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

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

      = \(\frac{7}{9}\)

Câu 2 tương tự như câu 1 

A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{154}\right)\)

A= 4                    -\(\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{11.14}\right)\)

Bạn tự làm tiếp 

Ta có : 

\(\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+...+\frac{1119}{1120}\)

\(=\)\(\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{40}\right)+\left(1-\frac{1}{88}\right)+...+\left(1-\frac{1}{1120}\right)\)

\(=\)\(\left(1+1+1+...+1\right)-\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{1120}\right)\)

\(=\)\(\left(1+1+1+...+1\right)-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{32.35}\right)\)

\(=\)\(11-\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{32}-\frac{1}{35}\right)\)

\(=\)\(11-\left(\frac{1}{2}-\frac{1}{35}\right)\)

\(=\)\(11-\frac{33}{70}\)

\(=\)\(\frac{737}{70}\)

Chúc bạn học tốt ~ 

a: \(150+50:5+2\cdot9\)

=150+10+18

=178

2+3+4+5+6+7+8+9+10+...+1119 ( có 1118)

=(2+1119).1118:2

Bằng bao nhiêu bn tự tính nha

Lời giải:
$D=79(1119+59)-1119.29-179.59$

$=79.1119+79.59-1119.29-179.59$

$=(79.1119-1119.29)-(179.59-79.59)$

$=1119(79-29)-59(179-79)$

$=1119.50-59.100=55950-5900=50050$