`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`

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 

oki

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\(\frac{2\cdot2+1}{2\cdot2\cdot2+1}=\frac{5}{9}\ne\frac{2}{3}\))

bài 1:

ssh của A là:

(151-3):2+1=75

A=(151+3)x75:2=5775

đáp số: 5775

Ta có:

\(A=\frac{4-7^{2020}}{7^{2020}}+\frac{5+7^{2021}}{7^{2021}}\) và \(B=\frac{1}{7^{2019}}\)

Ta xét 2 trường hợp:

\(TH1:\frac{4-7^{2020}}{7^{2020}}=\frac{-7^{2020}+4}{7^{2020}}=-1+\frac{4}{7^{2020}}\)

\(TH2:\frac{5+7^{2021}}{7^{2021}}=1+\frac{5}{7^{2021}}\)

\(\Rightarrow\left(-1+\frac{4}{7^{2020}}\right)+\left(1+\frac{5}{7^{2021}}\right)\)

\(\Rightarrow\frac{4}{7^{2020}}+\frac{5}{7^{2021}}\)

\(Do:\)

\(\frac{4}{7^{2020}}>\frac{1}{7^{2019}}\)

\(\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)

Nên:\(\frac{4}{7^{2020}}+\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)

\(\Rightarrow A>B\)

\(\frac{7}{11}>\frac{7}{7}>\frac{4}{7}\) \(\Rightarrow\frac{7}{11}>\frac{4}{7}\)

\(\frac{2018}{2020}< \frac{2018}{1}< 2019\) \(\Rightarrow\frac{2018}{2020}< 2019\)

\(\frac{145}{141}< \frac{145}{135}< \frac{132}{135}\) \(\Rightarrow\frac{145}{141}< \frac{132}{135}\)

\(\frac{1975}{1972}< \frac{1975}{194}< \frac{197}{194}\) \(\Rightarrow\frac{1975}{1972}< \frac{197}{194}\)

\(\frac{37}{39}>\frac{37}{43}>\frac{36}{43}\Rightarrow\frac{37}{39}>\frac{36}{43}\)

\(\frac{2000}{4000}=\frac{2}{4}=\frac{1}{2}\Rightarrow\frac{1}{2}=\frac{2000}{4000}\)

Ta có:

\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)

\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)

Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)

=> B>A