ta thấy 2 phân số 2017/2018 và 2019/2020 đều là phân số nhỏ hơn 1 nên 1 trong 2 phân số sẽ có 1 phân số nhỏ nhất. 

phần này bạn tự so sánh,2017/2018<2019/2020

tiếp theo bạn so sánh 2 phân số còn lại , 2018/2017>2020/2019

vậy 2017/2018<2019/2020<2018/2017<2020/2019

chúc bạn học tốt

bài này là 100

sai hết rồi

\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)và \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)

Xét \(A=\frac{2019^{2020}+1}{2019^{2021}+1}\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)

Xét \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}=1+\frac{2018}{2019^{2019}+1}\)

Vì \(1+\frac{2018}{2019^{2021}+1}< 1+\frac{2018}{2019^{2019}+1}\Rightarrow\frac{2019^{2020}+1}{2019^{2021}+1}< \frac{2018^{2019}+1}{2019^{2019}+1}\)

\(\Rightarrow A< B\)

Ta có:

\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)

\(\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}\)

\(\Rightarrow2019A=1+\frac{2019}{2019^{2021}+1}\)

\(\Rightarrow A=1+\frac{2019}{2019^{2021}+1}:2019\)

Ta lại có:

\(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)

\(\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}\)

\(\Rightarrow2019B=1+\frac{2019}{2019^{2019}+1}\)

\(\Rightarrow B=1+\frac{2019}{2019^{2019}+1}:2019\)

Do \(2019^{2021}+1>2019^{2019}+1\)

\(\Rightarrow\frac{2019}{2019^{2021}+1}< \frac{2019}{2019^{2019}+1}\)

\(\Rightarrow1+\frac{2019}{2019^{2021}+1}:2019< 1+\frac{2019}{2019^{2019}+1}:2019\)

\(\Rightarrow A< B\)

Vậy \(A< B.\)

ta có :\(E=\frac{2019^{2019}+1}{2019^{2020}+1}\Leftrightarrow2019\cdot E=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2019}{2019^{2020}+1}\)

\(F=\frac{2019^{2020}+1}{2019^{2021}+1}\Leftrightarrow2019\cdot F=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)

vì \(\frac{2019}{2019^{2020}+1}>\frac{2019}{2019^{2021}+1}\) nên E>F

E=2019 x 2019 x 2019 x ........ x 2019 x2019 +1 /2019 x 2019 x 2019 x.........x 2019 x 2019 + 1

E=1+1/2019+1

E=2/2020

E=1/1010

F=2019 x 2019 x 2019 x .......... x 2019 x 2019 +1 / 2019 x 2019 x 2019 x ....... x 2019 x 2019 +1

F= 1+1/2019+1

F=2/2020

F=1/1010

từ đó ta có E=F(=1/1010)

nghỉ sao rút gọn được vậy có dấu + mà(ví dụ\(\frac{2\cdot2+1}{2\cdot2\cdot2+1}=\frac{5}{9}\ne\frac{2}{3}\))

`a,`

`5/6=1-1/6`

`7/8=1-1/8`

Mà `1/6>1/8 -> 5/6<7/8`

`b,`

`9/5=(9 \times 2)/(5 \times 2)=18/10`

`3/2=(3 \times 5)/(2 \times 5)=15/10`

`18/10 > 15/10 -> 9/5 > 3/2`

`c,`

`2017/2018 = 1-1/2018`

`2019/2020=1-1/2020`

`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`

`d,`

`2018/2017 = 1+1/2017`

`2020/2019 = 1+1/2019`

`1/2017 > 1/2019 -> 2018/2017>2020/2019`

Bạn tham khảo link tại đây nhé :v

https://olm.vn/hoi-dap/detail/217907126396.html

bn vào câu hỏi tương tự

https://olm.vn/hoi-dap/detail/98028519014.html

Bạn tham khảo link trên nhé~~

Chúc bạn học tốt~~
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9/5 > 3/2

2017/2018 = 2019/2020

2018/2017 =  2020/2019

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Lời giải:
\(9B=\frac{9^{2019}+9}{9^{2019}+1}=1+\frac{8}{9^{2019}+1}> 1+\frac{8}{9^{2020}+1}=\frac{9^{2020}+9}{9^{2020}+1}=9A\)

$\Rightarrow B>A$