so sánh :
A=2015/2016+2016/2017+2017/2015 với 3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Hay \(A>B\)
Đặt 2015.2016+2016=n
suy ra A=(n+1)/n và B=(n+2)/(n+1)
Ta có A - B=(n+1)/n -(n+2)/(n+1)=((n+1)2-n(n+2))/n(n+1)=(n2+2n+1-n2-2n)/n(n+1)=1/n(n+1)
Vì A-B lớn hơn 0 nên A>B
Ta có
1 - A = 1 - 2014/2015 = 1/2015
1 - B = 1 - 2015/ 2016 = 1/2016
Vì 1/2015 > 1/2016 => 1 - 2014/2015 > 1 - 2015 / 2016
Hay 1 - A > 1 -B => A < B
A=2015/2016+2016/2017+2017/2018>2015/2018+2016/2018+2017/2018
=6048/2018>1
B=2015+2016+2017/2016+2017+2018=6048/6051<1
=>A>B
Có: B = 2015 + 2016 + 2017/2016 + 2017 + 2018
B= 2015 / (2015 + 2016+2017) + 2016/(2016+2017+2018) + 2017/(2016 + 2017 + 2018)
vì 2015/2016 > 2015/(2016 + 2017+2018) ; 2016/2017>2016/(2016+2017+2018) ; 2017/2018 > 2017/(2016+2017+2018)
=> A>B
ta có 2015/2016+2016/2017+2017/2015=(1-1/2016)+(1-1/2017)+(2+1/2015)
=4-(1/2016+1/2017-1/2015)
1/2016<1; 1/2017<1 nên 1/2016+1/2017<2 suy ra 1/2016+1/2017-1/2015<1(vì 1/2015<1)
4-(1/2016+1/2017-1/2015)>4-1=3
2015/2016+2016/2017+2017/2015>3
cho mik nhé