bài 1 so sánh
a) \(\dfrac{45}{105}va\dfrac{84}{147}\)
b) \(\dfrac{-12}{19}va\dfrac{-14}{17}\)
c) \(\dfrac{13}{19}va\dfrac{47}{53}\)
d) \(\dfrac{-29}{38}va\dfrac{-13}{22}\)
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Bài 1:
\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)
\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\)
Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\)
\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\)
\(\dfrac{-18}{31}\)
\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\)
Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)
Bài 2:
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
a, Ta có:
\(\dfrac{-13}{39}=\dfrac{-1}{3}\) và \(-\dfrac{21}{63}=\dfrac{-1}{3}\)
Vì \(\dfrac{-1}{3}=\dfrac{-1}{3}\) nên \(\dfrac{-13}{39}=-\dfrac{21}{63}\)
b, Ta có:
\(\dfrac{1}{234567}>0\) (số hữu tỉ dương) và \(-\dfrac{2}{14}< 0\) (số hữu tỉ âm)
=> \(\dfrac{1}{234567}>-\dfrac{2}{14}\)
c\(\dfrac{1}{2012}>-\dfrac{1}{14}\), Ta có:
\(\dfrac{-39}{65}=\dfrac{-3}{5}\) và \(-\dfrac{21}{35}=\dfrac{-3}{5}\)
mà \(\dfrac{-3}{5}=\dfrac{-3}{5}\) nên \(\dfrac{-39}{65}=-\dfrac{21}{35}\)
d,Ta có:
\(\dfrac{1}{2012}>0\) (số hữu tỉ dương) và \(-\dfrac{1}{14}< 0\) (số hữu tỉ âm)
Vậy suy ra: \(\dfrac{1}{2012}>-\dfrac{1}{14}\)
\(1-\dfrac{13}{19}=\dfrac{6}{19}\\ 1-\dfrac{47}{53}=\dfrac{6}{53}\\ \dfrac{6}{19}>\dfrac{6}{53}\Rightarrow\dfrac{13}{19}< \dfrac{47}{53}\)
9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)
\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)
10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
a: \(\dfrac{12}{49}< \dfrac{13}{49}< \dfrac{13}{47}\)
b: \(\dfrac{12}{47}>\dfrac{19}{47}>\dfrac{19}{77}\)
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