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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 \(\frac{1991.1999}{1995.1995}< 1.\)

Hai bài này bạn tính ra là xong mà

Cần gì phải hỏi

Dễ mà

13 tháng 8 2017

\(A< 4\)

\(B< 3\)

là đáp án đúng

17 tháng 6 2019

\(\frac{2016}{2017}\)\(\frac{2017}{2018}\)\(\frac{2018}{2016}\)< 3 

17 tháng 6 2019

2016/2017 + 2017/2018 + 2018/2016 > 3

Hok tốt

24 tháng 8 2019

Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

11 tháng 8 2016

bằng nhau

a=b

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

mà : 1 < 3

vậy:\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}< 3\)

26 tháng 6 2017

Giải: Ta có:

\(\frac{2016}{2017}=\frac{2017}{2017}-\frac{1}{2017}=1-\frac{1}{2017}\)

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

\(\frac{2018}{2016}=\frac{2016}{2016}+\frac{2}{2016}=1+\frac{2}{2016}\)

\(\Rightarrow3+\frac{-1}{2017}+\frac{-1}{2018}+\frac{2}{2016}=3+\frac{2}{2016}>3\)
 

2 tháng 9 2019

3.000000737

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

25 tháng 7 2018

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25 tháng 7 2018

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