Ta có b2 = ac
=> b.b = a.c
=> \(\frac{a}{b}=\frac{b}{c}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\Rightarrow\hept{\begin{cases}a=bk\\b=ck\end{cases}}\)
Khi đó \(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(bk\right)^2+b^2}{\left(ck\right)^2+c^2}=\frac{b^2k^2+b^2}{c^2k^2+c^2}=\frac{b^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=\frac{b^2}{c^2}=\frac{\left(ck\right)^2}{c^2}=k^2\)(1)
; \(\frac{a}{c}=\frac{bk}{c}=\frac{ckk}{c}=k^2\)(2)
Từ (1)(2) => \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
\(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)
=> đpcm
