Giải:

Dễ thấy:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(.................\)

\(\dfrac{1}{8^2}=\dfrac{1}{8.8}< \dfrac{1}{7.8}\)

Cộng vế theo vế ta được:

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\)

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

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

Vậy \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{8^2}< 1\) (Đpcm)

Ta có:

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

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

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

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

\(\Rightarrow B< 1\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=\frac{2-1}{1.2}+......+\frac{8-7}{7.8}\)

\(=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}=1-\frac{1}{8}< 1\)

ta có điều phải chứng minh

Ta có : 1/2^2 < 1/1.2

             1/3^2 < 1/2.3

             1/4^2 < 1/3.4

              ...

              1/8^2 < 1/7.8

=> B < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/7.8

B < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8

B < 1 - 1/8 < 1

=> B < 1 (đpcm)

Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{8^2}< \frac{1}{7\cdot8}\)

\(\Rightarrow B< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{7\cdot8}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow B< 1-\frac{1}{8}< 1\left(đpcm\right)\)

Ta có 

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

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...............

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

=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)

B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))

Vậy.....

Ta có 1/2²<1/1.2

         1/3²<1/2.3

         1/4²<1/3.4

         ................

        1/8²<1/7.8

=>B<1/1.2+1/2.3+1/3.4+...+1/7.8

=>B<1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8

=>B<1-1/8

Vậy B < 1

ad a zwe zxdb WE4RBTa

Ta có 1/22<1/1.2

         1/32<1/2.3

         1/42<1/3.4

         ................

        1/8²<1/7.8

=>B<1/1.2+1/2.3+1/3.4+...+1/7.8

=>B<1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8

=>B<1-1/8

Vậy B < 1 ai đồng tình với mình ko

b=1/2²+1/3²+1/4²+...+1/8²<1/1.2+1/2.3+1/3.4+......+1/7.8

b=1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8

b=1-1/8

b=7/8

<=>b<1

k cho mink nha

b=1/22+1/32+1/42+...+1/82<1/1.2+1/2.3+1/3.4+......+1/7.8

b=1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8

b=1-1/8

b=7/8

<=>b<1
owo

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

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 

\(\Rightarrowđpcm\)

\(tíchnhaminhftchlaij\)