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S = \(\frac{1}{20}+\frac{1}{21}...+\frac{1}{199}+\frac{1}{200}\) ( có 181 phân số )
=> S > \(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}+\frac{1}{200}\)
=> S > \(\frac{1}{200}.181\)
=> S > \(\frac{181}{200}\)> \(\frac{180}{200}\)= \(\frac{9}{10}\)
Vậy S > 9 / 10
Ta có :
\(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
\(A=\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{200}\right)\)
\(A>\left(\frac{1}{150}+\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\right)+\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)\)
\(A>50.\frac{1}{150}+50\frac{1}{200}\)
\(A>\frac{50}{150}+\frac{50}{200}\)
\(A>\frac{1}{3}+\frac{1}{4}\)
\(A>\frac{7}{12}\)
Vậy \(A>\frac{7}{12}\)
Chúc bạn học tốt ~
Ta có:\(\frac{1}{101}>\frac{1}{200}\)
\(\frac{1}{102}>\frac{1}{200}\)
\(\frac{1}{103}>\frac{1}{200}\)
A=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}.100\)
hay A>\(\frac{7}{12}\)
A=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}.100\)
hay A>\(\frac{5}{8}\)
mình ko biết có đúng ko bạn xem kĩ nhé
Ta có:
\(\frac{1}{101}\)>\(\frac{1}{200}\)
\(\frac{1}{102}\)>\(\frac{1}{200}\)
\(\frac{1}{103}\)>\(\frac{1}{200}\)
...
\(\frac{1}{200}\)=\(\frac{1}{200}\)
\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)>\(\frac{1}{200}\)+\(\frac{1}{200}\)+..+\(\frac{1}{200}\)(100 số hạng)=\(\frac{1}{2}\)
\(\Rightarrow\)\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)>\(\frac{1}{2}\)
Ta có
\(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};...;\frac{1}{149}>\frac{1}{150}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}=\frac{1}{150}.50=\frac{1}{3}\)
Ta lại có
\(\frac{1}{151}>\frac{1}{200};\frac{1}{152}>\frac{1}{200};...;\frac{1}{199}>\frac{1}{200}\)
\(\Rightarrow\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}=\frac{1}{200}.50=\frac{1}{4}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\RightarrowĐPCM\)
ta có
1/101 > 1/150
1/102> 1/150
...>1/150
1/150 = 1/150
=> 1/101 + 1/102 + .... + 1/150 > 1/150 +1/150+....+1/150(50 số hạng )= 1/3
ta có
1/151 >1/200
1/152 > 1/200
..>1/200
1/200 = 1/200
=> 1/151 + 1/152+....+1/200 > 1/200+1/200+ ...+1/200( 50 số hạng) = 1/4
==> 1/101 + 1/102+....+1/200 > 1/3 +1/4
==> A > 7/12
Tách A thành 2 nhóm A1 , A2
A1 = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\)
A2 = \(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{200}>\frac{1}{200}.50=\frac{1}{4}\)
\(\Rightarrow\)A = A1 + A2 > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)