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

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)

\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)

\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)

\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)

Vậy \(A>\frac{1}{10}\)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)

\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)

\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)

\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)

\(VayA>\frac{1}{100}=B\)

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

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

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

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

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

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

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

 \(=\frac{1}{2}-\frac{1}{9}\)

            * LÀM NỐT *

                              #Louis

Đặt C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)\(\left(C>0\right)\)

Và D = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)\(\left(D>0\right)\)

Ta có :

C .D = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}\)\(=\frac{1}{10001}\)\(\left(1\right)\)

Mặt khác :

\(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(.....\)

\(\frac{9999}{10000}< \frac{10000}{10001}\)

Nhân tất cả vế theo vế - - - > C < D - - - > C² < C . D \(\left(2\right)\)

\(\left(1\right),\left(2\right)\)- - - >C² < \(\frac{1}{10001}\)- - - >  C < căn \(\frac{1}{10001}\)< căn \(\frac{1}{10000}\)= \(\frac{1}{100}\)( đpcm )

P1 có 2 thừa số âm nhân với nhau nên P1 >0

P2 có 3 thừa số âm nhân với nhau nên P2 <0

P3 vì trong dấu ba chấm có thừa số 0/10 nên P3 =0

Vậy P2 < P3 < P1

KO khó lắm đâu. Mong bạn hiểu để bài sau tương tự thì làm được.

CHúc bạn học tốt.

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