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)
=>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
1/101+1/102+1/103+...+1/299+1/300>2/3
Chứng tỏ rằng 1/101+1/102+....+1/299+1/300>2/3
chung to rang 1/101+1/102+...+1/299+1/300>2/3
Chứng minh:
1/101+1/102+1/103+...+1/299+1/300>2/3
chứng tỏ rằng 1/101+1/102+...+1/299+1/300>2/3
Chứng minh rằng:
1/101+1/102+...+1/299+1/300>2/3
Chứng tỏ rằng: 1/101+1/102+....+1/299+1/300 > 2/3
Tính tích A = 3/4.8/9.15/16....899/900
tính A/B biết :
A=1/1*300+1/2*301+1/3*302+...+1/101*400
B=1/1*102+1/2*103+1/3*104+...+1/299*400
Chứng tỏ rằng\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
