So sánh 2011/2012+2012/2013+2013/2011 với 3
K
Khách
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Những câu hỏi liên quan
DH
1
HT
27 tháng 3 2016
Có : \(\frac{2011}{2012}=\frac{2012-1}{2012}=1-\frac{1}{2012}\)
Có : \(\frac{2012}{2013}=\frac{2013-1}{2013}=1-\frac{1}{2013}\)
Có : \(\frac{2013}{2011}=\frac{2011+2}{2011}=1+\frac{2}{2011}\)
Cộng vế với vế ta có : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{2}{2011}=1+1+1-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)=3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}>0\) nên \(3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)<3\)
Vậy \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}<3\)
MT
1
27 tháng 3 2016
\(\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Vì \(\frac{2010}{2011+2012+2013}<\frac{2010}{2011};\frac{2011}{2011+2012+2013}<\frac{2011}{2012};\frac{2012}{2011+2012+2013}<\frac{2012}{2013}\)
nên phép dưới nhỏ hơn phép trên
TG
0
BV
3
7 tháng 6 2021
Ta có :
B = \(\dfrac{2011}{2012}\) + \(\dfrac{2012}{2013}\) .
\(\dfrac{2011}{2012}\) > \(\dfrac{2011}{2012+2013}\) .
\(\dfrac{2012}{2013}\) > \(\dfrac{2012}{2012+2013}\) .
\(\Rightarrow\) A < B .
ND
1
H
21
14 tháng 6 2015
bạn tham khảo:
2010/2011+2012+2013 > 2010+2011+2012/2011+2012+2013
2011/2011+2012+2013 > 2010+2011+2012/2011+2012+2013
2012/2011+2012+2013 > 2010+2011+2012/2011+2012+2013
=> 2010/2011+2011/2012+2012/2013 > 2010+2011+2012/2011+2012+2013
14 tháng 6 2015
2010/2011+2012+2013 > 2010+2011+2012/2011+2012+2013
2011/2011+2012+2013 > 2010+2011+2012/2011+2012+2013
2012/2011+2012+2013 > 2010+2011+2012/2011+2012+2013
=> 2010/2011+2011/2012+2012/2013 > 2010+2011+2012/2011+2012+2013
LH
4
T
24 tháng 3 2018
Bài nãy sai rồi, cho mình làm lại nha:
\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}\)
\(=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{1}{2011}\)
Vì: \(\frac{1}{2011}>\frac{1}{2012}>\frac{1}{2013}\Rightarrow\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2012}+\frac{1}{2012}>0\)
\(\Rightarrow\frac{2012-1}{2012}+\frac{2013-1}{2013}+\frac{2011+1+1}{2011}>3\)
Nên \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)
DT
2
TT
18 tháng 5 2021
\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}+\dfrac{2013}{2011}\)
=1-\(\dfrac{1}{2011}\)+1\(-\dfrac{1}{2012}\)+1-\(\dfrac{1}{2013}\)+1-\(\dfrac{1}{2011}\)
=4-(\(\dfrac{2}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\)) < 4
18 tháng 5 2021
m=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}+\dfrac{2013}{2011}\)
=\(1-\dfrac{1}{2011}+1-\dfrac{1}{2012}+1-\dfrac{1}{2013}+1+\dfrac{2}{2011}\)
=4+\(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\)
vì:
do \(\dfrac{1}{2011}< 1\)
\(\dfrac{1}{2012}< 1\)
\(\dfrac{1}{2013}< 1\)
nên \(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}< 1-1-1=-1\)
hay \(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}< 0\)
nên 4+\(\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}< 4\)
vậy tổng m <4
bài này mình tưởng phải lên cấp 2 mới có thế mà mấy em lớp 4 đã phải làm á
Ta có : \(\frac{2011}{2012}=1-\frac{1}{2012}\)
\(\frac{2012}{2013}=1-\frac{1}{2013}\)
\(\frac{2013}{2011}=1+\frac{2}{2011}\)
Ta có : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)+\left(1+\frac{2}{2011}\right)\)
= \(\left(1+1+1\right)+\left(\frac{2}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)\)
= \(3+\frac{2}{2011}-\left(\frac{1}{2012}+\frac{1}{2013}\right)\)
Ta có :
\(\frac{1}{2012}+\frac{1}{2013}< \frac{1}{2012}+\frac{1}{2012}=\frac{2}{2012}\)
mà : \(\frac{2}{2012}< \frac{2}{2011}=>\frac{1}{2012}+\frac{1}{2013}< \frac{2}{2011}\)
=> \(\frac{2}{2011}-\left(\frac{1}{2012}+\frac{1}{2013}\right)>0\)
Vậy : \(3+\frac{2}{2011}-\left(\frac{1}{2012}+\frac{1}{2013}\right)>3\)
Vậy : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}>3\)
ủng hộ mik nhá các bạn ơiii ^_^"