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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 \(A< B\)
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 \(\frac{1991.1999}{1995.1995}< 1.\)
Ta có:\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
Vậy biểu thức \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
1/2 x 2/3 x 3/4 x ... x 2018/2019 x 2019/2020
= 1 x 2 x 3 x ... x 2018 x 2019 / 2 x 3 x 4 x ... x 2019 x 2020
Khử loại đi ta còn lại phân số 1/2020
Hok tốt ^^
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 \(\frac{17}{53}< \frac{19}{50}\)
\((2,5\cdot x+2017)\cdot2018=(7,5+2017)\cdot2018\)
\(\Rightarrow(2,5\cdot x+2017)\cdot2018=4085441\)
\(\Rightarrow2,5\cdot x+2017=2024,5\)
\(\Rightarrow2,5x=7,5\)
\(\Rightarrow x=7,5:2,5=3\)
\(3\frac{1}{5}+\frac{2}{5}\left[x+\frac{1}{3}\right]=\frac{21}{5}\)
\(\Rightarrow\frac{16}{5}+\frac{2}{5}\left[x+\frac{1}{3}\right]=\frac{21}{5}\)
\(\Rightarrow\frac{2}{5}\left[x+\frac{1}{3}\right]=\frac{21}{5}-\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}\left[x+\frac{1}{3}\right]=1\)
\(\Rightarrow x+\frac{1}{3}=\frac{5}{2}\)
\(\Rightarrow x=\frac{13}{6}\)
Sao các bn cứ tk sai mk vô cớ thế nhỉ , mk đã lm j sai , mk chỉ nói là mk ko bít bài 2 thui mak tự nhiên tk ngta sai , bn nào tk mk sai rồi các bn sẽ biết hậu quả thôi :PPP
\(\frac{2019}{2001}=2019:2001\)
\(\frac{2017}{2003}=2017:2003\)
ta có 2019:2001=1,008995502 và 2017:2003=1,006989516
ta có 1,008995502>1,006989516
\(\Rightarrow\)\(\frac{2019}{2001}>\frac{2017}{2003}\)
CHÚC BẠN HỌC TỐT
ta có 2019/2001=2019:2001 và 2017/2003=2017:2003
vì 2019:2001 >2017:2003
nên 2019/2001 >2017/2003
`a,`
`5/6=1-1/6`
`7/8=1-1/8`
Mà `1/6>1/8 -> 5/6<7/8`
`b,`
`9/5=(9 \times 2)/(5 \times 2)=18/10`
`3/2=(3 \times 5)/(2 \times 5)=15/10`
`18/10 > 15/10 -> 9/5 > 3/2`
`c,`
`2017/2018 = 1-1/2018`
`2019/2020=1-1/2020`
`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`
`d,`
`2018/2017 = 1+1/2017`
`2020/2019 = 1+1/2019`
`1/2017 > 1/2019 -> 2018/2017>2020/2019`