So sánh hai phân số :
A = \(\frac{2016^{2016}+1}{2016^{2017}+1}\)và B = \(\frac{2016^{2015}+1}{2016^{2016}+1}\)
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
PH
1
8 tháng 4 2017
TA có :\(\frac{2015.2016-1}{2015.2016}=\frac{2015.2016}{2015.2016}-\frac{1}{2015.2016}=1-\frac{1}{2015.2016}\)
Ta có:\(\frac{2016.2017-1}{2016.2017}=\frac{2016.2017}{2016.2017}-\frac{1}{2016.2017}=1-\frac{1}{2016.2017}\)
Vì \(2015.2016< 2016.2017\)
\(\Rightarrow\frac{1}{2015.2016}>\frac{1}{2016.2017}\)
\(\Rightarrow1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)
\(\Rightarrow\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)
Vậy \(\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)
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
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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
CH
Cô Hoàng Huyền
Admin
VIP
8 tháng 5 2017
Ta có
\(2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=\frac{2016^{2017}+1}{2016^{2017}+1}+\frac{2015}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)
\(2016B=\frac{2016^{2016}+2016}{2016^{2016}+1}=\frac{2016^{2016}+1}{2016^{2016}+1}+\frac{2015}{2016^{2016}+1}=1+\frac{2015}{2016^{2016}+1}\)
Do \(\frac{2015}{2016^{2017}+1}< \frac{2015}{2016^{2016}+1}\Rightarrow2016A< 2016B\Rightarrow A< B.\)
8 tháng 5 2017
B = \(\frac{2016^{2015}+1}{2016^{2016}+1}\)< A =\(\frac{2016^{2016}+1}{2016^{2017}+1}\)
21 tháng 9 2017
Ta có :
\(x=\frac{2016^{2017}+1}{2016^{2016}+1}\)
\(\frac{1}{2016}x=\frac{2016^{2017}+1}{2016^{2017}+2016}=\frac{2016^{2017}+2016-2015}{2016^{2017}+2016}\)
\(\Rightarrow\frac{1}{2006}x=1-\frac{2015}{2016^{2017}+2016}\)
Ta lại có :
\(y=\frac{2016^{2016}+1}{2016^{2015}+1}\)
\(\Rightarrow\frac{1}{2016}y=\frac{2016^{2016}+1}{2016^{2016}+2016}=\frac{2016^{2016}+2016-2015}{2016^{2016}+2016}\)
\(\Rightarrow\frac{1}{2016}y=1-\frac{2015}{2016^{2016}+2016}\)
Mà \(\frac{2015}{2016^{2017}+2016}< \frac{2015}{2016^{2016}+2016}\)(so sánh mẫu)
\(\Rightarrow1-\frac{2015}{2016^{2017}+2016}>1-\frac{2015}{2016^{2016}+2016}\)
\(\Rightarrow\frac{1}{2016}x>\frac{1}{2016}y\)
\(\Rightarrow x>y\)
21 tháng 9 2017
DÀI QUÁ KHÔNG TÍNH ĐƯỢC. CÁI NÀY CÓ MÀ ĐI HỎI THẦN ĐỒNG VỀ MÔN TOÁN ĐI
23 tháng 4 2018
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
23 tháng 4 2018
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
10 tháng 5 2017
Ta có 2A= 2(2^2015 + 1)/ 2^2016 + 1 = 2^2016 +2 / 2^2016 +1 = 2^2016+1/2^2016+1 + 1/2^2016 +1= 1 + 1/2^2016
2B= 2( 2^2016 + 1/ 2^2017+ 1) = 2^2017 +2 / 2^2017 +1 = 2^2017+1/2^2017+1 + 1/2^2017 +1 = 1 + 1/2^2017
Do 1/2^2016 > 1/2^2017 => 2A>2B => A>B
10 tháng 5 2017
10.A=\(10.A=\frac{10.\left(2^{2015+1}\right)}{2^{2016}+1}=\frac{2^{2016+10}}{2^{2016}+1}=1+\frac{2016}{2^{2016}+1}\)
\(10.B=\frac{10.\left(2^{2016}+1\right)}{2^{2017}+1}=\frac{2^{2017}+10}{2^{2017}+1}=1+\frac{2016}{2^{2017}+1}\)
Ta có:\(\frac{2016}{2^{2016}+1}>\frac{2016}{2^{2017}+1}\)
TO
3
24 tháng 2 2017
Vì 20162016 + 1 < 20162017 + 1
\(\Rightarrow B< \frac{2016^{2016}+1+2015}{2016^{2017}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}=\frac{2016^{2015}+1}{2016^{2016}+1}=A\)
Vậy A > B
Vì \(2016^{2016}+1< 2016^{2017}+1\) nên \(\frac{2016^{2016}+1}{2016^{2017}+1}< 1\)
\(\Rightarrow A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2017}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)Vậy A < B