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a) 1995/1996 và 1996/1997
Ta có: 1995/1996=1 - 1/1996
1996/1997=1 - 1/1997
Vì 1/1996 > 1/1997 nên 1995/1996 < 1996/1997
b) Ta có: \(\frac{2121}{3737}=\frac{2121:101}{3737:101}=\frac{21}{37}\); \(\frac{212121}{373737}=\frac{212121:10101}{373737:10101}=\frac{21}{37}\)
Vì 21/37=21/37 nên 2121/3737 = 212121/373737
a, Ta có: 1-1995/1996=1/1996 ; 1-1996/1997=1/1997
do 1/1996 > 1/1997 nên 1995/1996 < 1996 1997
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\(\frac{3737}{4747}\)\(=\frac{37}{47}\)
\(\frac{3737:11=37}{4747:11=47}\)
Hok tốt
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 \(\frac{1991.1999}{1995.1995}< 1.\)
theo mk là
A thì = tất cả các phân số có tử bé hơn mẫu lên cho là bé hơn 1
B = 3
vậy B > A
Tính làm sao cũng được
tùy theo cách tính ( tự tìm A)
theo tui tính
A=3
B=3
=> A=B
thich choi lien minh ha so lo ko chap 5 mang ta choi yasuo cho 2q
trả lời
tui trả lời rui mà
chúc bà học tốt
nhớ k tui nha
cám ơn các bn
Ta có:1-1995/1996=1/1996
1-1996/1997=1/1997
Vì 1/1996>1/1997=>1995/1996<1996/1997
b, 2121/3737=2121/101=21
3737/101=37
212121/373737=212121/10101=21
373737/10101=37
=>2121/3737=212121/373737
a) \(\dfrac{1995}{1996}\) < \(\dfrac{1996}{1997}\)
b) \(\dfrac{2121}{3737}\) = \(\dfrac{212121}{373737}\)