Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2020}{2027}=1-\frac{7}{2027}\)
\(\frac{2027}{2034}=1-\frac{7}{2034}\)
Vì \(\frac{7}{2027}>\frac{7}{2034}\)
Nên \(1-\frac{7}{2027}< 1-\frac{7}{2034}\)
Vậy \(\frac{2020}{2027}< \frac{2027}{2034}\)
\(\text{Ta có :}\)
\(\frac{2020}{2027}=1-\frac{7}{2027}\)
\(\frac{2027}{2034}=1-\frac{7}{2034}\)
\(\text{VÌ }\frac{7}{2027}>\frac{7}{2034}\text{ nên }1-\frac{7}{2027}< 1-\frac{7}{2034}\)
\(\text{Vậy }\frac{2020}{2027}< \frac{2027}{2034}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2020}{2019}\)bé hơn \(\frac{2021}{2020}\)
vì 2020 bé hơn 2021
2019 nhỏ hơn 2020
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1001}{1000}\)và \(\frac{1002}{1003}\)
Giải
Vì
\(\frac{1001}{1000}\)\(>1\)
\(\frac{1002}{1003}\)\(< 1\)
Nên
\(\frac{1001}{1000}\)\(>\frac{1002}{1003}\)
Hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{2000}{-2001}=-\frac{2000}{2001}=-\left(\frac{2001-1}{2001}\right)=-\left(\frac{2001}{2001}-\frac{1}{2001}\right)=-\left(1-\frac{1}{2001}\right)=-1+\frac{1}{2001}\)
\(-\frac{2003}{2002}=-\left(\frac{2002+1}{2002}\right)=-\left(\frac{2002}{2002}+\frac{1}{2002}\right)=-\left(1+\frac{1}{2002}\right)=-1-\frac{1}{2002}\)
Vì \(\frac{1}{2001}>-\frac{1}{2002}\) nên \(-1+\frac{1}{2001}>-1-\frac{1}{2002}\)
hay \(\frac{2000}{-2001}>-\frac{2003}{2002}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d: -11/17=-22/34
22/31>22/34
=>-22/31<-22/34
e: -23/91=-1587/91*69
-17/69=-1547/91*69
mà -1587<-1547
nên -23/91<-17/69
b: -1989/1991>-1>-2001/2000
c: 1/1000>0>-120/157
i: 2021/2020=1+1/2020
2022/2021=1+1/2021
mà 1/2020>1/2021
nên 2021/2020>2022/2021
f: 91/87>1>102/104
a)>
b)<
c)>
d)>
e)>
f)>
g)>
h)>
i)>