B = \(\frac{1}{2^2.3-5^2.7}\)= \(\frac{-1}{163}\)

Đặt A = \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{105^2}\)

Vì A > 0 (các số hạng của A đều > 0)

Mà B < 0

=> A > B

-Ta có:A= 1/101^2+1/102^2+1/103^2+1/104^2+1/105^2 
         A>1/(100x101)+1/(101x102)+1/(102x103)+... 

-Vì cùng tử mẫu nhỏ hơn thì lớn hơn
         A>1/100-1/101+1/101-1/102+1/102-1/103+... 
         A>1/100-1/105=1/2100=1/(2^2.3.5^2.7)=B 
=>Vậy A>B

Ta có: \(A=\frac{1}{101^2}+\frac{1}{102^2}+......\frac{1}{105^2};\frac{1}{2^2.3.5^2.7}\)

\(A>\frac{1}{\left(101.101\right)}+\frac{1}{\left(101.102\right)}+\frac{1}{\left(102.103\right)}+......\frac{1}{\left(104.105\right)}\)

Ta thấy mỗi mẫu đều < thì => sẽ lớn hơn

\(A>\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+........\)

\(A>\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{\left(2^2.3.5^2.7\right)}=B\)

=> gọi vế \(\frac{1}{\left(2^2.2.5^2.7\right)}\) là B

=> A>B

\(\text{Ta có :}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100.101}+\frac{1}{101.102}+.....+\frac{1}{105.106}\)

                \(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+....+\frac{1}{105}-\frac{1}{106}\)\

               \(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{105}\)

              \(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2100}\)

             \(\text{Mà :}\)\(\frac{1}{2100}=\frac{1}{2^2.3.5^2.7}\)

             \(\text{Nên:}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2^2.3.5^2.7}\)

\(A=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)

\(A< \frac{1}{100\cdot101}+\frac{1}{101\cdot102}+\frac{1}{102\cdot103}+\frac{1}{103\cdot104}+\frac{1}{104\cdot105}\)

\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)

\(=\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{2^2\cdot3\cdot5^2\cdot7}=B\)

Vậy \(A< B\)

A = \(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)< \(\frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\) =\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\) 

= \(\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}\)= \(\frac{1}{2^2.3.5^2.7}\)= B

Vậy A < B

\(A

\(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)

\(< \frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\)

\(< \frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)

\(< \frac{1}{100}-\frac{1}{105}=\frac{1}{2100}\)

\(< \frac{1}{2^2.3.5^2.7}\)

\(B=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}<\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{103.104}\)

Tính VP ra là được 

A<1/100.101+1/101.102+..+1/104.105

=> A<1/100-1/105=1/2100

Ma B=1/2100

=> A<B

khó quá