2011×2010-1/2009×2011+2010
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=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)
\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)
Có : \(2009+2010>\dfrac{2009}{2010}\) ; \(2011+2012>\dfrac{2011}{2012}\)
\(\dfrac{2011}{2010}>1\) ; \(\dfrac{2010}{2011}< 1\) \(\Rightarrow\dfrac{2011}{2010}>\dfrac{2010}{2011}\)
Ta có : \(2009+2010+\dfrac{2011}{2010}+2011+2012>\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2011}{2012}\)
\(\Leftrightarrow B>A\)
Hay \(A< B\)
Dễ thấy:
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
=>\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Hay \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
Vậy A > B
`(2011xx2020-1)/(2009xx2011+2010)`
`=((2009+1)xx2011-1)/(2009xx2011+2010)`
`=(2009xx2011+2011-1)/(2009xx2011+2010)`
`=(2009xx2011+2010)/(2009xx2011+2010)`
`=1`
\(\dfrac{2011.2010-1}{2009.2011+2010}\)
= \(\dfrac{2011.2009+2011-1}{2009.2011+2010}\)
= \(\dfrac{2011.2009+2010}{2009.2011+2010}\)
= 1
Trả lời:
\(\frac{2011\times2010-1}{2009\times2011+2010}\)
\(=\frac{2011\times\left(2009+1\right)-1}{2009\times2011+2010}\)
\(=\frac{2009\times2011+2011-1}{2009\times2011+2010}\)
\(=\frac{2009\times2011+2010}{2009\times2011+2010}\)
\(=1\)
\(\frac{2011\times2010-1}{2009\times2011+2010}\)
\(=\frac{2011\times2010-1}{2009\times2011+2011-1}\)
\(=\frac{2011\times2010-1}{2010\times2011-1}\)
\(=1\)