Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tính A và B rồi ta đi so sánh:
A = \(\frac{2016}{2017}\) + \(\frac{2017}{2018}\) = \(1.999008674\)
B = \(\frac{2016+2017}{2017+2018}\) = \(0.9995043371\)
Mà 1.999008674 > 0.9995043371
Nên: A > B
B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)
Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.
Vậy A > B .
Bạn Dont look at me
Bạn nên làm theo bạn ấy
Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng
Theo mk là vậy
Ta có:2015/2016>2015/2016+2017+2018
2016/2017>2016/2016+2017+2018
2017/2018>2017/2016+2017+2018-Mình áp dụng so sánh phân số cùng tử đấy.
Suy ra2015/2016+2016/2017+2017/2018>(2015+2016+2017)/(2016+2017+2018)=B
a, Bn quy đồng rồi làm nha
b,Có A=2017^2017+1/2017^2018+1
--> 2017A=2017^2018+2017/2017^2018+1
2017A=2017^2018+1/2017^2018+1 + 2016/2017^2018+1
2017A=1+ 2016/2017^2018+1
Có B=2017^2016+1/2017^2017+1
--> 2017B=2017^2017+2017/2017^2017+1
2017B=2017^2017+1/2017^2017+1 + 2016/2017^2017+1
2017B=1+2016/2017^2017+1
Vì 1+2016/2017^2018+1 < 1+2016/2017^2017+1
nên 2017A<2017B
-->A<B
A.Ta có :
\(A=-\frac{15}{46}>-\frac{15}{45}=-\frac{51}{153}>-\frac{51}{151}=B\)
\(\Rightarrow A>B\)
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 \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
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 \(A>B\)
Có: \(A>\frac{2016}{2016}+\frac{2017}{2017}=2\)
Có: \(B=\frac{4035}{4033}< 2\)
\(\Rightarrow A>B.\)
\(B=\frac{2017+2018}{2016+2017}=\frac{2017}{2016+2017}+\frac{2018}{2016+2017}\)
Ta có
\(\frac{2017}{2016+2017}< \frac{2017}{2016}\) ;
\(\frac{2018}{2017}< \frac{2018}{2017}\)
\(\Rightarrow B< \frac{2017}{2016}+\frac{2018}{2017}=A\)
Vậy B<A