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\(\dfrac{2010c-2011b}{2009}=\dfrac{2011a-2009c}{2010}=\dfrac{2009b-2010a}{2011}\)

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:

\(\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}=\dfrac{cxy-bxz+ayz-cxy+bxz-ayz}{x^2+y^2+z^2}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}cy=bz\Leftrightarrow\dfrac{b}{y}=\dfrac{c}{z}\\az=cx\Leftrightarrow\dfrac{a}{x}=\dfrac{c}{z}\\bx=ay\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\Leftrightarrow\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\left(đpcm\right)\)

\(\Leftrightarrow\left(a+2009\right)\left(b-2010\right)=\left(a-2009\right)\left(b+2010\right)\)

=>ab-2010a+2009b-2009x2010=ab+2010a-2009b-2009x2010

=>-4020a=-4018b

=>a/2009=b/2010

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)

\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)

\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)

thầy ơi bài này làm rồi

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)

Vậy:

\(\frac{a\cdot c}{b\cdot d}=\frac{bk\cdot dk}{b\cdot d}=\frac{k^2\cdot\left[b\cdot d\right]}{b\cdot d}=k^2\)

và

\(\frac{2009a^2+2010c^2}{2009b^2+2010d^2}=\frac{2009\left[bk\right]^2+2010\left[dk\right]^2}{2009b^2+2010d^2}=\frac{2009\cdot b^2k^2+201d^2k^2}{2009b^2+2010d^2}=\frac{k^2\left[2009b^2+2010d^2\right]}{2009b^2+2010d^2}=k^2\)Vậy khi \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{ac}{bd}=\frac{2009a^2+2010c^2}{2009b^2+2010d^2}\)

Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{2009a^2}{2009b^2}=\frac{2010c^2}{2010d^2}=\frac{2009a^2+2010c^2}{2009b^2+2010d^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{c}{d}=\frac{ac}{bd}\)

Vậy \(\frac{ac}{bd}=\frac{2009a^2+2010c^2}{2009b^2+2010d^2}\)