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\(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)
\(< \frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\)
\(< \frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)
\(< \frac{1}{100}-\frac{1}{105}=\frac{1}{2100}\)
\(< \frac{1}{2^2.3.5^2.7}\)
\(A=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)
\(A< \frac{1}{100\cdot101}+\frac{1}{101\cdot102}+\frac{1}{102\cdot103}+\frac{1}{103\cdot104}+\frac{1}{104\cdot105}\)
\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)
\(=\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{2^2\cdot3\cdot5^2\cdot7}=B\)
Vậy \(A< B\)
\(B=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}<\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{103.104}\)
Tính VP ra là được
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
\(\text{Ta có :}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100.101}+\frac{1}{101.102}+.....+\frac{1}{105.106}\)
\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+....+\frac{1}{105}-\frac{1}{106}\)\
\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{105}\)
\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2100}\)
\(\text{Mà :}\)\(\frac{1}{2100}=\frac{1}{2^2.3.5^2.7}\)
\(\text{Nên:}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2^2.3.5^2.7}\)
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
an vo cai nay la vo tra loi
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