\(\frac{2n+1}{\left[n\left(n+1\right)\right]^2}=\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

\(\Rightarrow C=\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

\(\Rightarrow C=1-\frac{1}{\left(n+1\right)^2}\)

Cảm ơn bạn! Chúc bạn học tốt!

M=(1.3.5.7.....99)/(2.4.6.8.....100)

số số hạng của tử = (99-1)/2 +1 = 50 -> 1.3.5.7....99= (99+1)*50/2 =2500

số số hạng của mẫu =  (100-2)/2+1 =50 -> 2.4.6.8....100= (100+2)*50/2 =2550

-->  M= 2500/2550 =50/51

Làm tương tự với N ta có kq N=51/52 ->M/N= 2600/2601 -> M<N

bấm phân số kiểu j z bạn

$#Nqocc$

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{29\cdot31}\)

`= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/29 - 1/31`

`= 1 - (1/3 - 1/3) - (1/5 - 1/5) - (1/7 - 1/7) - ... - (1/29 - 1/29) - 1/31`

`= 1- 1/31`

`= 30/31`

1 - 1/31 sao bằng 32/31>?

\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)

\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

 \(=\frac{1}{2}-\frac{1}{9}\)

            * LÀM NỐT *

                              #Louis

Mk ko biết lm nhưng cứ k thoải mái nha

SORRY

a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)

\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)

\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)

\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)

b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)

c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)

\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)

\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)

d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)

\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)

e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)

\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)

g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)

\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)

\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)

\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)

thank you,very well

\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)

\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)

Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:

\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)

\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)