Задача 23. Решите систему уравнений \(\left\{ {\,\begin{array}{*{20}{c}} {\dfrac{{x + y}}{{x\,y\,z}} = \dfrac{7}{{12}}\,} \\ {\dfrac{{y + z}}{{x\,y\,z}} = \dfrac{5}{{12}}\,} \\ {\dfrac{{z + x}}{{x\,y\,z}} = \dfrac{1}{3}\,\,\,\,} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {\,3;\,4;\,1} \right),\) \(\left( {\,-3;\,-4;\,-1} \right)\).
Решение
\(\left\{ {\begin{array}{*{20}{c}}{\dfrac{{x + y}}{{xyz}} = \dfrac{7}{{12}},}\\{\dfrac{{y + z}}{{xyz}} = \dfrac{5}{{12}},}\\{\dfrac{{z + x}}{{xyz}} = \dfrac{1}{3}\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{1}{{yz}} + \dfrac{1}{{xz}} = \dfrac{7}{{12}},}\\{\dfrac{1}{{xz}} + \dfrac{1}{{xy}} = \dfrac{5}{{12}},}\\{\dfrac{1}{{xy}} + \dfrac{1}{{yz}} = \dfrac{1}{3}.\,\,}\end{array}} \right.\)
Пусть \(\dfrac{1}{{yz}} = a,\,\,\,\dfrac{1}{{xz}} = b,\,\,\,\dfrac{1}{{xy}} = c.\) Тогда:
\(\left\{ {\begin{array}{*{20}{c}}{a + b = \dfrac{7}{{12}},}\\{b + c = \dfrac{5}{{12}},}\\{c + a = \dfrac{1}{3}\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a + b = \dfrac{7}{{12}},}\\{a-c = \dfrac{1}{6},\,\,\,}\\{c + a = \dfrac{1}{3}\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a + b = \dfrac{7}{{12}},}\\{a-c = \dfrac{1}{6},\,\,\,}\\{2a = \dfrac{1}{2}\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a = \dfrac{1}{4},}\\{b = \dfrac{1}{3},}\\{c = \dfrac{1}{{12}}.}\end{array}} \right.} \right.\)
Вернёмся к прежним переменным:
\(\left\{ {\begin{array}{*{20}{c}}{\dfrac{1}{{yz}} = \dfrac{1}{4},}\\{\dfrac{1}{{xz}} = \dfrac{1}{3},}\\{\dfrac{1}{{xy}} = \dfrac{1}{{12}}}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{yz = 4,}\\{xz = 3,}\\{xy = 12}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{yz = 4,}\\{\dfrac{y}{x} = \dfrac{4}{3},}\\{xy = 12}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,} \right.\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{yz = 4,\,\,\,\,\,\,}\\{y = \dfrac{4}{3}x,\,\,\,}\\{\dfrac{4}{3}{x^2} = 12}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{yz = 4,\,\,\,}\\{y = \dfrac{4}{3}x,}\\{\left[ {\begin{array}{*{20}{c}}{x = 3,\,}\\{x = -3}\end{array}\,} \right.}\end{array}\,\,} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 3,}\\{y = 4,}\\{z = 1,\,\,}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -3,}\\{y = -4,}\\{z = -1.}\end{array}} \right.}\end{array}} \right.\)
Ответ: \(\left( {3;4;1} \right),\,\,\,\left( {-3;-4;-1} \right).\)