\(\overline{abbc}=\overline{ab}.\overline{ac}.7\)

\(\Leftrightarrow100\times\overline{ab}+\overline{bc}=7\times\overline{ab}\times\overline{ac}\)

\(\Leftrightarrow\overline{ab}\times\left(7\times\overline{ac}-100\right)=\overline{bc}\)

\(7\times\overline{ac}-100=\frac{\overline{bc}}{\overline{ab}}\)

Vì \(0< \frac{\overline{bc}}{\overline{ab}}< 10\Rightarrow0< 7\times\overline{ac}-100< 10\)

\(\Rightarrow100< 7\times\overline{ac}< 110\)

\(14< \frac{100}{7}< \overline{ac}< \frac{110}{7}< 16\)

\(\Rightarrow\overline{ac}=15\Rightarrow\overline{a}=1,\overline{c}=5\)

Thay \(\overline{ac}=15\)ta được: \(\overline{1bb5}=15\times\overline{1b}\times7\)

\(\Rightarrow5\times\overline{b}=45\Rightarrow\overline{b}=\frac{45}{5}=9\)

Vậy \(a=1,b=9,c=5\ne0\left(tm\right)\)

Ta có:

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

\(\Rightarrow\dfrac{10a}{b}+\dfrac{b}{b}=\dfrac{10b}{c}+\dfrac{c}{c}=\dfrac{10c}{a}+\dfrac{a}{a}\)

\(\Rightarrow\dfrac{10a}{b}+1=\dfrac{10b}{c}+1=\dfrac{10c}{a}+1\)

\(\Rightarrow\dfrac{10a}{b}=\dfrac{10b}{c}=\dfrac{10c}{a}\)

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

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

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

\(\Rightarrow\left(\overline{abc}\right)^{123}=\left(\overline{aaa}\right)^{123}\)(1)

\(\Rightarrow c=111^{123}.a^{40}.a^{41}.a^{42}=111^{123}.a^{123}=\left(111.a\right)^{123}=\left(\overline{aaa}\right)^{123}\)(2)

Từ (1) và (2) suy ra: \(\left(\overline{abc}\right)^{123}=111^{123}.a^{40}.b^{41}.c^{42}\)

120 km nha@  nhớ tk cho nha,điểm mainhf đang âm@  `_'

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

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

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

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