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+)Đặt \(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)
\(A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...+\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+...+\dfrac{1}{200}\right)\)\(A>\dfrac{1}{125}.25+\dfrac{1}{150}.25+\dfrac{1}{175}.25+\dfrac{1}{200}.25=\dfrac{533}{840}>\dfrac{5}{8}\)
+)\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)
\(A=\left(\dfrac{1}{101}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...+\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...+\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...+\dfrac{1}{200}\right)\)\(A< \dfrac{1}{100}.20+\dfrac{1}{120}.20+\dfrac{1}{140}.20+\dfrac{1}{160}.20+\dfrac{1}{180}.20=\dfrac{1879}{2520}< \dfrac{3}{4}\)
Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Lại có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)
Vậy ...
Những dãy trên đều có 100 số hạng.
Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
.............................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.........+\dfrac{1}{\sqrt{99}}+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+.....+\dfrac{1}{10}=\dfrac{100}{10}=10\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)
caanf
cái j đấy nhỉ????mk k hiểu?