c) \(\frac{3065}{3054}=\frac{3054+11}{3054}=1+\frac{11}{3054}\)

\(\frac{4097}{4086}=\frac{4086+11}{4086}=1+\frac{11}{4086}\)

Mà: \(\frac{11}{3054}>\frac{11}{4086}\)

\(\Rightarrow1+\frac{11}{3054}>1+\frac{11}{4086}\)

\(\Rightarrow\frac{3065}{3054}>\frac{4097}{4086}\)

a,  1996.1999 < 1997.1998

b, 7001/7002 > 9003/9005 đặt phét tính r nhìn đuôi nhân vs nhau là biết ngay

c, 3065/3054 > 4097/4086 nhân 2 số đuôi 

a) \(1996.1999=\left(1997-1\right)1999=1999.1997-1999\)

\(1997.1998=1997.\left(1999-1\right)=1997.1999-1997\)

Mà: \(\hept{\begin{cases}1999.1997=1997.1999\\-1999< -1997\end{cases}}\)

\(\Rightarrow1999.1997-1999< 1997.1999-1997\Rightarrow1996.1999< 1997.1998\)

2017/2018>2018/2019

Lại nè

Xét phần bù tới 1 ta có:

1-2017/2018=1/2018 ; 1-2018/2019=1/2019

Ta thấy 1/2018>1/2019 nên 2017/2018<2018/2019

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

Co j góc lệch cùng trời cuối đất cú mèo

\(\dfrac{58}{21}=2\dfrac{16}{21}=2+\dfrac{16}{21}=2+\left(1-\dfrac{5}{21}\right)=3-\dfrac{5}{21}\)

\(\dfrac{61}{22}=2\dfrac{17}{22}=2+\dfrac{17}{22}=2+\left(1-\dfrac{5}{22}\right)=3-\dfrac{5}{22}\)

Ta có

\(\dfrac{5}{21}>\dfrac{5}{22}\Rightarrow3-\dfrac{5}{21}< 3-\dfrac{5}{22}\Rightarrow\dfrac{58}{61}< \dfrac{61}{22}\)

\(\dfrac{8}{9}\) < \(\dfrac{7}{5}\) , vì :

\(\dfrac{8}{9}\) < 1 , \(\dfrac{7}{5}\) > 1

8/9 < 7/5 

Vì \(\dfrac{8}{9}\) 75 nên \(\dfrac{8}{9}< \dfrac{7}{5}\)

2001/2000 > 2002/2001

j

o

2001/2000=1+1/2000

2002/2001=1+1/2001

Mà 1/2000>1/2001

=>1+1/2000>1+1/2001

hay 2001/2000>2002/2001

Ta có : 

1 - 2000/2001 = 1/2001 

1 - 2001/2002 = 1/2002 

Vì 1/2001 > 1/2002 nên 2000/2001 > 2001/2002