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12 tháng 8 2019

Trả lời

So sánh cái nào vs cái nào ạ

sao chỉ thấy có 1 vế ạ !

12 tháng 8 2019

vi 2018/2019<1

   2019/2020<1

   2020/2021<1

nen 2018/2019 + 2019/2020 + 2020/2021<1+1+1=3

26 tháng 4 2019

Ta có :

\(N=\frac{2018+2019+2020}{2019+2020+2021}\)

\(=\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)

Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)

\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)

\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)

\(\Leftrightarrow M>N\)

28 tháng 7 2020

Trả lời:

Ta có: 

\(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)

\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)

\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)

\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}>\frac{2018+2019+2020}{2019+2020+2021}\)

hay \(M>N\)

Vậy \(M>N\)

29 tháng 3 2020

Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)

12 tháng 8 2019

\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)

\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)

\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)

\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)

\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)

\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)

\(=\frac{0}{2019\times2018}\)

\(=0\)

Vậy A = 0 

12 tháng 8 2019

ta có

A=2020*2018/2019*2018-2019*2019/2018*2019+1/2018*2019

=>A*(2018*2019)=2020*2018-2019*2019+1

=>A*(2018*2019)=(2019+1)*2018-(2018+1)*2019+1

=>A*(2018*2019)=(2019*2018+2018)-(2018*2019+2019)+1

=>A*(2018*2019)=2019*2018+2018-2018*2019-2019+1

=>A*(2018*2019)=2018-2019+1

=>A*(2018*2019)=2018+1-2019

=>A*(2018*2019)=0

=>A=0/(2018*2019)

=>A=0

AH
Akai Haruma
Giáo viên
14 tháng 5 2019

Lời giải:

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

\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

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

\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)

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

---------------------

Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)

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

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

15 tháng 5 2019

thầy giải hay quá

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

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

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

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

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

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

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

AH
Akai Haruma
Giáo viên
14 tháng 5 2019

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Câu hỏi của Nguyễn Thị Yến Nhi - Toán lớp 6 | Học trực tuyến

14 tháng 5 2019

bạn ơi liệu nhầm lẫm gì ko

đây là toán lớp 6 á

NV
7 tháng 5 2019

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

\(\Rightarrow B< A\)