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\(\dfrac{19}{19}\) = 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 \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\)
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\(\frac{72}{73}< \frac{98}{99}\)
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
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+17}{17^{19}+17}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{17}+1}{17^{18}+1}=B\)
=> A < B
A,Ta có:\(\frac{19}{18}=1+\frac{1}{18};\frac{2011}{2010}=1+\frac{1}{2010}\)
Vì \(\frac{1}{18}>\frac{1}{2010}\Rightarrow\frac{19}{18}>\frac{2011}{2010}\)
B,ta có:\(1-\frac{72}{73}=\frac{1}{3};1-\frac{98}{99}=\frac{1}{99}\)
Vì \(\frac{1}{3}>\frac{1}{99}\Rightarrow\frac{72}{73}< \frac{98}{99}\)
C,Vì \(\frac{7}{9}< 1< \frac{19}{17}\Rightarrow\frac{7}{9}< \frac{19}{17}\)
a 19/18>2005/2004
b 72/73<98/99
a19/18>2005/2004
b72/73<98/99