Lời giải:
$b^2=ac\Rightarrow \frac{b}{a}=\frac{c}{b}$
Đặt $\frac{b}{a}=\frac{c}{b}=k\Rightarrow b=ak; c=bk$
Khi đó:
$\frac{a^{2022}+b^{2022}}{b^{2022}+c^{2022}}=\frac{a^{2022}+(ak)^{2022}}{b^{2022}+(bk)^{2022}}$
$=\frac{a^{2022}(1+k^{2022})}{b^{2022}(1+k^{2022})}=\frac{a^{2022}}{b^{2022}} (1)$
Và:
$(\frac{a+b}{b+c})^{2022}=(\frac{a+ak}{b+bk})^{2022}$
$=[\frac{a(k+1)}{b(1+k)}]^{2022}=(\frac{a}{b})^{2022}=\frac{a^{2022}}{b^{2022}}(2)$
Từ $(1); (2)$ ta có đpcm.