link nà:https://olm.vn/hoi-dap/tim-kiem?q=so+s%C3%A1nh+:+A=2017%5E2017/2018%5E2017+1B=2017%5E2016+1/2017%5E2017+1+&id=862033

Thanks

\(A=\frac{2017^{2018+1}}{2017^{2018-3}}\)và \(B=\frac{2017^{2018-1}}{2017^{2018-5}}\)

Có \(A=\frac{2017^{2019}}{2017^{2015}}\)và \(B=\frac{2017^{2017}}{2017^{2013}}\)

Mà\(\frac{2017^{2019}}{2017^{2015}}>\frac{2017^{2018}}{2017^{2015}}\)và\(\frac{2017^{2017}}{2017^{2013}}>\frac{2017^{2017}}{2017^{2015}}\)

Vì \(\frac{2017^{2018}}{2017^{2015}}>\frac{2017^{2017}}{2017^{2015}}\)

Vậy A>B

Ta có 

A= \(\frac{2017^{2018}-3+4}{2017^{2018}-3}=1+\frac{4}{2017^{2018}-3}\)

B= \(1+\frac{4}{2017^{2018}-5}\)

vậy A > B

Mk đang cần gấp !!! Giúp mk nha các bn!!!

Vì phân số A\(=\frac{2016^{2017}+1}{2017^{2018}+1}< 1\) mà B\(=\frac{2017^{2018}+1}{2017^{2017}+1}>1\)

\(\Rightarrow\frac{2016^{2017}+1}{2017^{2018}+1}< 1< \frac{2017^{2018}+1}{2017^{2017}+1}\)

Vậy A<B

a<1<b

=>A<b

Nguyễn Thị Kim Oanh

a<1<b

=>a>b

Explosion !

Vì A < 1

\(\Rightarrow A< \frac{2017^{2018}+1+2016}{2017^{2019}+1+2016}=\frac{2017^{2018}+2017}{2017^{2019}+2017}=\frac{2017\left(2017^{2017}+1\right)}{2017\left(2017^{2018}+1\right)}=\frac{2017^{2017}+1}{2017^{2018}+1}=B\)

Vậy A < B