B= \(\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2009}\). CM B>3
K
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N
26 tháng 2 2019
\(\frac{b-2011}{c-2010}:\frac{2011-b}{2010-c}=\frac{b-2011}{c-2010}\cdot\frac{-\left(c-2010\right)}{-\left(b-2011\right)}=1\)
\(\frac{a-2009}{b-2011}=\frac{2010-c}{2009-a}=\frac{-\left(c-2010\right)}{-\left(a-2009\right)}=\frac{c-2010}{a-2009}=1\Rightarrow a-2009=c-2010=b-2011\)
\(\Rightarrow a=c-1=b-2\Rightarrow c=b-1\Rightarrow\frac{b}{c}=\frac{b}{b-1}\)=.=' ko chắc lăm
L
30 tháng 5 2019
Bài giải
Theo bài ra :
\(A=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}\)
\(B=\frac{2009+2010+2011}{2010+2011+2012}=\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
Ta có :
\(\frac{2009}{2010}>\frac{2009}{2010+2011+2012}\)
\(\frac{2010}{2011}>\frac{2010}{2010+2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}>\frac{2009}{2010+2011+2012}+\frac{2010}{2010+2011+2012}+\frac{2011}{2010+2011+2012}\)
\(\Rightarrow\text{ }A>B\)
10 tháng 6 2020
\(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
22 tháng 2 2020
\(\frac{x-2010-2011}{2009}+\frac{x-2009-2011}{2010}+\frac{x-2009-2010}{2011}=3\)
\(\Leftrightarrow\left(\frac{x-2010-2011}{2009}-1\right)+\left(\frac{x-2009-2011}{2010}-1\right)+\left(\frac{x-2009-2010}{2011}-1\right)=0\)
\(\Leftrightarrow\frac{x-6030}{2009}+\frac{x-6030}{2010}+\frac{x-6030}{2011}=0\)
\(\Leftrightarrow\left(x-6030\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\)
\(\Leftrightarrow x-6030=0\)(vì \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}>0\))
\(\Leftrightarrow x=6030\)
Vậy ................
\(B=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2009+1+1}{2009}=\frac{2009}{2010}+\frac{2010}{2011}+1+\frac{1}{2009}+\frac{1}{2009}\)
\(B=\frac{2009}{2010}+\frac{1}{2009}+\frac{2010}{2011}+\frac{1}{2009}+1\)
\(B>\frac{2009}{2010}+\frac{1}{2010}+\frac{2010}{2011}+\frac{1}{2011}+1=3\)
: B = \(\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2009}\)
=> \(\frac{2009}{2010}+\frac{2010}{2011}+1+\frac{1}{2019}+\frac{1}{2019}\)
ma : + 1 - \(\frac{2009}{2010}=\frac{1}{2010}\) /// \(\frac{1}{2019}>\frac{1}{2010}\) => \(\frac{2009}{2010}+\frac{1}{2009}>1\)
+ \(1-\frac{2010}{2011}=\frac{1}{2011}\) //// \(\frac{1}{2019}>\frac{1}{2011}\) => \(\frac{1}{2019}+\frac{2010}{2011}>1\)
=> \(\left(\frac{2009}{2010}+\frac{1}{2009}\right)+\left(\frac{2010}{2011}+\frac{1}{2019}\right)+1\)
( >1 + >1 + 1 ) > 3
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