10}{99}\)

A=\(\frac{4}{11}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{99}-\frac{4}{101}\)

\(A=4\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=4.\left(\frac{1}{11}-\frac{1}{101}\right)\)

A=4. 90/1111=360/1111

Co dung khong vay Cuong Lucha DT

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

Nhưng bn ơiX là x^2 hay tách biệt nếu tách biệt thì là 9/49 còn nếu là x^2 thì là 3/7 nhé

\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)xX=\frac{9}{7} \)\(=\left(\frac{1}{3}-\frac{1}{21}\right)xX=\frac{9}{7}\)\(=\frac{2}{7}xX=\frac{9}{7}\)

\(X=\frac{9}{7}:\frac{2}{7}\)

\(X=\frac{9}{2}\)

1/5*7 + 1/7*9 + 1/9*11 + ... + 1/13*15

= 1/2(2/5*7 + 2/7*9 + 2/9*11 + ... + 2/13*15)

= 1/2(1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15)

= 1/2(1/5 - 1/15)

= 1/2.2/15

= 1/15

Bài giải

\(\text{Đặt }A=\frac{1}{5\text{ x }7}+\frac{1}{7\text{ x }9}+\frac{1}{9\text{ x }11}+\frac{1}{11\text{ x }13}+\frac{1}{13\text{ x }15}\)

\(A=\frac{1}{2}\left(\frac{2}{5\text{ x }7}+\frac{2}{7\text{ x }9}+\frac{2}{9\text{ x }11}+\frac{2}{11\text{ x }13}+\frac{2}{13\text{ x }15}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{15}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{15}\)

\(A=\frac{1}{15}\)

Bài giải

\(\text{Đặt }A=\frac{1}{5\text{ x }7}+\frac{1}{7\text{ x }9}+\frac{1}{9\text{ x }11}+\frac{1}{11\text{ x }13}+\frac{1}{13\text{ x }15}\)

\(A=\frac{1}{2}\left(\frac{2}{5\text{ x }7}+\frac{2}{7\text{ x }9}+\frac{2}{9\text{ x }11}+\frac{2}{11\text{ x }13}+\frac{2}{13\text{ x }15}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{15}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{15}\)

\(A=\frac{1}{15}\)

=4x(\(\frac{1}{11x13}\)+\(\frac{1}{13x15}\)+.......+\(\frac{1}{99x101}\))

=4x(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\))

4x(\(\frac{1}{11}\)-\(\frac{1}{101}\))

=4x \(\frac{90}{1111}\)

=\(\frac{360}{1111}\)

\(\frac{4}{11\times13}+\frac{4}{13\times15}+\frac{4}{15\times17}+...+\frac{4}{99\times101}\)

\(=\frac{4}{11}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+\frac{4}{15}-\frac{4}{17}+...+\frac{4}{99}-\frac{4}{101}\)

\(=\frac{4}{11}-\frac{4}{101}\)

\(=\frac{360}{1111}\)

Đặt biểu thức là A . Ta co 

\(A=\frac{4}{11.13}+\frac{4}{13.15}+....+\frac{4}{99.101}\)

\(\Rightarrow2\left(\frac{2}{11.13}+\frac{2}{13.15}+....+\frac{2}{99.101}\right)\)

\(\Rightarrow A=2\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+.....+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow A=2\left(\frac{1}{11}-\frac{1}{101}\right)\)

\(\Rightarrow A=\frac{2.90}{1111}=\frac{180}{1111}\)

  \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\)\(...+\frac{2}{8.9}+\frac{2}{9.10}\)

Đặt \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

      \(B=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}+\frac{2}{9.10}\)

              Ta có:

\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(A=\frac{1}{3}-\frac{1}{15}\)

\(A=\frac{4}{15}\)

    \(B=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}+\frac{2}{9.10}\)

    \(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

     \(B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

    \(B=2\left(1-\frac{1}{10}\right)\)

    \(B=2.\frac{9}{10}\)

    \(B=\frac{9}{5}\)

\(\Rightarrow A+B=\frac{4}{15}+\frac{9}{5}\)

                   \(=\frac{31}{15}\)

   Vậy biểu thức trên có giá trị là \(\frac{31}{15}\)

=2/5-2/7+ 2/7-2/9+2/9-2/11+2/11-2/13+2/13-2/15
=2/5-(2/7-2/7)-(2/9-2/9)-(2/11-2/11)-(2/13-2/13)-2/15

=2/5-0-0-0-0-2/15

=2/5-2/15

4/15