So sánh:
a) \(\dfrac{-9}{4}\) và \(\dfrac{1}{3}\).
b) \(\dfrac{-8}{3}\) và \(\dfrac{4}{-7}\).
c) \(\dfrac{9}{-5}\) và \(\dfrac{7}{-10}\).
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a) \(< \)
b) \(>\)
c) \(< \)
d) \(>\)
e) \(< \)
g) \(>\)
h) \(>\)
k) \(>\)
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: \(-\dfrac{2}{5}< \dfrac{1}{3}\)
a) \(\dfrac{4}{9}< \dfrac{4}{8}=\dfrac{1}{2}\)
b) \(\dfrac{5}{8}=\dfrac{15}{24}>\dfrac{14}{24}=\dfrac{7}{12}\)
a) \(\dfrac{2}{9}+\dfrac{5}{9}=\dfrac{7}{9};\dfrac{5}{9}+\dfrac{2}{9}=\dfrac{7}{9}\)
Vậy: \(\dfrac{2}{9}+\dfrac{5}{9}=\dfrac{5}{9}+\dfrac{2}{9}\)
b) \(\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{7}{25}=\dfrac{14}{25};\dfrac{3}{25}+\dfrac{7}{25}+\dfrac{4}{25}=\dfrac{14}{25}\)
Vậy: \(\dfrac{3}{25}+\dfrac{4}{25}+\dfrac{7}{25}=\dfrac{3}{25}+\dfrac{7}{25}+\dfrac{4}{25}\)
a, \(\dfrac{3}{4}\) = \(\dfrac{3\times5}{4\times5}\) = \(\dfrac{15}{20}\) \(\dfrac{5}{7}\) = \(\dfrac{5\times3}{7\times3}\) = \(\dfrac{15}{21}\)
Vì \(\dfrac{15}{20}\) > \(\dfrac{15}{21}\) nên \(\dfrac{3}{4}\) > \(\dfrac{5}{7}\)
b, \(\dfrac{2}{7}\) = \(\dfrac{2\times2}{7\times2}\) = \(\dfrac{4}{14}\) < \(\dfrac{4}{9}\)
Vậy \(\dfrac{2}{7}\) < \(\dfrac{4}{9}\)
c, \(\dfrac{5}{8}\) < 1 < \(\dfrac{8}{5}\)
Vậy \(\dfrac{5}{8}\) < \(\dfrac{8}{5}\)
a) \(\dfrac{5}{9}< \dfrac{7}{9}\)
b) \(\dfrac{7}{6}>\dfrac{6}{6}\)
c) \(\dfrac{3}{14}< \dfrac{5}{14}\)
d) \(\dfrac{5}{8}< \dfrac{9}{8}\)
em trả lời ccaua này hi vọng thầy còn nhớ em
a) -9/4<`1/3
c) 9/-5<7/-10