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Ta có:\(\frac{2016}{2017}< 1\)
\(\frac{2017}{2018}< 1\)
\(\frac{2018}{2019}< 1\)
\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}>1+1+1=3\)
Vậy ......
Tham khảo nha \(https://www.olm.vn/hoi-dap/question/1216047.html\)
\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\)
\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1+\frac{2}{2015}\right)\)
\(=\left(1+1+1\right)-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{1}{2015}\)
\(=3-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}\)
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
Ta có:
\(2016^{10}+2016^9=2016^9.2016+2016^9=2016^9(2016+1)=2017.2016^9\)
\(2017^{10}=2017.2017^9\)
Xét thấy: \(2016<2017\Rightarrow 2016^9<2017^9\Rightarrow 2017.2016^9<2017.2017^9\)
\(\Rightarrow 2016^{10}+2016^9<2017^{10}\)
Có: \(A>\frac{2016}{2016}+\frac{2017}{2017}=2\)
Có: \(B=\frac{4035}{4033}< 2\)
\(\Rightarrow A>B.\)
\(B=\frac{2017+2018}{2016+2017}=\frac{2017}{2016+2017}+\frac{2018}{2016+2017}\)
Ta có
\(\frac{2017}{2016+2017}< \frac{2017}{2016}\) ;
\(\frac{2018}{2017}< \frac{2018}{2017}\)
\(\Rightarrow B< \frac{2017}{2016}+\frac{2018}{2017}=A\)
Vậy B<A
\(B=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}\)
\(B=1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{2}{2016}\)
\(B=\left(1+1+1\right)-\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)
\(B=3-\left(...\right)< 3\)
P/s :
\(\left(...\right)la`\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)
quên ^^