K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 6 2018

B = 2017 x 2019 = (2018 - 1)(2018 + 1) = 2018 x 2018 + 2018 - 2018 - 1 = 2018 x 2018 - 1 = A - 1

==> A - 1 = B   => A > B

4 tháng 6 2018

A > B

A = 2018 x (2019 - 1) = 2018 x 2019 - 2018

B = 2019 x (2018 - 1) = 2019 x 2018 - 2019

Vì hai vế đều có 2018 x 2019 nên ta trượt tiêu. Còn 2018 và 2019 ta so sánh. Vì 2019 > 2018 nên B < A hay A > B ( Vì hai 2018 x 2019 đều bằng nhau nếu hai phép bằng nhau trừ cho số nào nhỏ hơn thì sẽ lớn hơn.)

Chúc bạn hok tốt nha!

11 tháng 12 2023

b.\(\dfrac{1}{2019.2018}\)

11 tháng 12 2023

b nhé 

nhiên 5a1 dúng ko

 

7 tháng 3 2022

c bn nhé

13 tháng 3 2022

thank bạn

6 tháng 4 2023

`a,`

`5/6=1-1/6`

`7/8=1-1/8`

Mà `1/6>1/8 -> 5/6<7/8`

`b,`

`9/5=(9 \times 2)/(5 \times 2)=18/10`

`3/2=(3 \times 5)/(2 \times 5)=15/10`

`18/10 > 15/10 -> 9/5 > 3/2`

`c,`

`2017/2018 = 1-1/2018`

`2019/2020=1-1/2020`

`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`

`d,`

`2018/2017 = 1+1/2017`

`2020/2019 = 1+1/2019`

`1/2017 > 1/2019 -> 2018/2017>2020/2019`

8 tháng 6 2019

#)Giải :

\(Q=2+\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)

Ta thấy : \(2>\frac{2016}{2017};2>\frac{2017}{2018};2>\frac{2018}{2019}\left(1\right)\)

\(\frac{2016}{2017+2018+2019}< \frac{2016}{2017}\left(2\right)\)

\(\frac{2017}{2017+2018+2019}< \frac{2017}{2018}\left(3\right)\)

\(\frac{2018}{2017+2018+2019}< \frac{2018}{2019}\left(4\right)\)

Từ (1) (2) (3) (4) \(\Rightarrow P>Q\)

25 tháng 6 2018

a) \(\frac{1995}{1997}\)và \(\frac{1995}{1996}\)

Ta có : \(\frac{1995}{1996}=\frac{1995\times2}{1996\times2}=\frac{3990}{3992}\)

\(1-\frac{1995}{1997}=\frac{2}{1997};1-\frac{3990}{3992}=\frac{2}{3992}\)

Vì \(\frac{2}{1997}>\frac{2}{3992}\)nên \(\frac{1995}{1997}< \frac{3990}{3992}\)hay \(\frac{1995}{1997}< \frac{1995}{1996}\).

b) \(\frac{2016}{2017}\)và \(\frac{2017}{2018}\)

Ta có : \(1-\frac{2016}{2017}=\frac{1}{2017};1-\frac{2017}{2018}=\frac{1}{2018}\)

Vì \(\frac{1}{2017}>\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\).

c) \(\frac{2018}{2019}\)và \(\frac{2017}{2016}\).

Vì \(\frac{2018}{2019}< 1;1< \frac{2017}{2016}\)nên \(\frac{2018}{2019}< \frac{2017}{2016}\).

~ HOK TỐT ~

25 tháng 6 2018

a)\(\frac{1995}{1997}\)<   \(\frac{1995}{1996}\)

b)\(\frac{2016}{2017}\)<    \(\frac{2017}{2018}\)

c)\(\frac{2018}{2019}\)<     \(\frac{2017}{2016}\)

11 tháng 6 2018

Bài 1:

Ta có:

\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)

Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)

                                                     \(\Leftrightarrow N< M\)

Vậy \(M>N.\)

Bài 2:

Ta có:

\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)

\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)

Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)

\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)

                                                                     \(\Leftrightarrow A>B\)

Vậy \(A>B.\)

Bài 3:

\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)

                                                                \(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)

                                                                \(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)

Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)

\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm

\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)

Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)

Bài 4:

\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)

Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)

\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)

\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)

Vậy \(\frac{1991.1999}{1995.1995}< 1.\)

15 tháng 7 2018

\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}=\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)+\left(1-\frac{1}{2019}\right)\)

\(A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)< 3\)

15 tháng 7 2018

Ta có :

2016/2017 < 1

2017/2018 < 1

2018/2019 < 1

Mà 2016/2017 + 2017/2018 + 2018/2019 < 1 + 1 + 1 = 3

Nên A < 3

15 tháng 7 2018

\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)

Ta có:

 \(\frac{2016}{2017}< 1\)

\(\frac{2017}{2018}< 1\)

\(\frac{2018}{2019}< 1\)

\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}< 1+1+1=3\)

\(\Rightarrow A< 3\)

Vậy \(A< 3\)

Tham khảo nhé

15 tháng 7 2018

\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)

\(=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}\)

\(=\left(1+1+1\right)-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

\(=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)< 3\)

Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}< 3\left(đpcm\right)\)