Ta đi so sánh \(\frac{2017.2018+1}{2017.2018}\)với\(\frac{2018.2019+1}{2018.2019}\)có :

\(\frac{2017.2018+1}{2017.2018}=\frac{2017.2018}{2017.2018}+\frac{1}{2017.2018}=1+\frac{1}{2017.2018}\left(\cdot\right)\)

\(\frac{2018.2019+1}{2018.2019}=\frac{2018.2019}{2018.2019}+\frac{1}{2018.2019}\left(\cdot\cdot\right)\)

\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\left(\cdot\cdot\cdot\right)\)Từ \(\left(\cdot\right);\left(\cdot\cdot\right)\&\left(\cdot\cdot\cdot\right)\Rightarrow\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)

\(\Leftrightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}.\)

#)Trả lời :

\(\frac{2017\times2018}{2017\times2018+1}=\frac{0}{1}=0\)

\(\frac{2018\times2019}{2018\times2019+1}=\frac{0}{1}=0\)

\(\Rightarrow\frac{2017\times2018}{2017\times2018+1}=\frac{2018\times2019}{2018\times2019+1}\)

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

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

Có \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow A< B\)

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

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

Do  \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)nên  \(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

Vậy  \(A< B\)

Ta có:

\(\frac{2017.2019}{2018.2018}\)

\(=\frac{2017.\left(2018+1\right)}{\left(2017+1\right).2018}\)

\(=\frac{2017.2018+2017}{2017.2018+2018}\)

Vì \(2017.2018+2017< 2017.2018+2018\)( tử nhỏ hơn mẫu )

\(\Rightarrow\frac{2017.2018+2017}{2017.2018+2018}< 1\)

Vậy \(\frac{2017.2019}{2018.2018}< 1\)

        ( Mk nghĩ vậy )

                          ~~~~~~~Hok tốt~~~~~~~

\(\frac{2017.2019}{2018.2018}=\frac{2017.\left(2018+1\right)}{2018.\left(2017+1\right)}=\frac{2017.2018+2017}{2018.2017+2018}\)

\(2017< 2018\Rightarrow2017.2018+2017< 2018.2017+2018\Rightarrow\frac{2017.2018+2017}{2018.2017+2018}< 1\Rightarrow\frac{2017.2019}{2018.2018}< 1\)

a) ta có: \(A=\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)

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

\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

=> A < B

a)A= 2017*2018/2017*2018-1/2017*2018=1-1/2017*2018

    B = 2018*2019/2018*2019-1/2018*2019=1-1/2018*2019

vì 1/2017*2018>1/2018*2019=> A<B

b)

ta có: \(Q=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)

\(Q=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+1\)

\(Q=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)

\(Q=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{P}{Q}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{2}+...+\frac{1}{47}+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\right)}=\frac{1}{50}\)

Đề là gì vậy bạn !

Ta xét : \(B=\left(2017\right).2019=\left(2018-1\right)\left(2018+1\right)\)

\(B=2018.2018+2018-2018-1\)

\(B=2018.2018-1\)

Mà : \(A=2018.2018\)

\(Dođó:A>B\)

A = 2018.2018

= (2019-1).2018

=2019.2018-2018(1)

B = 2017.2019

= (2018-1).2019

= 2018.2019-2019(2)

Từ (1)và(2)=>A>B

Nếu hay cho mình 5 sao ạ cảm ơn

\(\frac{2019.2020-4038}{2017.2019+2019}\)

\(=\frac{2019.2020-2.2019}{2019\left(2017+1\right)}=\frac{2019\left(2020-2\right)}{2019.2018}=\frac{2019.2018}{2019.2018}=1\)

\(A=\frac{2019.2020-4038}{2017.2019+2019}\)

   \(=\frac{2019\left(2020-2\right)}{2019\left(2017+1\right)}\)

   \(=\frac{2019.2018}{2019.2018}=1\)

Vậy \(A=1.\)

Mà lớp 5 làm gì đã học đến dấu \(.\)(dấu nhân lớp 5 viết kiểu này cơ: x )

Chúc em học tốt.

\(A=\frac{2019.2020-4038}{2017.2019+2019}\)

\(A=\frac{2019.2020-2.2019}{2017.2019+2019}\)

\(A=\frac{2019.\left(2020-2\right)}{2019.\left(2017+1\right)}\)

\(A=\frac{2018}{2018}\)

\(A=1\)

Tham khảo nhé~

vì\(\frac{2016}{2017}< 1\)

và\(\frac{2017}{2018}< 1\)

cho nên\(\frac{2016}{2017}+\frac{2017}{2018}< 1\)

A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)

Vậy \(A=\frac{2018}{2019}\)

Chúc bn học tốt

\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)

\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)

\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)

\(=2018.\frac{2}{2019}\)

\(=\frac{4036}{2019}\)

\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)

\(A=\frac{2018}{2019}\)