\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)

\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\) 

\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)

Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\) 

1) 19/19 < 2005/2004

2)72/73 > 98/99

a,19/7=5/7 +2

2>7/9 => 19/7>7/9

b, 72/73=1- 1/73

98/99=1- 1/99

1/73>1/99

c,19/18=1+ 1/18

2005/2004=1+ 1/2004

1/18>1/2004

d, 72/73=(58+14)/73=58/73 + 14/73

58/73>58/99

=> 72/73>58/99

a, \(\frac{-11}{12}>\frac{17}{-18}\)

b\(\frac{2}{5}< \frac{5}{7}\)

c\(\frac{-3}{4}>\frac{-6}{7}\)

d\(\frac{19}{18}>\frac{2005}{2004}\)

e\(\frac{72}{73}< \frac{98}{99}\)

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Bài 1:

Ta thấy A < 1

=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy A < B

Bài 2:

Ta thấy C < 1

=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)

Vậy C < D

Ta có :

  19/18 = 19/19 - 18/19 = 1/19

   2005/2004 = 2005/2005 - 2004/2005 = 1/2005

Ta thấy 1/19 > 1/2005 Vậy 19/18 < 2005/2004

19/18 >2005/2004

\(\frac{19}{18}\)và \(\frac{2005}{2004}\)

\(\frac{19}{18}-1=\frac{1}{18};\frac{2005}{2004}-1=\frac{1}{2004}\)

Vì 18 < 2004 nên \(\frac{1}{18}>\frac{1}{2004}\)nên \(\frac{19}{18}>\frac{2005}{2004}\)

Ủng hộ nhé ! 

Ta có :

\(\frac{72}{73}=\frac{72\times99}{73\times99}=\frac{7128}{7227}\)

\(\frac{98}{99}=\frac{98\times73}{99\times73}=\frac{7154}{7227}\)

Vì : 7128/7227 < 7154/7227

Nên : 72/73 < 98/99

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+17}{17^{19}+17}\)

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{17}+1}{17^{18}+1}=B\)

=> A < B

(98^99-1)/(98^98-1)