Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
TA có :\(\frac{2015.2016-1}{2015.2016}=\frac{2015.2016}{2015.2016}-\frac{1}{2015.2016}=1-\frac{1}{2015.2016}\)
Ta có:\(\frac{2016.2017-1}{2016.2017}=\frac{2016.2017}{2016.2017}-\frac{1}{2016.2017}=1-\frac{1}{2016.2017}\)
Vì \(2015.2016< 2016.2017\)
\(\Rightarrow\frac{1}{2015.2016}>\frac{1}{2016.2017}\)
\(\Rightarrow1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)
\(\Rightarrow\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)
Vậy \(\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)
Xét :
\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)
\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)
Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm )
...
1-1/2015 + 1-1/2016
=(1+1)-(1/2016+1/2015)
=2-4031/4062240
2015+2016/2016+2017
=4031/4033
2-4031/4062240>1
4031/4033<1
vậy 2014/2015 + 2015/2016 > 2015+2016/2016+2017