So sánh 2 biểu thức sau:
A= \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
B=\(\frac{2015+2016}{2016+2017}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
5 tháng 5 2018
Ta có : \(B=\frac{2015+2016+2017}{2016+2017+2018}\) \(=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)
Cộng vế theo vế, ta có :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
TM
4 tháng 5 2018
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Ta có:
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Cộng vế theo vế, ta có:
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Vậy A > B
28 tháng 5 2021
Bạn có nhầm không, tớ thấy cả hai đều giống nhau mà,
Hai cái bằng nhau
2 tháng 5 2016
2014+2015+2016/2015+2016+2017<2014/2015+2015/2016+2016/2017
26 tháng 4 2018
\(Q=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\)\(\frac{2017}{2016+2017+2018}\)
ta có :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
nên \(P>Q\)
26 tháng 4 2018
Q=2015+2016+2017/2016+2017+2018=+2018+2016/2016+2017+2018+2017/2016+2017+2018
vì 2015/2016>2015/2016+2017+2018[1]
2016/2017>2016+2017+2018[2]
2017/2018>2016+2017+2018[3]
từ [1] [2] [3] suy ra P>Q
DL
5 tháng 5 2018
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 .
5 tháng 5 2018
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
LP
0
14 tháng 4 2016
dễ thấy B=\(\frac{2015+2016}{2016+2017}\)<1
A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)=1-\(\frac{1}{2016}\)+1-\(\frac{1}{2017}\)=(1+1)-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))=2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))
vì (\(\frac{1}{2016}\)+\(\frac{1}{2017}\))<0,5+0,5=1 suy ra 2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))>1 mà b<1suy ra A>B
15 tháng 4 2016
Ta thấy: B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)
A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
Mà\(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)
Suy ra: \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)>\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)=\(\frac{2015+2016}{2016+2017}\)
Hay A>B
17 tháng 4 2017
Ta có:
\(B=\frac{2015+2016}{2016+2017}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
vì: \(\frac{2015}{2016}>\frac{2015}{2016+2017}\)VÀ \(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
\(\Rightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Vậy: \(A>B\)
T
0
bang nhau
\(\frac{2015+2016}{2016+2017}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
\(\frac{2015}{2016}>\frac{2015}{2016+2017}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(A>B;\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)