Giải:

a) Ta có:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

1/5²=1/5.5 < 1/4.5

1/6²=1/6.6 < 1/5.6

1/7²=1/7.7 < 1/6.7

1/8²=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

   B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

   B<1/1-1/8

   B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

   S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

   S=1/1-1/46

   S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!

a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)

\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)

Vậy B < 1 (đpcm)

Đặt B=122+132+...+182B=122+132+...+182A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8

=1−18<1(2)=1−18<1(2)

Từ  ta có: 

c1:A=B

c2:A=11

c3:B=1\20

c4:mk k bit

A = 1/2^2 + 1/3^2 +.. + 1/8^2 < 1/1.2 + 1/2.3  +... + 1/7.8 = 1 - 1/2 + 1/2 -1/3 +...  + 1/7 - 1/8

=  1 - 1/8 < 1 

\(\Rightarrowđpcm\)

\(tíchnhaminhftchlaij\)

b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)

                 \(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)

                 \(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)

                 \(=1+1,5+2+...+8,5\)

                 \(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)

B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)                                                                               

 B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)

Vậy 1/8<2 hay 1/8<16/8