so sánh 2 số hữu tỉ -2009/2010 và 2010/-2009
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\(\frac{-2009}{2010}và\frac{2010}{-2009}=>\frac{-2009}{2010}và\frac{-2009}{2010}=>\frac{-2009}{2010}=\frac{-2009}{2010}\)
Vậy \(\frac{-2009}{2010}=\frac{2010}{-2009}\)
vi \(\frac{-2009}{2010}>\frac{-2010}{1010}=1\)
\(\frac{2010}{-2009}=\frac{-2010}{2009}<\frac{-2009}{2009}=-1\)
=> x<y
violympic vòng 2
\(\dfrac{2009}{2010}=\dfrac{2009\cdot10001}{2010\cdot10001}=\dfrac{20092009}{20102010}\)
Ta có :
\(N=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)
\(=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}\)
\(=\frac{2009^{2009}+1}{2009^{2010}+1}=M\)
Vậy \(M>N\)
Ta có: \(B< 1\)
\(\Rightarrow B< \frac{2009^{2010}-2+3}{2009^{2011}-2+3}=\frac{2009^{2010}+1}{2009^{2011}+1}\left(1\right)\)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}< 1\)
\(\Rightarrow\frac{2009^{2010}+1}{2009^{2011}+1}< \frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}=A\left(2\right)\)
Từ (1) và (2) suy ra A > B
Ta có :
�=20092010−220092011−2<1B=20092011−220092010−2<1
⇔�<20092010−2+201120092011−2+2011=20092010+200920092011+2009=2009(20092009+1)2009(20092010+1)=20092009+120092010+1=�⇔B<20092011−2+201120092010−2+2011=20092011+200920092010+2009=2009(20092010+1)2009(20092009+1)=20092010+120092009+1=A
⇔�>�⇔A>B
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
Ta so sánh 2009/2010 và 2010/2009
Ta có 2009/2010<1<20010/2009
=>2009/2010<2010/2009 => -2009/2010 > 2010/- 2009