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

=>10ac+bc=10b^2+bc

=>ac=b^2

=>a/b=b/c=k

=>a=bk; b=ck

=>a=ck^2; b=ck

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{c^2k^4+c^2k^2}{c^2k^2+c^2}=k^2\)

\(\dfrac{a}{c}=\dfrac{ck^2}{c}=k^2\)

=>\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)

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

\(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}\Rightarrow\dfrac{10a+b}{10b+c}=\dfrac{a}{c}=\dfrac{9a+b}{10b}\\ =\dfrac{111...11\left(9a+b\right)}{111...11.10b}\)(có n chữ số 1 trong 111...11)

\(\dfrac{999...99a+111...11b}{111.110b}\\ =\dfrac{999...99a+a+111...11}{111.10b+c}=\dfrac{abbb...bb}{bbb...bc}=\dfrac{a}{c}\)(đpcm)

1+1=3

1234567

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

=>10ac+bc=10b^2+cb

=>10ac=10b^2

=>ac=b^2

=>a/b=b/c=k
=>a=bk; b=ck

=>a=ck*k=k^2*c

\(\dfrac{a}{c}=\dfrac{k^2c}{c}=k^2\)

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{b^2k^2+b^2}{c^2k^2+c^2}=\dfrac{b^2}{c^2}=\dfrac{c^2k^2}{c^2}=k^2\)

=>ĐPCM

Ta có: \(\dfrac{\overline{ab}}{a+b}=\dfrac{\overline{bc}}{b+c}\)

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

\(\Rightarrow ab^2+abc=abc+b^2c\)

\(\Rightarrow ab^2=b^2c\)

\(\Rightarrow ab=bc\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\rightarrowđpcm.\)

Ta có:

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

\(\Rightarrow\overline{ab}.\left(b+c\right)=\overline{bc}.\left(a+b\right)\)

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

\(\Rightarrow10ab+10ac+b^2+bc=10ab+10b^2+ac+bc\)

\(\Rightarrow10ac+b^2=10b^2+ac\) (bớt mỗi bên đi \(10ab+bc\))

\(\Rightarrow10ac-ac=10b^2-b^2\Rightarrow9ac=9b^2\)

\(\Rightarrow ac=b^2\) (chia mỗi bên cho 9)

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

Chúc bạn học tốt!!!