Ta có \(A=\frac{2017-2018}{2017+2018}=\frac{\left(2017-2018\right)\left(2017+2018\right)}{\left(2017+2018\right)^2}=\frac{2017^2-2018^2}{2017^2+2018^2+2.2017.2018}< \frac{2017^2-2018^2}{2017^2+2018^2}=B\)

Vậy A<B

A=24783,14746B=49566,29188

Vậy A<B

Ta thấy \(A=\frac{2018-2017}{2018+2017}=\frac{2018^2-2017^2}{\left(2018+2017\right)^2}=\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}\)

Mà \(2018^2+2.2018.2017+2017^2>2018^2+2017^2\)

\(\Rightarrow\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}< \frac{2018^2-2017^2}{2018^2+2017^2}\)

Vậy A<B

Biểu thức M lớn hơn biểu thức N

N = \(\frac{2016+2017}{2017+2018}=\frac{2016}{2017+2018}+\frac{2017}{2017+2018}\)

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

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

Nên M > N

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

Vì : \(\frac{2016}{2017}\)>\(\frac{2016}{2017+2018}\)

\(\frac{2017}{2018}\)>\(\frac{2017}{2017+2018}\)

Cộng vế với vế ta được : \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)> \(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)

Hay M > N

Vậy M > N

Chúc bạn hok tốt !!

id nhu 1 tro dua

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 \(\frac{1991.1999}{1995.1995}< 1.\)