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DL
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 2019
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\(\frac{2018}{2019}>\frac{2019}{2020}\)
XO
15 tháng 7 2019
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 \(\frac{17}{53}< \frac{19}{50}\)
LM
2
KT
15 tháng 8 2018
\(A=\frac{2017\times2108}{2017\times2018+1}=1-\frac{1}{2017\times2018+1}\)
\(B=\frac{2018\times2019}{2018\times2019+1}=1-\frac{1}{2018\times2019+1}\)
Nhận thấy:\(2017\times2018< 2018\times2019\)
=> \(2017\times2018+1< 2018\times2019+1\)
=> \(\frac{1}{2017\times2018+1}>\frac{1}{2018\times2019+1}\)
=> \(A< B\)
DD
1
DH
1
22 tháng 7 2015
15/27 < 31/54
1965/1967 > 1973/1975 tick đúng nha Dao Huong Giang
11 tháng 7 2019
Ta có:
\(A=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
mà 2019+2020 >2019>2020 \(\Rightarrow\frac{2018}{2019+2020}< \frac{2018}{2019};\frac{2019}{2019+2020}< \frac{2019}{2020}\)
\(\Rightarrow\frac{2018}{2019+2020}+\frac{2019}{2019+2020}< \frac{2018}{2019}+\frac{2019}{2020}\)hay \(A< B\)
DH
1
22 tháng 7 2015
1212/1313=12/13
2424/2525=24/25
phần bù của 12/13 là:1-12/13=1/13
phần bù của 24/25: 1-24/25=1/25
vì phần bù 1/13>1/25 nên 1212/1313>2424/2525
2019/2020<15/14
2019/2020<1
mà 15/14 >1
nên suy ra 2019/2020<15/14
Ta có: \(\frac{2019}{2020}< 1< \frac{15}{14}\)
Vậy \(\frac{2019}{2020}< \frac{15}{14}\)