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- \(\frac{2011}{2012}-\frac{2012}{2011}\Rightarrow\frac{2011^2-2012^2}{2012.2011}< 0\) Vì hai lũy thừa cùng cơ số và 20011<2012 nên \(\frac{2011}{2012}< \frac{2012}{2011}\)
- \(\frac{2000}{2013}-\frac{2011}{2012}=\frac{2000.2012-2011.2013}{2013.2011}< 0\)vì 2000<2011,2012<2013 nên 2000.2012<2011.2013 nên\(\frac{2000}{2013}< \frac{2011}{2012}\)
1, Ta thấy \(\frac{2011}{2012}< \frac{2012}{2012}=1\)(1)
Ta có :\(\frac{2012}{2011}>\frac{2011}{2011}=1\) (2 )
Từ (1) và (2) suy ra \(\frac{2011}{2012}< 1< \frac{2012}{2011}\)hay \(\frac{2011}{2012}< \frac{2012}{2011}\)
Vậy \(\frac{2011}{2012}< \frac{2012}{2011}\)
2, Ta thấy \(\frac{2000}{2013}=\frac{2013-3}{2013}=\frac{2013}{2013}-\frac{3}{2013}=1-\frac{3}{2013}\)
\(\frac{2011}{2012}=\frac{2012-1}{2012}=\frac{2012}{2012}-\frac{1}{2012}=1-\frac{1}{2012}\)
Ta có \(\frac{1}{2012}=\frac{3}{6036}< \frac{3}{2013}\)nên \(\frac{1}{2012}< \frac{3}{2013}\)suy ra \(1-\frac{3}{2013}< 1-\frac{1}{2012}\)
Hay \(\frac{2000}{2013}< \frac{2011}{2012}\)
Vậy \(\frac{2000}{2013}< \frac{2011}{2012}\)
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a, \(\frac{2011}{2012}\)và \(\frac{2012}{2011}\)
Vì \(\frac{2011}{2012}\)có Tử số bé hơn Mẫu số nên phân số đó < 1 ; \(\frac{2012}{2011}\)có Tử số lớn hơn Mẫu số nên phân số đó > 1
=> \(\frac{2011}{2012}< \frac{2012}{2011}\)
b, \(\frac{2000}{2013}\)và \(\frac{2011}{2012}\)
Ta có:
\(\frac{2000}{2013}=\frac{2000}{2013}+\frac{13}{2013}\) ; \(\frac{2011}{2012}=\frac{2011}{2012}+\frac{1}{2012}\)
Ta thấy \(\frac{13}{2013}>\frac{1}{2012}\)
\(\Rightarrow\frac{2000}{2013}< \frac{2011}{2012}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
B = \(\dfrac{2011}{2012}\) + \(\dfrac{2012}{2013}\) .
\(\dfrac{2011}{2012}\) > \(\dfrac{2011}{2012+2013}\) .
\(\dfrac{2012}{2013}\) > \(\dfrac{2012}{2012+2013}\) .
\(\Rightarrow\) A < B .
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a) \(\left\{{}\begin{matrix}\dfrac{2011}{2012}< 1\\\dfrac{2012}{2011}>1\end{matrix}\right.\Rightarrow\dfrac{2011}{2012}< \dfrac{2012}{2011}\)
b)\(\left\{{}\begin{matrix}\dfrac{2000}{2013}\\\dfrac{2011}{2012}\end{matrix}\right.\left\{{}\begin{matrix}2000< 2011\\2013>2012\end{matrix}\right.\Rightarrow\left\{{}\dfrac{2000}{2013}}< \dfrac{2011}{2012}}}\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{2011}{2012}< \dfrac{2012}{2012}=1\\\dfrac{2012}{2011}>\dfrac{2011}{2011}=1\end{matrix}\right.\) => \(\dfrac{2011}{2012}< \dfrac{2012}{2011}\)
b) sửa đề: \(\dfrac{2010}{2013}\) và \(\dfrac{2011}{2012}\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{2010}{2013}< \dfrac{2010}{2012}\\\dfrac{2011}{2012}>\dfrac{2010}{20122}\end{matrix}\right.\) => \(\dfrac{2010}{2013}< \dfrac{2010}{2012}< \dfrac{2011}{2012}\)
=> \(\dfrac{2010}{2013}< \dfrac{2011}{2012}\)
P/s: Thật ra câu b ko cần sửa đề cx đc nhg mà thấy sai sai nên sửa thôi. Hjhj![leuleu leuleu](https://hoc24.vn/media/cke24/plugins/smiley/images/leuleu.png)