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Chứng minh:\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{149}+\frac{1}{150}>\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy tổng trên có 50 số hạng .
Ta có:
1/101>1/150
1/102>1/150
...
1/149>1/150
1/150=1/150
=>1/101+1/102+...+1/149+1/150>1/150+1/150+...+1/150
---50 số hạng 1/150-------
=>1/101+1/102+...+1/149+1/150>1/150.50
=>1/101+1/102+...+1/149+1/150>50/150
=>1/101+1/102+...+1/149+1/150>1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\)(50 phân số)
=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\)(50 phân số)
=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50\)
=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{3}\)(Đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
1/101 + 1/102 + ... + 1/149 + 1/150 > 1/150 + 1/150 + ... + 1/150 + 1/150
50 phân số 1/150
> 50.1/150
> 1/3
=> đpcm
ta có
\(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};....;\frac{1}{150}=\frac{1}{150}\)
=>tổng>\(\frac{1}{150}.50=\frac{1}{3}\)
=>đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
RÚT GỌN:
\(\frac{3.13-13.18}{15.40-80}=\frac{13.\left(3-18\right)}{15.40-40.2}=\frac{13.\left(-15\right)}{40\left(15-2\right)}=\frac{13.\left(-15\right)}{40.13}=-\frac{15}{40}=-\frac{3}{8}\)
CHỨNG MINH:
Ta thấy \(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};...;\frac{1}{149}>\frac{1}{150}\)
=>\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}.\left(150-101+1\right)=\frac{1}{150}.50=\frac{50}{150}=\frac{1}{3}\)(đpcm)
TÍNH HỢP LÝ:
B=\(\frac{5}{13}+\frac{-5}{7}-\frac{20}{41}+\frac{8}{13}+\frac{-21}{41}=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-21}{41}-\frac{20}{41}\right)+\frac{-5}{7}=1+\left(-1\right)+\frac{-5}{7}=0+\frac{-5}{7}=\frac{-5}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{1}{101}>\frac{1}{200}\)
\(\frac{1}{102}>\frac{1}{200}\)
\(\frac{1}{103}>\frac{1}{200}\)
\(..........\)
\(\frac{1}{200}=\frac{1}{200}\)
Cộng vế với vế ta được :
\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\) (có 100 số \(\frac{1}{200}\) )\(=\frac{100}{200}=\frac{1}{2}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+......+\frac{1}{200}>\frac{1}{2}\) (đpcm)
Ta có:
1/101>1/200
1/102>1/200
...
1/199>1/200
=>1/101+1/102+...+1/103>1/200+1/200+...+1/200(100 số 1/200)
=1/200.100=1/2
Vậy 1/101+1/102+1/103+...+1/200>1/2
Ta thấy 1/101>1/150 ; 1/102>1/150 ; .... ; 1/149>1/150 ; 1/150=1/150
suy ra 1/101+1/102+1/103+.....+1/149+1/150>50.1/150
1/101+1/102+1/103+.....+1/148+1/150>1/3