Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)

Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)

Vậy B < 1

Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{8^2}<\frac{1}{7.8}\) 

<=> B<\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\) 

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

<=> B<\(1-\frac{1}{8}\) 

<=> B<\(\frac{7}{8}\) <1

Ta có 

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

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

đúng không đấy mình thấy cứ sai sai thế nào ý

mũ hai nhưng \(8^2\) mới lập lại 1 lần

Thank nhưng mình cần kiểm tra lại 

Ta có :  \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{8^2}< \frac{1}{7.8}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

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

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

\(\Rightarrow B< \frac{7}{8}\)

\(\Rightarrow B< \frac{8}{8}=1\)

Vậy \(B< 1\left(Đpcm\right)\)

Chúc bạn học tốt !!! 

nhan xet1/2^2<1/1.2=1/1-1/2

1/3^2<1/2.3=1/2-1/3

1/4^2<1/3.4=1/3-1/4

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

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

1/1-1/8=8/8-1/8=7/8<1 vay B<1

Ta có

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.......;\frac{1}{10^2_{ }}< \frac{1}{9.10}\)

Suy ra

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

Hay B < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

B<\(1-\frac{1}{10}\)

B<1(Vì 1/10 >0)

Học tốt nhé 

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

B < \(1-\frac{1}{8}\)mà 1 - 1/8 < 1

=> B < 1 ( dpcm )

Vậy ...

\(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}< 1-\frac{1}{8}=\frac{7}{8}< 1\)

Vậy B<1

Hok tốt

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

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}\)

vì \(\frac{1}{2^2}>\frac{1}{1\cdot2}\)

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

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

\(...\)

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

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

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

\(B=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{8-7}{7\cdot8}\)

\(B=\left(\frac{2}{1\cdot2}-\frac{1}{1\cdot2}\right)+\left(\frac{3}{2\cdot3}-\frac{2}{2\cdot3}\right)+...+\left(\frac{8}{7\cdot8}-\frac{1}{7\cdot8}\right)\)

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

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

\(B=\frac{7}{8}< 1\)    (2)

(1)(2) \(\Rightarrow A< B< 1\)

\(\Rightarrow A< 1\) (đpct)

bé hơn mấy???

Hình như hơi thiếu đề :(

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 = 1/2^2 + 1/3^2 + 1/4^2 + ... +1/8^2 < 1/1.2 + 1/2.3 + 1/3.4 +....+ 1/7.8 = 1 - 1/2 +1/2 - 1/3 + 1/3 - 1/4 +...+1/7 - 1/8 = 1 - 1/8 < 1

=> B < 1 (ĐPCM)