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NV
7 tháng 5 2019

\(B=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}< \frac{2018}{2019}+\frac{2019}{2020}=A\)

\(\Rightarrow B< A\)

27 tháng 4 2019

Ta có:\(\frac{2018}{2019}\)<1\(\Rightarrow\)\(\frac{2018}{2019}\)>\(\frac{2018}{2019+2020}\)

           \(\frac{2019}{2020}\)<1\(\Rightarrow\)\(\frac{2019}{2020}\)>\(\frac{2019}{2019+2020}\)

\(\Rightarrow\)\(\frac{2018}{2019}\)+\(\frac{2019}{2020}\)>\(\frac{2018}{2019+2020}\)+\(\frac{2019}{2019+2020}\)=\(\frac{2018+2019}{2019+2020}\)

\(\Rightarrow\)A>B

Vậy A>B

27 tháng 4 2019

Ta có :\(A=\frac{2018}{2019}+\frac{2019}{2020}\)

\(B=\frac{2018+2019}{2019+2020}\)

\(B=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)

Ta thấy :

\(\frac{2018}{2019}>\frac{2018}{2019+2020}\left(2019< 2019+2020\right)\)

\(\frac{2019}{2020}>\frac{2019}{2019+2020}\left(2020< 2019+2020\right)\)

 \(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)

Vậy \(A>B\)

~ Thiên Mã ~

B= 1/1.2+1/2.3+...+1/2019.2020

B=1/1-1/2+1/2-1/3+...+1/2019-1/2020

B=1-1/2020=2020/2020-1/2020=2019/2020

14 tháng 5 2019

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\) 

Với  :   \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\) 

Và   :   \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\) 

             \(B=1-\frac{1}{2020}< 1< A\)

5 tháng 3 2019

A=B chắc vậy.

Để mik tìm cách làm r gửi cho nha!!!!!

5 tháng 3 2019

B>A nha

22 tháng 5 2019

đặt 22018 = a ; 32019 = b ; 52020 = c

Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)

\(B< \frac{3}{4}\)

\(\Rightarrow A>1>\frac{3}{4}>B\)

22 tháng 5 2019

Mình chỉ biết cách tính B thôi, đây nhé:

B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)

B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)

30 tháng 4 2019

b,

          \(B=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)

  Ta thấy :

\(\frac{2018}{2019}>\frac{2018}{2019+2020}\)

\(\frac{2019}{2020}>\frac{2019}{2019+2020}\)

Từ đó , suy ra :

\(\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)

                                       Vậy...

                                                        #Louis

30 tháng 4 2019

Ta có :

\(\frac{205}{321}< \frac{205}{315}\)

\(\frac{214}{315}>\frac{205}{315}\)

\(\Leftrightarrow\frac{205}{321}< \frac{214}{315}\)

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