Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)

đặt A=1/1×2+1/2×3+1/3×4+

bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2

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

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

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}\)

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

\(=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{1}-\frac{1}{9}\)

\(=\frac{8}{9}\)

\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}....\frac{1}{9x10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)

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

1/1*2+1/2*3+1/3*4+...+1/9*10

=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

=1-1/10

=9/10

nho k cho minh voi nhe

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ......... + \(\frac{1}{7.8}\)+ \(\frac{1}{8.9}\)+ \(\frac{1}{9.10}\)

\(=\)\(1\)\(-\)\(\frac{1}{10}\)

\(=\)\(\frac{9}{10}\)

9/10 nha

Bài 1:

Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)

\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)

\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)

\(A=\frac{2x19}{20}=\frac{19}{10}\)

Bài 2:

Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)

Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)

\(Bx100=\frac{9}{10}x100=90\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)

=>  \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)

bài 1 đáp án là:19/10

2:147/50

đáp án bài 1 là: 19/10

đáp án bài 2 là 147/50

1/1x2+1/2x3+1/3x4+..+1/9x10

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...-1/10

=1-1/10

=9/10