2019/2000 + 2019/2001 + .....................+ 2019/ 2019

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12 tháng 6 2020

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

15 tháng 5 2019

\(\frac{2019}{210}+\frac{2019}{280}+\frac{2019}{360}+\frac{2019}{450}+\frac{2019}{550}\)

\(=\frac{673}{70}+\frac{2019}{280}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)

\(=\left[\frac{673}{70}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)

\(=\left[\frac{2692}{280}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)

\(=\frac{673}{40}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)

\(=\left[\frac{673}{40}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)

\(=\left[\frac{2019}{120}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)

\(=\frac{673}{30}+\frac{673}{150}+\frac{2019}{550}\)

\(=\left[\frac{673}{30}+\frac{673}{150}\right]+\frac{2019}{550}\)

\(=\frac{673}{25}+\frac{2019}{550}=\frac{14806}{550}+\frac{2019}{550}=\frac{16825}{550}=\frac{673}{22}\)

P/S : Các a chị check dùm em ạ

3 tháng 3 2020

T gợi ý nhé

Bạn nhân cả E vs F vs 2019 r so sánh 2019E và 2019 F là suy ra đc mà

Tham khảo https://olm.vn/hoi-dap/detail/243936177029.html

VC
1 tháng 10 2019

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

10 tháng 11 2017

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....................

10 tháng 11 2017

tui ko bt

20 tháng 3 2019

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

20 tháng 3 2019

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