So sánh : a, \(\dfrac{29}{60}và\dfrac{15}{28}\) b, \(\dfrac{13}{30}và\dfrac{23}{42}\) c, \(\dfrac{13}{36}và\dfrac{14}{45}\) d, \(\dfrac{1919}{9595}và\dfrac{1111}{5050}\)
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a) \(\frac{29}{60}< \frac{1}{2}< \frac{15}{28}\)
b)\(\frac{13}{30}< \frac{1}{2}< \frac{23}{42}\)
c)\(\frac{13}{36}>\frac{1}{3}>\frac{14}{45}\)
d)\(\frac{1919}{9595}=\frac{1}{5}< \frac{11}{50}=\frac{1111}{5050}\)
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a, Ta có: \(\frac{29}{60}< \frac{30}{60}=\frac{1}{2}=\frac{14}{28}< \frac{15}{28}\) . Vậy \(\frac{29}{60}< \frac{15}{28}\)
b, \(\frac{13}{30}< \frac{15}{30}=\frac{1}{2}=\frac{21}{42}< \frac{23}{42}\).Vậy \(\frac{13}{30}< \frac{23}{42}\)
c, \(\frac{13}{36}>\frac{12}{36}=\frac{1}{3}=\frac{15}{45}>\frac{14}{45}\).Vậy \(\frac{13}{36}>\frac{14}{45}\)
d, \(\frac{1919}{9595}=\frac{1}{5}=\frac{1010}{5050}< \frac{1111}{5050}.\)Vậy \(\frac{1919}{9595}< \frac{1111}{5050}\)
a) \(\dfrac{23}{24}< 1\)
\(\dfrac{24}{23}>1\)
\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)
b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)
c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)
d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)
a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)
a: \(\dfrac{11}{42}=\dfrac{11\cdot1}{42\cdot1}=\dfrac{11}{42}\)
\(\dfrac{5}{6}=\dfrac{5\cdot7}{6\cdot7}=\dfrac{35}{42}\)
b: \(\dfrac{40}{7}=\dfrac{40\cdot4}{7\cdot4}=\dfrac{160}{28}\)
\(\dfrac{10}{28}=\dfrac{10\cdot1}{28\cdot1}=\dfrac{10}{28}\)
c: \(\dfrac{4}{15}=\dfrac{4\cdot4}{15\cdot4}=\dfrac{16}{60}\)
\(\dfrac{14}{60}=\dfrac{14\cdot1}{60\cdot1}=\dfrac{14}{60}\)
a)\(\dfrac{11}{42}=\dfrac{35}{42}\)
b) \(\dfrac{160}{28}=\dfrac{10}{28}\)
c) \(\dfrac{16}{60}=\dfrac{14}{60}\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\\ =28\cdot\dfrac{-23}{21}-8=-\dfrac{92}{3}-8=-\dfrac{116}{3}\)
\(C=\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}-\dfrac{29}{42}:\dfrac{1}{28}-8\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\)
\(=28\left(\dfrac{9}{42}-\dfrac{26}{42}-\dfrac{29}{42}\right)-8\)
\(=28\cdot\dfrac{-46}{42}-8\)
\(=\dfrac{-116}{3}\)
c: \(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}\right)\)
\(=28\cdot\left(\dfrac{9}{42}-\dfrac{26}{42}\right)\)
\(=28\cdot\dfrac{-17}{42}=\dfrac{-34}{3}\)
Theo quy ước với mọi phân số lớn hơn 0 thì ta có:
\(\dfrac{a}{b}>0=>\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N;n\ne0\right)\)
Áp dụng với bài trên ta => ĐPCM
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