\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}>1\)

Thank you bạn dcv new ^ ^

Trl:

Đây ko phải là bài lp 5 bn nhé.

Hok tốt!

trời đất toán lớp 5 khó bằng toán 6 lun á

 Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)

\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)

\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)

Mà \(\frac{1}{9}>\frac{1}{10}\)

\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)

Vậy : M > N

tổng của A là

1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 223/140

=> 223/140 > 6/7

k mk nha

tính dần rồi so sánh tkoi bạn 

Ta có 

A= 1,066018877

=> A > 2/3

tớ tính máy tính ra A = 1,066018877

1) a) \(\frac{5454}{5757}-\frac{171717}{191919}=\frac{18\times3\times101}{19\times3\times101}-\frac{17\times10101}{19\times10101}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)

b) \(\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times....\times\frac{2021}{2020}=\frac{6\times7\times8\times...\times2021}{5\times6\times7\times...\times2020}=\frac{2021}{5}\)

2) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}=2\times\frac{1}{6}+2\times\frac{1}{12}+2\times\frac{1}{20}+...+2\times\frac{1}{90}\)

\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=2\times\left(\frac{1}{2}-\frac{1}{10}\right)=2\times\frac{2}{5}=\frac{4}{5}\)

b)Vì \(a-1< a+1\)

=> \(\frac{1}{a-1}>\frac{1}{a+1}\)

1/ a x b = 1/a - 1/b