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Cho: (2010c-2011b)/2009= (2011a-2009c)/2010= (2009b-2010a)/2011
Chứng minh rằng: a/2009=b/2010=c/2011
\(\dfrac{2010c-2011b}{2009}=\dfrac{2011a-2009c}{2010}=\dfrac{2009b-2010a}{2011}\)
Đặt: \(\left\{{}\begin{matrix}2009=x\\2010=y\\2011=z\end{matrix}\right.\) Ta có:
\(\dfrac{cy-bz}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}\)
\(\Leftrightarrow\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}=\dfrac{cxy-bxz+ayz-cxy+bxz-ayz}{x^2+y^2+z^2}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}cy=bz\Leftrightarrow\dfrac{b}{y}=\dfrac{c}{z}\\az=cx\Leftrightarrow\dfrac{a}{x}=\dfrac{c}{z}\\bx=ay\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\Leftrightarrow\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\left(đpcm\right)\)
\(\frac{b-2011}{c-2010}:\frac{2011-b}{2010-c}=\frac{b-2011}{c-2010}\cdot\frac{-\left(c-2010\right)}{-\left(b-2011\right)}=1\)
\(\frac{a-2009}{b-2011}=\frac{2010-c}{2009-a}=\frac{-\left(c-2010\right)}{-\left(a-2009\right)}=\frac{c-2010}{a-2009}=1\Rightarrow a-2009=c-2010=b-2011\)
\(\Rightarrow a=c-1=b-2\Rightarrow c=b-1\Rightarrow\frac{b}{c}=\frac{b}{b-1}\)=.=' ko chắc lăm
\(B=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2009+1+1}{2009}=\frac{2009}{2010}+\frac{2010}{2011}+1+\frac{1}{2009}+\frac{1}{2009}\)
\(B=\frac{2009}{2010}+\frac{1}{2009}+\frac{2010}{2011}+\frac{1}{2009}+1\)
\(B>\frac{2009}{2010}+\frac{1}{2010}+\frac{2010}{2011}+\frac{1}{2011}+1=3\)
: 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
Dung 100%