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

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:

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

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

Ta có: (a + 2009)(b - 2010) = (a - 2009)(b + 2010) 

=> ab - 2010a + 2009b - 2009.2010 = ab + 2010a - 2009b - 2009.2010

=> ab - 2010a + 2009b - 2009.2010 - ab - 2010a + 2009b + 2009.2010 = 0

=> -2010a + 2.2009b = 0

=> 2010a = 2.2009b

Đề sai

Ta có: (a + 2009)(b - 2010) = (a - 2009)(b + 2010)

=> ab - 2010a + 2009b - 2009.2010 = ab + 2010a - 2009b - 2009.2010

=> ab - 2010a + 2009b - 2009.2010 - ab - 2010a + 2009b + 2009.2010 = 0

=> -2010a + 2.2009b = 0

=> 2010a = 2.2009b Đề sai 

Làm ơn giúp mk!!

\(\frac{b-2011}{c-2010}:\frac{2011-b}{2010-c}=\frac{b-2011}{c-2010}\cdot\frac{-\left(c-2010\right)}{-\left(b-2011\right)}=1\)

\(\frac{a-2009}{b-2011}=\frac{2010-c}{2009-a}=\frac{-\left(c-2010\right)}{-\left(a-2009\right)}=\frac{c-2010}{a-2009}=1\Rightarrow a-2009=c-2010=b-2011\)

\(\Rightarrow a=c-1=b-2\Rightarrow c=b-1\Rightarrow\frac{b}{c}=\frac{b}{b-1}\)=.=' ko chắc lăm

Thanks!!! Nhưng xin lỗi mặc dù phải là -1. Cảm ơn bạn