So sánh hai phân số bằng cách dùng số trung gian:

$\begin{array}{l}a)\frac{5}{-17}và\frac{2}{7};\\ \\ b)\frac{41}{-73}và\frac{-67}{-33};\\ \\ c)\frac{23}{21}và\frac{21}{23};\\ \\ d)\frac{19}{26}và\frac{21}{25};\\ \\ e)\frac{19}{40}và\frac{41}{80};\\ \\ f)\frac{9}{23}và\frac{34}{97};\end{array}$

a) Do $A=\frac{{98}^{99}+1}{{98}^{89}+1}>1$ nên

$A=\frac{{98}^{99}+1}{{98}^{89}+1}>\frac{{98}^{99}+1+97}{{98}^{89}+1+97}=\frac{98({98}^{98}+1)}{98({98}^{88}+1)}=\frac{{98}^{98}+1}{{98}^{88}+1}=B$

Vậy A > B

b) Do C = $\frac{{100}^{2008}+1}{{100}^{2018}+1}$ < 1 nên

C= $\frac{{100}^{2008}+1}{{100}^{2018}+1}>\frac{{100}^{2008}+1+99}{{100}^{2018}+1+99}=\frac{100({100}^{2007}+1)}{100({100}^{2017}+1)}=\frac{{100}^{2007}+1}{{100}^{2017}+1}=D$

Vậy C > D.

a) Ta có: $1-\frac{26}{27}=\frac{1}{27};1-\frac{96}{97}=\frac{1}{97}$ . Vì $\frac{1}{27}>\frac{1}{97}$   nên  $\frac{26}{27}<\frac{96}{97}$

b) Ta có: $1-\frac{102}{103}=\frac{1}{103};1-\frac{103}{105}=\frac{2}{105}$.   Vì  $\frac{1}{103}=\frac{2}{206}<\frac{2}{105}nên\frac{102}{103}>\frac{103}{105}$

c) Ta có :

$\begin{array}{l}\frac{2017}{2016}=1+\frac{1}{2016};\frac{2019}{2018}=1+\frac{1}{2018} . \\ Vì \frac{1}{2016}>\frac{1}{2018} nên \frac{2017}{2016}>\frac{2019}{2018}\\ \end{array}$

d) $\frac{73}{64}>\frac{51}{45}$

$Tacó: \frac{73}{64}=1+\frac{9}{64} ;\frac{51}{45}=1+\frac{6}{45}.$

$\begin{array}{l} Vì \frac{9}{64}=\frac{18}{128} >\frac{6}{45}=\frac{18}{135} nên \frac{73}{64}>\frac{51}{45}\\ \end{array}$