Đặt A=\(\frac{9}{10}+\frac{39}{40}+...+\frac{1119}{1120}\)

=>A=\(\frac{10-1}{10}+\frac{40-1}{40}+...+\frac{1120-1}{1120}\)

=>A=\(1-\frac{1}{10}+1-\frac{1}{40}+...+1-\frac{1}{1120}\)

=>A=\(11-\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\right)\)

Đặt B=\(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{1120}\)

=>3B=\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{1120}\)

=>3B=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)

=>3B=\(\frac{33}{70}\)

=>B=\(\frac{11}{70}\)

=>A=11-\(\frac{11}{70}\)

=>A=\(\frac{759}{70}\)

Ta có:  

=  

=\(\left(1+1+...+1\right)-\)\(\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{80}+...+\frac{1}{1120}\right)\)

=11\(-\left[\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...\frac{1}{35}\right).\frac{1}{3}\right]\)

=11\(-\left[\left(\frac{1}{2}-\frac{1}{35}\right).\frac{1}{3}\right]\)

=11\(-\left(\frac{33}{10}.\frac{1}{3}\right)\)

=11\(-\frac{11}{70}\)

9/10 + 39/40 + 87/88 + ... + 1119/1120

= ( 1 - 1/10 ) + ( 1 - 1/40 ) - ( 1 - 1/88 ) + .... + ( 1 - 1/1120 )

= \(\left(1-\frac{1}{2.5}\right)+\left(1-\frac{1}{5.8}\right)+...+\left(1-\frac{1}{32.35}\right)\)

= ( 1 + 1 +1 + ... +1 ) - ( \(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{32.35}\)

= 11 - ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/32 - 1/35 )

= 11 - ( 1/2  - 1/35 ) 

= 11 - 33/70

= \(10\frac{37}{70}\)

9/10 + 39/40 + 87/88 + ... + 1119/1120

= 1 - 1/10 + 1 - 1/40 + 1 - 1/88 + ... + 1 - 1/1120

= ( 1 + 1 + 1 + .. + 1 ) - ( 1/10 + 1/40 + 1/88 + ... + 1/1120 )

=  ( 1 + 1 + 1 + .. +1 ) - ( 1/2.5  + 1/5.8 + 1/8.11+...+1/32.35)

= 11 - 1/3.( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/32.35)

= 11 - 1/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/32 - 1/35)

= 11 - 1/3( 1/2 - 1/35)

= 11 - 1/3.33/70 = 11 - 11/90 =  979/90

Mình không hiểu quy luật của dãy số cho lắm! Xin lỗi.

(ngày mai bạn thi HSG môn Toán đúng không?)

Dài ngoằng ngoặc

979/90 bạn nhé

bạn ơi tách ra thừa số chung rồi làm như bình thường nha 

1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)

     =4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)

     = 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)

    =4-\(\left(1-\frac{1}{9}\right)\)

     = 4-\(\frac{8}{9}\)

      = \(\frac{7}{9}\)