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Đặt A = \(\frac{2019^{2019}+1}{2019^{2020}+1}\)
=> \(2019A=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2018}{2019^{2020}+1}\)
Đặt B = \(\frac{2019^{2020}+1}{2019^{2021}+1}\)
=> \(2019B=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2018}{2019^{2021}+1}\)
Vì \(\frac{2018}{2019^{2020}+1}>\frac{2018}{2019^{2021}+1}\Rightarrow1+\frac{2018}{2019^{2020}+1}>1+\frac{2018}{2019^{2021}+1}\Rightarrow10A>10B\Rightarrow A>B\)

\(\frac{2019}{210}+\frac{2019}{280}+\frac{2019}{360}+\frac{2019}{450}+\frac{2019}{550}\)
\(=\frac{673}{70}+\frac{2019}{280}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{70}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{2692}{280}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\frac{673}{40}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{40}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{2019}{120}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)
\(=\frac{673}{30}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{30}+\frac{673}{150}\right]+\frac{2019}{550}\)
\(=\frac{673}{25}+\frac{2019}{550}=\frac{14806}{550}+\frac{2019}{550}=\frac{16825}{550}=\frac{673}{22}\)
P/S : Các a chị check dùm em ạ

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

a,Vì 2001 chia 4 dư 1 nên 20012014 chia 4 dư 1
Đặt 20012014=4k+1
Ta có:20024k+1=(20024)ik.2002=(...............6)k.2002=.......................6.2002=.................................2
Vậy \(2002^{2001^{2014}}\) có tận cùng là 2
b,Cậu b tương tự câu a
Vì 81 chia 4 dư 1 nên \(81^{82^{83}}\) chia 4 dư 1
Đặt \(81^{82^{83}}\)=4k+1
.....................Bạn tự làm tiếp đi(tận cùng bằng 2)
c,Vì 2017 chia 4 dư 1 nên \(2017^{2018^{2019}}\) chia 4 dư 1
Đặt \(2017^{2018^{2019}}=4k+1\)
Ta có:20174k+1=(20174)k.2017=(............1)k.2017=...................1.2017=.........................7
Vậy....................

a ) Ta có :
\(\frac{450}{463}=1-\frac{13}{463}\) ( 1 )
\(\frac{123}{126}=1-\frac{3}{126}\)( 2 )
Từ ( 1 ) và ( 2 ) thấy 13/463 > 3/126 do đó 450/463 < 123/126
Vậy 450/463 < 123/126
b ) Ta có :
\(\frac{36}{53}=1-\frac{17}{53}\)( 1 )
\(\frac{58}{89}=1-\frac{31}{89}\)( 2 )
Từ 1 và 2 thấy 31/89 > 17/53 => 35/53 > 58/89
Vậy 35/53 > 58/89