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 vs lại -9 à

Nếu là âm thì:

\(\frac{13}{17}>\frac{-15}{19}\);\(\frac{12}{48}>\frac{-9}{36}\)

\(\frac{12}{7}>1\)

\(\frac{89}{1112}< 1\)

\(\Rightarrow\frac{12}{7}>\frac{89}{1112}\)  ( So sánh với 1)

Các so sánh ta có là so sánh tất cả các phân số với 1.

Ta có: \(\frac{12}{7}>1;\frac{89}{1112}< 1\) nên ta có \(\frac{12}{7}>\frac{89}{1112}\).

Vậy \(\frac{12}{7}>\frac{89}{1112}\).

~ Hok tốt ~

a) Ta có : \(\frac{12}{48}< \frac{12}{47}\); \(\frac{12}{48}< \frac{13}{48}\)

=> \(\frac{12}{48}< \frac{13}{47}\)

b) Ta có : \(\frac{7}{13}=1-\frac{6}{13}\)

               \(\frac{17}{23}=1-\frac{6}{23}\)

Mà \(-\frac{6}{13}< -\frac{6}{23}\)=> \(\frac{7}{13}< \frac{17}{23}\)

\(A=\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{90}\)

Tổng trên có số số hạng là: \(\left(90-32\right)\div1+1=59\)

\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{90}\)

\(>\frac{1}{45}+\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\)

\(=\left(\frac{1}{90}+\frac{1}{90}\right)+\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\)

\(=\frac{60}{90}=\frac{2}{3}\)

Đoàn Đức Hà:  Tại sao dòng số 4 phân số đầu tiên lại là \(\frac{1}{45}\)ạ?

\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right):15\frac{2}{3}=2008\)

\(2009-\left(\frac{41}{9}+x-\frac{133}{8}\right):\frac{47}{3}=2008\)

\(2009-\left(\frac{41}{9}+x-\frac{133}{8}\right)\times\frac{3}{47}=2008\)

\(2009-\frac{41}{9}\times\frac{3}{47}-x\times\frac{3}{47}+\frac{133}{8}\times\frac{3}{47}=2008\)

\(2009-\frac{41}{141}-x\times\frac{3}{47}+\frac{399}{376}=2008\)

\(2009+(\frac{399}{376}-\frac{41}{141})-x\times\frac{3}{47}=2008\)

\((2009+\frac{869}{1128})-x\times\frac{3}{47}=2008\)

\(x\times\frac{3}{47}=2009+\frac{869}{1128}-2008\)

\(x\times\frac{3}{47}=1\frac{869}{1128}\)

\(x\times\frac{3}{47}=\frac{1997}{1128}\)

\(x=\frac{1997}{1128}:\frac{3}{47}\)

\(x=\frac{1997}{72}\)

\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right):15\frac{2}{3}=\)2008

\(\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2009-2008\)

\(\left(\frac{41}{9}+x-\frac{133}{18}\right)=1.\frac{47}{3}=\frac{47}{3}\)

\(\frac{82}{18}+x-\frac{133}{18}=\frac{47}{3}\)

\(x=\frac{282}{18}-\frac{82}{18}+\frac{133}{18}\)

\(x=\frac{333}{18}=\frac{37}{2}\)

Đáp số \(x=\frac{37}{2}\)

xin lỗi bn dấu nhân nó bị trùng với x nên mk thay dấu nhân thành dấu "." theo cách lớp 6 nha.

Nếu có chỗ nào sai thì mk xin lỗi các bạn và mong các bạn góp ý 

*****Chúc bạn học giỏi*****

\(\frac{2017}{2018}\)và   \(\frac{2019}{2020}\)

Ta có : \(1-\frac{2017}{2018}=\frac{1}{2018};1-\frac{2019}{2020}=\frac{1}{2020}\)

Vì \(\frac{1}{2018}>\frac{1}{2020}\)nên \(\frac{2017}{2018}< \frac{2019}{2020}\)

Cái này là so sánh bằng phần bù của đơn vị nha bn !

Học tốt #

\(\frac{2017}{2018};\frac{2018}{2019};\frac{2019}{2020}\)

 \(\Rightarrow\frac{2017}{2018}< \frac{2019}{2020}\)

ta có

\(1-\frac{2018}{2019}=\frac{1}{2019}\)và\(1-\frac{2019}{2020}=\frac{1}{2020}\)

vì\(\frac{1}{2019}>\frac{1}{2020}\)vậy\(\frac{2018}{2019}>\frac{2019}{2020}\)

a) Ta có \(\frac{13}{7}=2-\frac{1}{7}\)

              \(\frac{21}{12}=2-\frac{1}{4}\)

Vì \(\frac{1}{7}< \frac{1}{4}\)\(\Rightarrow2-\frac{1}{7}>2-\frac{1}{4}\)\(\Rightarrow\frac{13}{7}>\frac{21}{12}\)

Vậy \(\frac{13}{7}>\frac{21}{12}\)

b) Ta có : \(\frac{2018}{2019}=1-\frac{1}{2019}\)

               \(\frac{2019}{2020}=1-\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\Rightarrow1-\frac{1}{2019}< 1-\frac{1}{2020}\Rightarrow\frac{2018}{2019}< \frac{2019}{2020}\)

Vậy \(\frac{2018}{2019}< \frac{2019}{2020}\)

c) Ta có :Vì  \(\frac{17}{53}< \frac{17}{50}< \frac{19}{50}\) \(\Rightarrow\frac{17}{53}< \frac{19}{50}\)

Vậy \(\frac{17}{53}< \frac{19}{50}\)