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
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 \(A>B\)
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 :
\(\frac{2016}{2017}>\frac{2016}{2017+2018+2019}\)
\(\frac{2017}{2018}>\frac{2017}{2017+2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2017+2018+2019}\)
\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}>\) \(\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)
\(\Rightarrow P>\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow P>Q\)
Chúc bạn học tốt !!!
vì P có các số bé hơn 1 còn Q có các số lớn hơn 1 =>P<Q
Vậy P<Q.
mình làm hơi tắt xin bạn thông cảm bạn tự viết các số có trong P;Q ra nhá
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
Ta có :
\(\dfrac{2016}{2017}>\dfrac{2016}{2018}\Rightarrow A>\dfrac{2016}{2018}+\dfrac{2017}{2018}\)\(\Rightarrow A>\dfrac{2016+2017}{2018}\)
\(B=\dfrac{2016+2017}{2017+2018}=\dfrac{2016+2017}{4035}\)
Vì \(\dfrac{2016+2017}{2018}>\dfrac{2016+2017}{4035}\)
\(\Rightarrow A>B\)
~ Chúc bn học tốt ~
B=\(\dfrac{2016+2017}{2017+2018}\)=\(\dfrac{2016}{2017+2018}+\dfrac{2017}{2017+2018}\)
Ta có:
\(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)\(^{\left(1\right)}\)
\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)\(^{\left(2\right)}\)
Cộng vế với vế của biểu thức (1), (2) suy ra:
\(\dfrac{2016}{2017+2018}+\dfrac{2017}{2017+2018}< \dfrac{2016}{2017}+\dfrac{2017}{2018}\)
Hay A<B