Giải:
Ta có:
\(\overline{ab}+\overline{bc}+\overline{ac}=\overline{abc}\)
\(\Rightarrow10a+b+10b+c+10a+c=100a+10b+c\)
\(\Rightarrow\left(10a+10a\right)+\left(b+10b\right)+\left(c+c\right)=100a+10b+c\)
\(\Rightarrow20a+11b+2c=100a+10b+c\)
\(\Rightarrow b+c=80a\)
mk chỉ biết giải tới đây thôi
\(\overline{ab}+\overline{bc}+\overline{ca}\)
\(\Rightarrow\left(\overline{a}.10+\overline{b}\right)+\left(\overline{b}.10+\overline{c}\right)+\left(\overline{c}.10+\overline{a}\right)=\overline{a}.100+\overline{b}.10.c\)
\(\Rightarrow\overline{a}.11+\overline{b}.11+\overline{c}.11=\overline{a}.100+\overline{b}.10+c\)
Cùng bớt \(\overline{a}.11+\overline{b}.10+c\) ở hai vế , ta có :
\(\overline{b}.1+\overline{c}.10=\overline{a}.89\)
\(\Rightarrow\overline{a}=1\)
\(\Rightarrow\overline{b}=9\)
