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

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

\(=\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}+...+\frac{15-13}{13x15}=\)

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

Ta có : \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{11.13}\)

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

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

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

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\)

\(=\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{11\times13}\)

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

\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

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

\(=\frac{31}{15}\)

              Bài làm :

Ta có :

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{31}{15}\)

\(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....-\frac{1}{15}\)

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

Tinh nhanh nhe

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

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

\(=\frac{1}{5}-\frac{1}{15}=\frac{3}{15}-\frac{1}{15}=\frac{3-1}{15}=\frac{2}{15}\)

P/s : Dấu chấm là nhân nhé!

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

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

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

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

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

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

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

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

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

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

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

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

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

2/(7 × 9) + 2/(9 × 11) + 2/(11 × 13) + ... + 2/(97 × 99)

= 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + ... + 1/97 - 1/99

= 1/7 - 1/99

= 92/693

Bài 1:

Nếu thêm 1 đơn vị vào số thứ hai ta được số mới và 1/3 số thứ nhất = số mới

Tổng số thứ nhất và số mới là

43+1=44

Chia số mới thành 1 phần thì số thứ nhất là 3 phần như thế

Tổng số phần bằng nhau là

1+3=4 phần

Giá trị 1 phần là

44:4=11

Số thứ nhất là

11x3=33

Số thứ hai là

43-33=10

Bài 2:

\(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{13x15}=\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}+...+\frac{15-13}{13x15}\)

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