Ta có \(\frac{2010c-2011b}{2009}=\frac{2011a-2009c}{2010}=\frac{2009b-2010a}{2011}\)

=> \(\frac{2009.2010c-2009.2011b}{2009^2}=\frac{2010.2011a-2009.2010c}{2010^2}=\frac{2009.2011b-2010.2011a}{2011^2}\)

= \(\frac{2009.2010c-2009.2011b+2010.2011a-2009.2010c+2009.2011b-2010.2011a}{2009^2+2010^2+2011^2}\)= 0

=> \(\hept{\begin{cases}2010c-2011b=0\\2011a-2009c=0\\2009b-2010a=0\end{cases}}\Rightarrow\hept{\begin{cases}2010c=2011b\\2011a=2009c\\2009b=2010a\end{cases}}\Rightarrow\hept{\begin{cases}\frac{c}{2011}=\frac{b}{2010}\\\frac{a}{2009}=\frac{c}{2011}\\\frac{b}{2010}=\frac{a}{2009}\end{cases}}\)

=> \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}\)(đpcm)

\(x\left(x+2\right)+x^2-5x=0\)

\(\Leftrightarrow x^2+2x+x^2-5x=0\)

\(\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\)

\(\Leftrightarrow x=0;\frac{3}{2}\)

X(x+2)+ x² -5x= 0

x² +2x +x² - 5x= 0

2x²- 3x= 0

x( 2x-3 )= 0

x= 0 hoặc 2x-3= 0

x=0 hoặc x= 1,5

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)

Khi đó 4(a - b)(b - c) = 4(2018k - 2019k)(2019k - 2020k)

= 4(-k).(-k) 

= 4k² (1)

Lại có (c - a)² = (2020k - 2018k)² = (2k)² = 4k² (2)

Từ (1)(2) => 4(a - b)(b - c) = (c - a)²