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

=>B<1-1/2+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<1-1/8=7/8<1

Vậy B<1

k cho mk nha

1: 

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

...

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)

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

Ta có B=1/2²+1/3²+...+1/8²<1/1.2+1/2.3+...+1/7.8=1/1-1/2+1/2-...+1/7-1/8=1/1-1/8=7/8<1

Vậy B<1

Ta có : \(B=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}\)

=> \(B<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)

                                                                                            \(=1-\frac{1}{8}\)

                                                                                            \(=\frac{7}{8}\)<1

Vậy B < 1

ta thay 1/2²<1/1.2

1/3²<1/2.3

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

1/8²<1/7.8

nen B < 1/1.2+1/2.3+1/3.4+.....+1/7.8

nen B < 1/1-1/8

B<1

dm mày ngu vừa thôi

Có 1/2^2 < 1/1.2

      1/3^2 <1/2.3

          ...

      1/8^2 < 1<7.8

...tự làm như các phép bình thường

Đặt A=1/1.2+1/2.3+....+1/7.8

Ta có:

B=1/2^2+1/3^2+....+1/8^2<A=1/1.2+1/2.3+....+1/7.8 (1)

Mà A=1/1.2+1/2.3+....+1/7.8

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

=1-1/8<1 (2)

Từ (1) và (2) ta có: B<A<1

=>B< (Đpcm)

bạn xét :1/2+1/3+1/4>1 
vậy 1/5+1/6+1/7+1/8...>1 
vậy nó >2 
cách khác. 
tính S62=31*[2*1/2-(62-1)*(-1/6)]>2

163-x=13\5

163-x=2dư3

\(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}\)

\(>\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)\)

\(>\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+...+\frac{1}{8}\right)+\left(\frac{1}{16}+...+\frac{1}{16}\right)\)

\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)

giúp mình nhé

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}+\dfrac{1}{64}\\ =\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\)

Ta thấy:

\(\dfrac{1}{3}\) lớn hơn \(\dfrac{1}{4}\)

\(\dfrac{1}{5};\dfrac{1}{6};\dfrac{1}{7}\) lớn hơn \(\dfrac{1}{8}\)

\(\dfrac{1}{9};\dfrac{1}{10};...;\dfrac{1}{15}\) lớn hơn \(\dfrac{1}{16}\)

\(\dfrac{1}{17};\dfrac{1}{18};...;\dfrac{1}{31}\) lớn hơn \(\dfrac{1}{32}\)

\(\dfrac{1}{33};\dfrac{1}{34};...;\dfrac{1}{63}\) lớn hơn \(\dfrac{1}{64}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)(ĐPCM)