Chứng tỏ rằng :
\(\frac{1}{101}\)+ \(\frac{1}{102}\)+ ...... + \(\frac{1}{299}\)+ \(\frac{1}{300}\)> \(\frac{2}{3}\)
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theo tớ nghĩ:
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}.200=\frac{2}{3}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}\)
Ta có:
\(\frac{1}{101}>\frac{1}{300}\)
\(\frac{1}{102}>\frac{1}{300}\)
..........................
\(\frac{1}{299}>\frac{1}{300}\)
\(\frac{1}{300}=\frac{1}{300}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)
\(\Rightarrow VT>200.\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\) (ĐPCM)
Ta đặt \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}\)
Vì \(\frac{1}{101}>\frac{1}{102}>...>\frac{1}{299}>\frac{1}{300}\)
\(\Rightarrow A=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)
\(\Rightarrow A>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)
\(\Rightarrow A>\left(\frac{1}{200}\cdot100\right)+\left(\frac{1}{300}\cdot100\right)\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow A>\frac{5}{6}>\frac{2}{3}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}\)>\(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)=\(\frac{1}{2}\)(có 200 c/s\(\frac{1}{200}\))
\(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\)>\(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)=\(\frac{2}{3}\)(có 200 c/s \(\frac{1}{300}\))
Vậy \(\frac{1}{101}+\frac{1}{102+}+....+\frac{1}{300}\)>\(\frac{1}{2}+\frac{2}{3}=\frac{2}{3}\) Đpcm
Hok tốt
ta có
\(\frac{1}{300}< \frac{1}{101}\); \(\frac{1}{300}< \frac{1}{102}\); \(\frac{1}{300}< \frac{1}{102}\)....\(\frac{1}{300}< \frac{1}{299}\)
\(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}< \frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)
\(\frac{200}{300}< \frac{1}{101}+\frac{1}{102}+...+\text{}\text{}\)
rút gọn là xong
Đặt A=1/101+1/102+1/103+...+1/300
vì 1/101>1/102>1/103>...>1/300
=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!)
=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100
=> A > 1/2+1/3
=> A > 5/6
Mà 5/6>2/3
=> A > 2/3
Vậy 1/101+1/102+1/103+...+1/300 >2/3
Vì : 1/101 > 1/300 ; 1/102 > 1/300 .... ; 1/299 >1/300 ; Do 1/101.....1/300 có 200 số
=>1/101+1/102+....+1/299+1/300 > 1/300 x 200
> 2/3
1-1/2+1/3-1/4+...+1/199-1/200
=(1+1/3+...+1/199)-(1/2+1/4+...+1/200)
=(1+1/2+1/3+...+1/199+1/200)-2(1/2+1/4+...+1/200)
=(1+1/2+1/3+...+1/199+1/200)-(1+1/2+...+1/100)
=1/101+1/102+...+1/200 (đpcm)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)( có 200 số )
Ta có
\(\frac{1}{101}>\frac{1}{300}\); \(\frac{1}{102}>\frac{1}{300}\); ...;\(\frac{1}{299}>\frac{1}{300}\)
=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}+\frac{1}{300}\)
=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}.200\)
=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{2}{3}\)( dpcm )
Ta có\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200.\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\Rightarrowđpcm\)
Đặt A=1/101+1/102+1/103+...+1/300
vì 1/101>1/102>1/103>...>1/300
=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!)
=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100
=> A > 1/2+1/3
=> A > 5/6
Mà 5/6>2/3
=> A > 2/3
Vậy 1/101+1/102+1/103+...+1/300 >2/3
Đặt A=1/101+1/102+1/103+...+1/300
vì 1/101>1/102>1/103>...>1/300
=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!)
=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100
=> A > 1/2+1/3
=> A > 5/6
Mà 5/6>2/3
=> A > 2/3
Vậy 1/101+1/102+1/103+...+1/300 >2/3
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< 1-\frac{1}{8}< 1\)
Ta có
\(\frac{1}{101}>\frac{2}{3}\)
\(\frac{1}{102}>\frac{2}{3}\)
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\(\frac{1}{300}>\frac{2}{3}\)
Vậy \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)