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a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)
Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)
a) 34=3x44x4=1216,67=6x27x2=121434=3x44x4=1216,67=6x27x2=1214
Do 16 > 14 => 1216<1214hay34<67
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a) − 4 9 = − 12 27 < − 12 52 = − 3 13
b) − 4 − 11 = 12 33 > 12 38 = − 6 − 19
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1. \(\frac{7}{39}=\frac{140}{780};\frac{11}{65}=\frac{132}{780};\frac{9}{52}=\frac{135}{780}\) và thứ tự tăng dần là : \(\frac{132}{780}< \frac{135}{780}< \frac{140}{780}\)
2. \(\frac{17}{20}=\frac{153}{180};-\frac{19}{30}=-\frac{114}{180};\frac{38}{45}=\frac{152}{180};-\frac{13}{10}=-\frac{130}{180}\) và thứ tự tăng dận là :
\(-\frac{130}{180}< -\frac{114}{180}< \frac{152}{180}< \frac{153}{180}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
-6=-66/11
1/11=1/11