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\(\dfrac{60}{81}< \dfrac{60}{80}\) và \(\dfrac{21}{25}>\dfrac{20}{25}\)
=> \(\dfrac{60}{80}=\dfrac{3}{4}\) và \(\dfrac{20}{25}=\dfrac{4}{5}\) => \(\dfrac{4}{5}>\dfrac{3}{4}\)
\(=>\dfrac{21}{25}>\dfrac{20}{25}>\dfrac{60}{80}>\dfrac{60}{81}\)
Vậy \(\dfrac{21}{25}>\dfrac{60}{81}\)
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\(\frac{21}{25};\frac{60}{81};\frac{19}{29}\)
Đúng 100% Chúc bạn zui ~^^
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k mik nha
Ta co:
21/25>21/27=63/81>60/81
19/29<19/27=57/81<60/81
suy ra ; 21/25>60/81>19/29
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Học cái này chưa bạn
\(\frac{a}{b}>\frac{a}{b+n}\left(n\in N^{\cdot}\right)\)và \(\frac{a}{b}< \frac{a+n}{b+n}\)
khỏi chứng minh nhé (nếu cần liên hệ https://www.facebook.com/tiendung.truong.370 )
ta so sánh từng cặp
ta có \(\frac{60}{81}< \frac{60}{80}\)
\(\frac{20}{25}< \frac{21}{25}\)
mà \(\frac{60}{80}=\frac{3}{4};\frac{20}{25}=\frac{4}{5};=>\frac{3}{4}< \frac{4}{5}\)
vậy
\(\frac{60}{81}< \frac{21}{25}\)
cặp tiếp theo
\(\frac{60}{81}vs\frac{19}{29}\)
ta có
\(\frac{60}{81}>\frac{60}{81+9};\frac{60}{81}>\frac{2}{3}\)
\(\frac{19}{29}< \frac{19+1}{29+1};\frac{19}{29}< \frac{2}{3}\)
vậy \(\frac{60}{81}>\frac{19}{29}\)
còn cặp cuối không cần so vì a>b; b>c suy ra a>c
vậy ta có \(\frac{21}{25}>\frac{60}{81}>\frac{19}{29}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cần nhớ:
\(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\left(n\inℕ^∗\right)\)
\(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\left(n\inℕ^∗\right)\)
Ta thấy:
\(\frac{19}{29}< 1\Rightarrow\frac{19}{29}< \frac{19+1}{29+1}=\frac{20}{30}=\frac{2}{3}\)
Ta lại có:
\(\frac{2}{3}=\frac{2.7}{3.7}=\frac{14}{21}< 1\Rightarrow\frac{14}{21}< \frac{14+6}{21+6}=\frac{20}{27}=\frac{20.3}{27.3}=\frac{60}{81}\)
\(\Rightarrow\frac{19}{29}< \frac{60}{81}\) (1)
Ta có:
\(\frac{60}{81}=\frac{20}{27}< 1\Rightarrow\frac{20}{27}< \frac{20+1}{27+1}=\frac{21}{28}< \frac{21}{25}\)
=>\(\frac{60}{81}< \frac{21}{25}\) (2)
Từ (1) và (2) => \(\frac{19}{29}< \frac{60}{81}< \frac{21}{25}\)
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\(\frac{21}{25}=\frac{49329}{58725}\)
\(\frac{60}{81}=\frac{43500}{58725}\)
\(\frac{19}{29}=\frac{38475}{58725}\)
\(\)vì \(\frac{49329}{58725}>\frac{43500}{58725}>\frac{38475}{58725}\)nên\(\frac{21}{25}>\frac{60}{81}>\frac{19}{29}\)
\(\frac{21}{25};\frac{60}{81};\frac{19}{29}\)
Ta có : \(\frac{60}{81}=\frac{60:3}{81:3}=\frac{20}{27}\); giữ nguyên \(\frac{21}{25};\frac{19}{29}\).
Vì \(\frac{21}{25}>\frac{20}{25}>\frac{20}{27}\)nên \(\frac{21}{25}>\frac{20}{27}\)
Vì \(\frac{20}{27}>\frac{20}{29}>\frac{19}{29}\)nên \(\frac{20}{27}>\frac{19}{29}\). Vậy :
\(\frac{21}{25}>\frac{20}{27}>\frac{19}{29}\)hay \(\frac{21}{25}>\frac{60}{81}>\frac{19}{29}\)