Đáp án:
$\begin{array}{l}
Do:\dfrac{x}{z} = \dfrac{z}{y}\\
\Rightarrow x.y = z.z\\
\Rightarrow x.y = {z^2}\\
Thay\,{z^2} = x.y\\
\Rightarrow \dfrac{{{x^2} + {z^2}}}{{{y^2} + {z^2}}} = \dfrac{{{x^2} + x.y}}{{{y^2} + x.y}} = \dfrac{{x.\left( {x + y} \right)}}{{y.\left( {y + x} \right)}} = \dfrac{x}{y}\\
Vậy\,\dfrac{{{x^2} + {z^2}}}{{{y^2} + {z^2}}} = \dfrac{x}{y}
\end{array}$