\[\boxed{\text{596}\text{\ (596)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[a^{2} = x;\ \ b^{2} = x + d;\ \ \]
\[c² = x + 2d;\]
\[\frac{1}{a + c} =\]
\[= \frac{1}{b + c} + \left( \frac{1}{a + c} - \frac{1}{b + c} \right) =\]
\[= \frac{1}{b + c} + \frac{b - a}{(a + c)(b + c)} =\]
\[= \frac{1}{b + c} + \frac{b^{2} - a^{2}}{(a + c)(b + c)(a + b)} =\]
\[= \left( \frac{1}{b + c} \right) + \frac{d}{(a + c)(b + c)(a + b)};\]
\[\frac{1}{a + b} =\]
\[= \frac{1}{b + c} + \left( \frac{1}{a + b} - \frac{1}{b + c} \right) =\]
\[= \frac{1}{b + c} + \frac{b + c - a - b}{(a + b)(b + c)} =\]
\[= \frac{1}{b + c} + \frac{c - a}{(a + b)(b + c)} =\]
\[= \frac{1}{b + c} + \frac{c^{2} - a^{2}}{(a + c)(b + c)(a + b)} =\]
\[= \frac{1}{b + c} + \frac{2d}{(a + c)(b + c)(a + b)}.\]
\[Что\ и\ требовалось\ доказать.\]