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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 \(\frac{1991.1999}{1995.1995}< 1.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}\)
\(A=\frac{2017\cdot2018-2+1}{2017\cdot2018-2}\)
\(A=\frac{2017\cdot2018-2}{2017\cdot2018-2}+\frac{1}{2017\cdot2018-2}\)
\(A=1+\frac{1}{2017\cdot2018-2}\)
Ta có phân số trung gian là 1. Ta có:
\(A>1\) ; \(B< 1\)
\(\Rightarrow A>1>B\)
\(\Rightarrow A>B\)
Vậy A>B
Chúc em học tốt!
\(\Rightarrow\text{❤️✔✨♕✨✔️❤ }\Leftarrow\)
\(\text{Ta có :}\)
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}=\frac{4070305}{4070304}=1\frac{1}{4070304}\)
\(B=\frac{2017}{2018}\)
\(\text{Vì : }1\frac{1}{4070304}>1\text{ mà }\frac{2017}{2018}< 1\text{ nên }1\frac{1}{4070304}>\frac{2017}{2018}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}\)= \(\frac{2016}{2016}=1\)
mà : 1 < 3
vậy:\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}< 3\)
Giải: Ta có:
\(\frac{2016}{2017}=\frac{2017}{2017}-\frac{1}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=\frac{2018}{2018}-\frac{1}{2018}=1-\frac{1}{2018}\)
\(\frac{2018}{2016}=\frac{2016}{2016}+\frac{2}{2016}=1+\frac{2}{2016}\)
\(\Rightarrow3+\frac{-1}{2017}+\frac{-1}{2018}+\frac{2}{2016}=3+\frac{2}{2016}>3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
không quy đồng phân số hãy so sánh 2 phân số sau: 2017/ 2018 và 2016/2017
Ta so sánh 1/2018 và 1/2017
1/2018<1/2017
=> 2017/2018>2016/2017
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2017}{2018}\)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:
\(\frac{63}{64}=\frac{63.2018}{64.2018}=\frac{127134}{129152}\)
\(\frac{2017}{2018}=\frac{2017.64}{2018.64}=\frac{129088}{129152}\)
Vậy \(\frac{63}{64}< \frac{2017}{2018}\)
Ta có 1 - 63/64=1/64
1 - 2017/2018=1/2018
(Ta so sánh phần tử số)
Vì 1/64>1/2018 nên 63/64>2017/2018