Giải:
Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\)
\(\Rightarrow a=bk,b=ck\)
Ta có: \(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(bk\right)^2+\left(ck\right)^2}{b^2+c^2}=\frac{b^2.k^2+c^2.k^2}{b^2.c^2}=\frac{\left(b^2+c^2\right).k^2}{b^2.c^2}=k^2\) (1)
\(\frac{a}{c}=\frac{bk}{c}=\frac{ckk}{c}=k^2\) (2)
Từ (1) và (2) \(\Rightarrow\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\left(đpcm\right)\)