Ta có: \(\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab\)
\(\Rightarrow\frac{a^2+c^2}{b^2+c^2}=\frac{a^2+ab}{b^2+ab}=\frac{a\left(a+b\right)}{b\left(b+a\right)}=\frac{a}{b}\left(Đpcm\right)\)
Cách khác:
Đặt \(\dfrac{a}{c}=\dfrac{c}{b}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=ck\\c=bk\end{matrix}\right.\)
Ta có:
\(\begin{array}{l} \dfrac{{{a^2} + {c^2}}}{{{b^2} + {c^2}}} = \dfrac{{{{\left( {ck} \right)}^2} + {{\left( {bk} \right)}^2}}}{{{b^2} + {{\left( {bk} \right)}^2}}} = \dfrac{{{k^2}\left( {{c^2} + {b^2}} \right)}}{{{b^2}\left( {{k^2} + 1} \right)}}\\ = \dfrac{{{k^2}\left[ {{{\left( {bk} \right)}^2} + {b^2}} \right]}}{{{b^2}\left( {{k^2} + 1} \right)}} = \dfrac{{{k^2}\left[ {{b^2}\left( {{k^2} + 1} \right)} \right]}}{{{b^2}\left( {{k^2} + 1} \right)}} = {k^2}\left( 1 \right)\\ \Rightarrow \dfrac{a}{b} = \dfrac{{ck}}{b} = \dfrac{{b.{k^2}}}{b} = {k^2}\left( 2 \right) \end{array}\)
Từ $(1)$ và $(2)$ suy ra: \(\dfrac{{{a^2} + {c^2}}}{{{b^2} + {c^2}}} = \dfrac{a}{b} \)
