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

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

Đến đây bạn làm tiếp nhé

Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)

         = \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)

         = \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)

         = \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)

         = 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)

         = \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)

1 ) 

m = 3 

n = 2 

biết vậy nhưng ko biết cách giải

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

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

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

A = \(1-\frac{1}{10}\)

A = \(\frac{9}{10}\)

1/2=1-1/2 ; 1/2.1/3=1/2-1/3 ; 1/3.1/4=1/3-1/4...v...v

Vậy A bằng: 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.............+1/8-1/9+1/9-1/10

                =1-1/10=9/10