\(\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Ta thấy      \(2017.2018< 2018.2019\)

nên      \(\frac{1}{2017.1018}>\frac{1}{2018.2019}\)

\(\Rightarrow\)\(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

Vậy      \(\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)

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

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

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

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

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

đặt 2²⁰¹⁸ = a ; 3²⁰¹⁹ = b ; 5²⁰²⁰ = c

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Trước tiên để tính diện tích hình thang chúng ta có công thức Chiều cao nhân với trung bình cộng hai cạnh đáy.
cach tinh dien h hinh thang vuong can khi biet do dai 4 canh cong thuc tinh 2
S = h * (a+b)1/2
Trong đó
a: Cạnh đáy 1
b: Cạnh đáy 2
h: Chiều cao hạ từ cạnh đấy a xuống b hoặc ngược lại(khoảng cách giữa 2 cạnh đáy)
Ví dụ: giả sử ta có hình thang ABCD với các cạnh AB = 8, cạnh đáy CD = 13, chiều cao giữa 2 cạnh đáy là 7 thì chúng ta sẽ có phép tính diện tích hình thang là:
S(ABCD) = 7 * (8+13)/2 = 73.5
cach tinh dien h hinh thang vuong can khi biet do dai 4 canh cong thuc tinh 3
Tương tự với trường hợp hình thang vuông có chiều cao AC = 8, cạnh AB = 10.9, cạnh CD = 13, chúng ta cũng tính như sau:
S(ABCD) = AC * (AB + CD)/2 = 8 * (10.9 + 13)/2 = 95.6

lien quan vai

bằng nhau

1

\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)

\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)

2

\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)

\(=\frac{100^{100}+1}{100^{99}+1}=N\)

A=B chắc vậy.

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

B>A nha

Na nissss

Uk chắc vậy

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

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

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

\(\Leftrightarrow M>N\)

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 \(M>N\)

Ta có :

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

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

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

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

\(\Rightarrow M>N\)