Bài tập: So sánh phân số

$a)\frac{1}{3}<\frac{2}{3}\begin{array}{cccc}& & & \end{array}b)\frac{3}{2}>\frac{3}{4}$

c) ta có $\frac{-2}{5}<0;\frac{3}{5}>0=>\frac{-2}{5}<\frac{3}{5}$

d) ta có $\frac{7}{3}>\frac{7}{4}=>\frac{-7}{3}<\frac{-7}{4}$

$a)\frac{1}{3}<\frac{3}{4}$

$b)\frac{4}{5}<\frac{4}{3}$

$c)\frac{4}{-6}<\frac{5}{6}$

$d)\frac{-8}{5}<\frac{-8}{7}$

a) $\frac{1}{3}=\frac{2}{6};\frac{2}{6}<\frac{5}{6}=>\frac{1}{3}<\frac{5}{6}$

$$\begin{gathered} b)\frac{4}{5}>\frac{3}{7} \\ c)\frac{-3}{11}>\frac{-4}{13} \end{gathered}$$

d)$\frac{-63}{70}=\frac{-9}{10}$

Qui đồng ta được : $\frac{-9}{10}=\frac{-27}{30};\frac{-5}{6}=\frac{-25}{30}=>\frac{-5}{6}>\frac{63}{-70}$

$\begin{array}{l}a)\frac{1}{2}<\frac{5}{6}\\ b)\frac{4}{7}>\frac{5}{9};\\ c)\frac{-3}{7}>\frac{-4}{9}\\ d)\frac{-7}{8}<\frac{35}{-42}\end{array}$

$$\begin{gathered} a)\frac{3}{4}=\frac{6}{8}<\frac{6}{7} \\ b)\frac{17}{-21}=\frac{51}{-63}>\frac{51}{-31} \end{gathered}$$

$$\begin{gathered} c)\frac{-4}{9}=\frac{12}{27}<\frac{-12}{52}=\frac{-3}{13} \\ d)\frac{-4}{-11}=\frac{12}{33}>\frac{12}{38}=\frac{-6}{-19} \end{gathered}$$

$\begin{array}{l}a)\frac{2}{3}<\frac{4}{5}\\ b)\frac{13}{-27}>\frac{39}{-37};\\ c)\frac{-3}{7}<\frac{-2}{9}\\ d)\frac{-2}{-7}<\frac{-5}{-17}\end{array}$

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

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

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

$\begin{array}{l}a)\frac{22}{23}>\frac{16}{77};\\ b)\frac{64}{65}>\frac{45}{47};\\ c)\frac{199}{198}>\frac{200}{199};\\ d)\frac{61}{58}>\frac{73}{72}.\end{array}$

$$\begin{gathered} a)\frac{16}{-19}<0<\frac{15}{17} \\ b)\frac{419}{-723}<0<\frac{-697}{-313} \end{gathered}$$

$\begin{array}{l}c)\frac{311}{256}>1>\frac{199}{203}\\ \\ d)\frac{30}{235}=\frac{6}{47}>\frac{6}{48}=\frac{8}{64}>\frac{8}{63}=\frac{168}{1323}\\ \\ e)\frac{19}{60}<\frac{20}{60}=\frac{30}{90}<\frac{31}{90}\\ \\ f)\frac{15}{23}>\frac{14}{23}=\frac{70}{115}>\frac{70}{117}\end{array}$

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

a) Thực hiện quy đồng $\frac{a}{b}=\frac{a(b+m)}{b(b+m)}=\frac{ab+am}{b^2+bm};$

$\frac{a+m}{b+m}=\frac{b(a+m)}{b(b+m)}=\frac{ab+bm}{b^2+bm}.$ Vì $\frac{a}{b}$ < 1=> a < b => ab +am < ab + bm

Từ đó thu được $\frac{a}{b}$ < $\frac{a+m}{b+m}$

b) $\frac{437}{564}<\frac{437+9}{564+9}=\frac{446}{573}.$

$\frac{237}{142}<\frac{237+9}{142+9}=\frac{246}{151}.$

a) $\frac{510}{714}=\frac{5}{7}>\frac{1717}{3535}=\frac{17}{35}.$

b) $\frac{-292929}{333333}=\frac{-29}{33}>\frac{-29}{32}>\frac{-30}{32}=\frac{-15}{16}=\frac{-16665}{17776}.$

a) $\frac{1734}{2346}=\frac{17}{23}<\frac{19}{23}=\frac{1919}{2323}$

b) $\frac{-15151515}{23232323}=\frac{-15}{23}>\frac{-188887}{211109}=\frac{-17}{19}.$

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) A < B              

b) C > D

$\begin{array}{l}a)\frac{5}{5}<\frac{6}{5}<\frac{7}{5}<\frac{8}{5};\\ \\ b)\frac{-15}{15}<\frac{-14}{15}<\frac{-13}{15}<\frac{-12}{15}\end{array}$