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Ta có: \(\dfrac{6}{{15}} = \dfrac{{6:3}}{{15:3}} = \dfrac{2}{5}\)
\(\begin{array}{l}BCNN\left( {20,5} \right) = 20\\\dfrac{2}{5} = \dfrac{{2.4}}{{5.4}} = \dfrac{8}{{20}}\end{array}\)
Vì 3 < 8 nên \(\dfrac{3}{{20}} < \dfrac{8}{{20}}\)
Suy ra \(\dfrac{3}{{20}} < \dfrac{6}{{15}}\)
a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)
\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)
mà -21>-22
nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)
b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)
\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)
mà -25>-28
nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)
c: \(\dfrac{-8}{7}< -1\)
\(-1< -\dfrac{2}{5}\)
Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)
d: \(-\dfrac{2}{5}< 0\)
\(0< \dfrac{1}{3}\)
Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)
b: \(M=\dfrac{53\cdot71-18}{71\cdot52+53}=\dfrac{52\cdot71+71-18}{71\cdot52+53}=1\)
\(N=\dfrac{53\cdot107+107-53}{53\cdot107+54}=1\)
\(P=\dfrac{134\cdot269+269-133}{134\cdot269+135}=1\)
=>M=N=P
a)\(\dfrac{-8}{9}< \dfrac{-7}{9}\\ \dfrac{6}{7}< \dfrac{11}{10}\)
Do \(\dfrac{{ - 11}}{8} < 0\) và \(\dfrac{1}{{24}} > 0\) nên \(\dfrac{{ - 11}}{8} < \dfrac{1}{{24}}\)
Ta có:
\(\dfrac{-49}{211}< 0;\dfrac{13}{1999}>0\)
⇒ \(\dfrac{-49}{211}< \dfrac{13}{1999}\)
\(\frac{18}{91}=\frac{558}{2821}\)
\(\frac{6}{31}=\frac{546}{2821}\)
=>\(\frac{18}{91}>\frac{6}{31}\)
Ta có : \(\frac{18}{91}< \frac{18}{90}\left(91>90\right)\)
\(\Rightarrow\frac{18}{91}< \frac{1}{5}\left(1\right)\)
\(\frac{6}{31}>\frac{6}{30}\left(31>30\right)\)
\(\Rightarrow\frac{6}{31}>\frac{1}{5}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{18}{91}< \frac{1}{5}< \frac{6}{31}\)
\(\Rightarrow\frac{18}{91}< \frac{6}{31}\)
Chúc bạn học tốt !!!