a, ta xét:

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

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

\(\frac{5}{6}< \frac{6}{7}\)

.....

\(\frac{99}{100}< \frac{100}{101}\)

=>\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)

hay:A<B(đpcm)

b,\(A.B=\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.....\frac{100}{101}\)

\(=\frac{1.2.3....100}{2.3.4....101}=\frac{1}{101}\)

c,vì A<B (theo phần a)

=>A.A<B.A

Mà B.A=\(\frac{1}{101}\)

=>A²<101

Mà A²=\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)

=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{101}\)<\(\frac{1}{100}=\frac{1}{10^2}\)

=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{10^2}\)

=>\(\frac{1}{2}.\frac{3}{4}....\frac{99}{100}< \frac{1}{10}\)

Hay A<\(\frac{1}{10}\)

a, Xét 1/2 < 2/3 ; 3/4<4/5 ; ............ ; 99/100<100/101

=> 1/2.3/4.......99/100 < 2/3.4/5.........100/101

=> M<N

b, M.N = 1/2.3/4.4/5......99/100.2/3.4/5.5/6......100/101

M.N = 1/2.2/3.3/4.4/5.............99/100.100/101

M.N = 1/101

c, Vì M<N nên M.M < M.N Hay M.M < 1/101 < 1/100

                                           hay M.M < 1/10 . 1/10

=> M < 1/10 (Đpcm)

 a) Ta có M.N = 1/2.2/3.3/4.4/5....99/10.10/101 = 1/101 
b) Xét M và N đều gồm 50 thừa số mà: 
1/2 < 2/3 
3/4 < 4/5 
............. 
99/100 < 100/101 
=> M < N 
c) Do M < N nên => M.M < M.N (Nhân 2 vế với M) 
=> M.M < 1/101 (Vì M.N = 1/101 theo cma) 
Mặt khác 1/101 < 1/100 
=> M.M < 1/100 = 1/10.1/10 
=> M < 1/10

CMR A = 1/5 + 1/6 + 1/7 + ... + 1/17 < 2

A = 1/5 + 1/6 + 1/7 + ... + 1/17

Vì 1/6 < 1/7 < 1/8 < 1/9 < 1/5 và 1/10 < 1/11 < 1/12 < 1/13 < 1/14 <1/15 < 1/16 < 1/17 < 1/8 nên:

A = (1/5 + 1/6 + 1/7 + 1/8 + 1/9) + (1/10 + 1/11 + 1/12 + 1/13 + 1/14+ 1/15 + 1/16 + 1/17)

A < (1/5 + 1/5 + 1/5 + 1/5 + 1/5) + (1/8 + 1/8 + 1/8 + 1/8 + 1/8+1/8 + 1/8 + 1/8)

A < 1 + 1

A < 2

Vậy: A < 2 (đpcm)