Пусть \(\sqrt {x + y} = a,\,\,\,\sqrt {y + z} = b,\,\,\,\sqrt {z + x} = c.\) Тогда:
\(\left\{ {\begin{array}{*{20}{c}}{a + b = 3,}\\{b + c = 5,}\\{c + a = 4\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a = 1,}\\{b = 2,}\\{c = 3.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left\{ {\begin{array}{*{20}{c}}{\sqrt {x + y} = 1,}\\{\sqrt {y + z} = 2,}\\{\sqrt {z + x} = 3\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x + y = 1,}\\{y + z = 4,}\\{z + x = 9\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = 3,\,\,\,}\\{y = -2,}\\{z = 6.\,\,\,}\end{array}} \right.} \right.\)
Ответ: \(\left( {3;-2;6} \right).\)