\(\frac{9}{11}\)nha kb đi

lấy (1/3 + 1/15 +1/10 + 1/21 ) + (1/36 + 1/28 + 1/6) + (1/45 + 1/55)

      =  (4/50 + 3/70) + 2/100

      = 7/120 + 2/100

      = 9/220

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)

\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)

\(\Rightarrow A=1-\frac{1}{2006}\)

\(\Rightarrow A=\frac{2005}{2006}\)

Sai đầu bài bạn ơi ! Bạn xem lại giúp mình sau đó thay đổi đầu bài mình giải cho

C=\(\frac{1}{3}+\frac{1}{10}+\frac{1}{15}+\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+\frac{1}{35}+\frac{1}{40}\)

=\(\frac{1}{3}+\frac{1}{15}+\frac{1}{10}+\frac{1}{20}+\frac{1}{30}+\frac{1}{40}+\frac{1}{25}+\frac{1}{35}\)

=\(\frac{5}{15}+\frac{1}{15}+\frac{4}{40}+\frac{2}{40}+\frac{1}{40}+\frac{1}{30}+\frac{1}{25}+\frac{1}{35}\)

=\(\frac{6}{15}+\frac{7}{40}+\frac{107}{1050}\)

= 2/2 + 2/6 + 2/12+...+2/90 = 2(1/2 +1/6 + 1/12 + ...+ 1/90) = 2(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10) = 2(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10) = 2(1 - 1/10) = 2 . 9/10 = 9/5

còn +\(\frac{1}{10}\)

giúp với ạ

giải dc nhưng mà hoi lâu

=(1+1+1+1+1+1+1+1)+(1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45)

Đặt A = 1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45

Ta có:

A x 1/2= 1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90

1/6=1/2x3=1/2-1/3

1/12=1/3x4=1/3-1/4

……………………

1/90=1/9x10=1/9-1/10

A x 1/2=1/2-1/3+1/3-1/4+1/4-1/5+…+1/9-1/10

A x 1/2=1/2-1/10=4/10

A=4/10:1/2=4/5

Vậy 4/3+7/6+11/10+16/15+22/21+29/28+37/36+46/45=1+1+1+1+1+1+1+1+4/5=8+4/5=44/5

\(\frac{4}{3}+\frac{7}{6}+\frac{11}{10}+...+\frac{46}{45}\)

\(=1+\frac{1}{3}+1+\frac{1}{6}+1+\frac{1}{10}+...+1+\frac{1}{45}\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)(8 chữ số 1)

\(=8+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)

  Đặt A = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)

  => \(\frac{1}{2}\)A = \(\frac{1}{2}\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)

              = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

              \(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

              \(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

Vậy A = \(\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)

    Do đó biểu thức trên là 8 + \(\frac{4}{5}\) = \(\frac{44}{5}\)

                                         Đáp số: \(\frac{44}{5}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{2}-\frac{1}{12}\)

\(=\frac{5}{12}\)

bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại

1/3 + 1/15 + 1/35+ 1/63 +...... + 1/195

=  1/3 + 1/3x5 + 1/5 x7 + 1/7x9 + ....+1/13x15

= 1/3+1/3-1/5+1/5-1/7+1/7-1/9+....+1/13-1/15 ( vì +- nên rút gọn )

= 1/3+1/3-1/15

=3/5

=1/1.3+1/3.5+1/5.7+...+1/13.15

=1/2.2(1/1.3+1/3.5+1/5.7+...+1/13.15)

=1/2(2/1.3+2/3.5+2/5.7+...+2/13.15)

=1/2(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)

=1/2[(1-1/15)+(1/3-1/3)+(1/5-1/5)+...+(1/13-1/15)]

=1/2[(1-1/15)+0+...+0=1/2(1-1/15)=1/2.14/15=14/30=7/15