ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 402

\[\boxed{\mathbf{402}.}\]

\[a = 60;\ \ b = 60;\ \ c = 51.\]

\[\left\{ \begin{matrix} x^{2} + y^{2} + z^{2} = 3600\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\ \\ (x - 12)^{2} + y^{2} + z^{2} = 3600\ \ \ \ \ \ \ \ \ \ \ (2)\ \\ (x - 6)^{2} + (y - 18)^{2} + z^{2} = 2601\ (3) \\ \end{matrix} \right.\ \]

\[(1) - (2):\]

\[x^{2} - (x - 12)^{2} = 0\]

\[x^{2} - x^{2} + 24x - 144 = 0\]

\[24x = 144\]

\[x = 6.\]

\[\left\{ \begin{matrix} 36 + y^{2} + z^{2} = 3600\ \ \ \\ (y - 18)^{2} + z^{2} = 2601 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y^{2} + z^{2} = 3564\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 36y + 324 + z^{2} = 2601 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} z^{2} = 3564 - y^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y^{2} - 36y + 3564 - y^{2} = 2277 \\ \end{matrix} \right.\ \]

\[- 36y = - 1287\]

\[y = 35,75.\]

\[z^{2} = 3564 - 1278 = 2286\]

\[z \approx 47,8.\]

\[Ответ:M(6;35,75;47,8).\]