Задача 22. Решите систему уравнений \(\left\{ {\,\begin{array}{*{20}{c}} {x + y + z = 4\,\,\,\,\,\,\,\,\,} \\ {x + 2y + 3z = 5\,\,\,\,} \\ {{x^2} + {y^2} + {z^2} = 14} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {\,\dfrac{{11}}{3};\,-\dfrac{1}{3};\,\dfrac{2}{3}} \right),\;\left( {\,2;\,3;\,-1} \right)\).
Решение
Вычтем из второго уравнения первое:
\(y + 2z = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,y = 1-2z.\)
Тогда первое уравнение примет вид:
\(x + y + z = 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x + 1-2z + z = 4\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = z + 3.\)
\(\left\{ {\begin{array}{*{20}{c}}{y = 1-2z,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = z + 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + {y^2} + {z^2} = 14}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 1-2z,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = z + 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {z + 3} \right)}^2} + {{\left( {1-2z} \right)}^2} + {z^2} = 14}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 1-2z,\,\,\,\,\,\,\,\,}\\{x = z + 3,\,\,\,\,\,\,\,\,\,\,\,}\\{3{z^2} + z-2 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 1-2z,}\\{x = z + 3,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{z = -1,}\\{z = \dfrac{2}{3}\,\,}\end{array}\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 2,\,\,}\\{y = 3,\,\,}\\{z = -1,}\end{array}\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{11}}{3},\,\,}\\{y = -\dfrac{1}{3},}\\{z = \dfrac{2}{3}.\,\,\,}\end{array}} \right.}\end{array}} \right.\)
Ответ: \(\left( {2;3;-1} \right),\,\,\,\left( {\dfrac{{11}}{3};-\dfrac{1}{3};\dfrac{2}{3}} \right).\)