Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Vì 20162016 + 1 < 20162017 + 1
\(\Rightarrow\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2016}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow\)A < B
Ta có :
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow A< B\)
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\)
\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1+\frac{2}{2015}\right)\)
\(=\left(1+1+1\right)-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{1}{2015}\)
\(=3-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}\)
Vì \(\frac{1}{2016}< \frac{1}{2015};\frac{1}{2017}< \frac{1}{2015}\)
=> \(\frac{1}{2016}+\frac{1}{2017}< \frac{2}{2015}\)
=> \(-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}>0\)
=> \(3-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{1}{2015}>3\)
trước tiên ta rút gọn 2 phân số 2015/2016+2016/2017
TA RÚT GỌN 2016 LẠI VỚI NHAU = 2015/1 +1/2017
sau đó ta rút gọn 2 phân số 1/2017 + 2017/2015
TA RÚT GỌN 2017 LẠI VỚI NHAU = 1/1 + 1/2015
TA CÓ: 2015/1 + 1/1 + 1/2015=2015/1 + 1 + 1/015=1/1 + 1 + 1/1= 1+1+1 = 3(VÌ TA RÚT GỌN 2015 LẠI VỚI NHAU)
VÌ: 3 = 3
Vậy:2015/2016 + 2016/2017 + 2017/2015 = 3
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
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
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 :
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
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
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 :")
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.
Ta có :
\(\frac{2015}{2016}>\frac{2015}{2016+2017}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt !!!
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015+2016}{4033}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015}{4033}+\frac{2016}{4033}\)
\(\Rightarrow A>B\)