So sánh hai phân số A=\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và B=\(\frac{2015+2016+2017}{2016+2017+2018}\)
Các bn giải giúp mk nha ! Mk cần gấp ! Thanks nhiều. ^-^
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
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
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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
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
HT
0
6 tháng 5 2019
có B=2015+2016+\(\frac{2017}{2016}\)+2017+2018
B=\(\frac{2015}{2015+2016+2017}\)+\(\frac{2016}{2016+2017+2018}\)+\(\frac{2017}{2016+2017+2018}\)
vì \(\frac{2015}{2016}\)>\(\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}\)>\(\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}\)>\(\frac{2017}{2016+2017+2018}\)
⇒A>B
Chúc bạn học tốt :")
N
6 tháng 5 2019
Dễ thấy B<1.
\(A=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)\)\(=3-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
Vậy A>2.
Vậy A>B.
1 tháng 5 2018
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
30 tháng 3 2018
Bài 1 : dễ bạn tự làm được :)
Bài 2 :
Ta có :
\(B=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
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 \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
30 tháng 3 2018
Ta có : B = 2016 + 2017 + 2018 2015 + 2016 + 2017 = 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 Vì : 2016 2015 > 2016 + 2017 + 2018 2015 2017 2016 > 2016 + 2017 + 2018 2016 2018 2017 > 2016 + 2017 + 2018 2017 Nên 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 ⇔ 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 ⇔A > B Vậy A > B Chúc bạn học tốt ~
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