Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)

\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)

\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)

Áp dụng t/c dãy tỉ số bằng nhau,ta có:

\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)

\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)

\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)

=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)

=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)

=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)

=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)

Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)

\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)

\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)

Sửa đề thành : (a-b ) .(b-c)

\(\frac{\overline{ab}}{b}=\frac{\overline{bc}}{c}=\frac{\overline{ca}}{a}=\frac{10a+b}{b}=\frac{10b+c}{c}=\frac{10c+a}{a}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{\overline{ab}}{b}=\frac{\overline{bc}}{c}=\frac{\overline{ca}}{a}=\frac{10a+b}{b}=\frac{10b+c}{c}=\frac{10c+a}{a}=\frac{11.\left(a+b+c\right)}{a+b+c}=11\)

\(\frac{10a+b}{b}=11\Rightarrow10a+b=11b\Rightarrow10a=10b\Rightarrow a=b\)(1)

\(\frac{10b+c}{c}=11\Rightarrow10b+c=11c\Rightarrow10b=10c\Rightarrow b=c\)(2)

\(\frac{10c+a}{a}=11\Rightarrow10c+a=11a\Rightarrow10c=10a\Rightarrow c=a\)(3)

từ (1), (2), (3)  => a=b=c (đpcm) 

Ta có:\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+b+a}\)

\(=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy giá trị của mỗi tỉ số là:\(\frac{1}{2}\)

Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{a}{a+b}.\)

\(\Rightarrow\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{1}{2}\)

Xét 2 trường hợp: Nếu a+b+c = 0

                    Và Nếu a+b+c = \(\frac{1}{2}\)

Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\),Xét 2 TH sau:

+Nếu a+b+c \(\ne\) 0 thì theo t/c dãy tỉ số=nhau:

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{\left(a+a\right)+\left(b+b\right)+\left(c+c\right)}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

+Nếu a+b+c = 0 thì a+b=-c ; b+c=-a;c+a=-b

\(=>\frac{a}{b+c}=\frac{a}{-a}=1;\frac{b}{a+c}=\frac{b}{-b}=-1;\frac{c}{a+b}=\frac{c}{-c}=-1\)

\(=>\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=-1\)

Vậy............

1./ Nếu a + b + c = 0 

\(\Rightarrow a=-\left(b+c\right)\Rightarrow\frac{a}{b+c}=-1\)

=> Giá trị các tỷ số đó = -1.

2./ Nếu a + b + c khác 0 thì:

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Giá trị các tỷ số đó = 1/2

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{b+a}\)

\(=\frac{a-b-c}{b+c-a-c-b-a}\)

\(=\frac{a-b-c}{-2a}\)

\(=>\frac{a}{b+c}=\frac{a-b-c}{-2a}\)

\(=>\frac{b}{a+c}=\frac{a-b-c}{-2a}\)

\(=>\frac{c}{b+a}=\frac{a-b-c}{-2a}\)

Từ \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

Áp dụng tính chất của dãy tỉ  số bằng nhau ta có 

\(\frac{a}{b+c}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Tương tự \(\frac{b}{c+a}=\frac{c}{a+b}=\frac{1}{2}\)

Vậy \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{1}{2}\)

\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{1}{2}\)(dãy tỉ số bằng nhau)

\(\Rightarrow2a=b+c\)
\(\Rightarrow2b=c+a\)

\(\Rightarrow2c=a+b\)

ta có hpt:

\(\hept{\begin{cases}2a=b+c\\2b=c+a\\2c=a+b\end{cases}\hept{\begin{cases}b=2a-c\\2b=c+a\\2c=a+b\end{cases}}}\)

thế b ta đc

\(\hept{\begin{cases}4a-2c=c+a\\2c=a+2a-c\end{cases}\hept{\begin{cases}3a-3c=0\\3c=3a=0\end{cases}\Rightarrow}}a=c\)

\(b=2a-c=a\)

\(\Rightarrow a=b=c\)vậy pt vô số nghiệm

cho 3 tỉ số bằng nhau thì làm sao hả Thỏ Nghịch Ngợm