Có : \(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}\Rightarrow\dfrac{10a+b}{10b+c}=\dfrac{a}{c}=\dfrac{9a+b}{10b}\)( áp dụng dãy tỉ số bằng nhau)

\(=\dfrac{111...11.\left(9a+b\right)}{111..11.10b}\)(có n chữ số 1 trong số 111..111)

\(\dfrac{999..99a+111..11b}{111..110b}=\dfrac{a}{c}=\dfrac{999..99a+a+111..11b}{111..110b+c}=\dfrac{100...000a+111...11b}{111..110b+c}\)=\(\dfrac{\overline{abbb...bb}}{\overline{bbb..bbc}}=\dfrac{a}{c}\)

Ta có:

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

Mà: \(\left\{\begin{matrix}\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{10a+b+10b+c}{a+b}=9a+10b+c\\\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{10b+c+10c+a}{b+c}=9b+10c+a\\\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{10c+a+10a+b}{c+a}=9c+10a+b\end{matrix}\right.\)

\(\Rightarrow9a+10b+c=9b+10c+a=9c+10a+b\)

\(\Rightarrow\left\{\begin{matrix}9a=9b=9c\\10b=10c=10a\\c=a=b\end{matrix}\right.\)\(\Rightarrow a=b=c\)

Vậy \(a=b=c\) (Đpcm)

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

\(\left(10a+b\right)\left(b+c\right)=\left(a+b\right)\left(10b+c\right)\)

\(10ab+b^2+10ac+bc=10ab+10b^2+ac+bc\)

\(9ac=9b^2\)

\(ac=b^2\)

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

\(\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)=\(1+\frac{9a}{a+b}=1+\frac{9b}{b+c}\)

\(\frac{9a}{a+b}=\frac{9b}{b+c}=>\frac{9a}{9b}=\frac{a+b}{b+c}\)

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

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

nếu đúng thì k nka

đừng k t nhé

Ngu người 

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

\(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}+\overline{bc}+\overline{ca}+\overline{ca}+\overline{ab}}{a+b+b+c+c+a}=\frac{2\left(\overline{ab}+\overline{bc}+\overline{ca}\right)}{2\left(a+b+c\right)}=\frac{\overline{ab}+\overline{bc}+\overline{ca}}{a+b+c}\)

\(=\frac{10a+b+10b+c+10c+a}{a+b+c}=\frac{11a+11b+11c}{a+b+c}=\frac{11\left(a+b+c\right)}{a+b+c}=11\)

Lại có : \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}\)

+) Nếu \(a+b+c=0\) : 

\(\Rightarrow\)\(a+b=-c\)

\(\Rightarrow\)\(b+c=-a\)

\(\Rightarrow\)\(a+c=-b\)

Thay \(a+b=-c\)\(;\)\(b+c=-a\) và \(a+c=-b\) vào \(\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}\) ta được : 

\(\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)

+) Nếu \(a+b+c\ne0\) : 

Do đó : 

\(\frac{\overline{ab}+\overline{bc}}{a+b}=11\)\(\Rightarrow\)\(10a+11b+c=11a+11b\)\(\Rightarrow\)\(c=a\)\(\left(1\right)\)

\(\frac{\overline{bc}+\overline{ca}}{b+c}=11\)\(\Rightarrow\)\(10b+11c+a=11b+11c\)\(\Rightarrow\)\(a=b\)\(\left(2\right)\)

\(\frac{\overline{ca}+\overline{ab}}{c+a}=11\)\(\Rightarrow\)\(10c+11a+b=11c+11a\)\(\Rightarrow\)\(b=c\)\(\left(3\right)\)

Từ (1), (2) và (3) suy ra : 

\(a=b=c\)

Suy ra : 

\(P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{b+b}{b}.\frac{c+c}{c}.\frac{a+a}{a}=\frac{2b}{b}.\frac{2c}{c}.\frac{2a}{a}=2.2.2=8\)

Vậy \(P=-1\) hoặc \(P=8\)

Chúc bạn học tốt ~ 

+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}-\overline{bc}-\overline{ca}+\overline{ca}+\overline{ab}}{a+b-b-c+c+a}=\frac{2\overline{ab}}{2a}=10+\frac{b}{a}\)

+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}+\overline{bc}+\overline{ca}-\overline{ca}-\overline{ab}}{a+b+b+c-c-a}=\frac{2\overline{bc}}{2b}=10+\frac{c}{b}\)

+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{-\overline{ab}-\overline{bc}+\overline{bc}+\overline{ca}+\overline{ca}+\overline{ab}}{-a-b+b+c+c+a}=\frac{2\overline{ca}}{2c}=10+\frac{a}{c}\)

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