ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1425

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\[1)\ \left\{ \begin{matrix} \frac{x}{y} - \frac{y}{x} = \frac{3}{2}\text{\ \ \ \ \ \ \ } \\ x^{2} + y^{2} = 20 \\ \end{matrix} \right.\ \]

\[Пусть\ z = \frac{x}{y},\ тогда:\]

\[z - \frac{1}{z} = \frac{3}{2}\ \ \ \ \ | \bullet 2z\]

\[2z^{2} - 2 = 3z\]

\[2z^{2} - 3z - 2 = 0\]

\[D = 9 + 16 = 25\]

\[z_{1} = \frac{3 - 5}{2 \bullet 2} = - \frac{1}{2};\]

\[z_{2} = \frac{3 + 5}{2 \bullet 2} = 2;\]

\[1)\ \left\{ \begin{matrix} \frac{x}{y} = - \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} + y^{2} = 20 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} x = - 0,5y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} - 20 = 0 \\ \end{matrix} \right.\ \]

\[( - 0,5y)^{2} + y^{2} - 20 = 0\]

\[0,25y^{2} + y^{2} = 20\]

\[1,25y^{2} = 20\]

\[y^{2} = 16\]

\[y = \pm 4;\]

\[x_{1} = - 0,5 \bullet ( - 4) = 2;\]

\[x_{2} = - 0,5 \bullet 4 = - 2.\]

\[2)\ \left\{ \begin{matrix} \frac{x}{y} = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} = 20 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} - 20 = 0 \\ \end{matrix} \right.\ \]

\[(2y)^{2} + y^{2} - 20 = 0\]

\[4y^{2} + y^{2} = 20\]

\[5y^{2} = 20\]

\[y^{2} = 4\]

\[y = \pm 2;\]

\[x_{1} = 2 \bullet ( - 2) = - 4;\]

\[x_{2} = 2 \bullet 2 = 4.\]

\[Ответ:\ \ (2;\ - 4);\ \ ( - 2;\ 4);\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }( - 4;\ - 2);\ \ (4;\ 2)\text{.\ }\]

\[2)\ \left\{ \begin{matrix} \frac{y}{x} + \frac{x}{y} = 3\frac{1}{3} \\ x^{2} - y^{2} = 8 \\ \end{matrix} \right.\ \]

\[z = \frac{x}{y}:\]

\[z + \frac{1}{z} = 3\frac{1}{3}\ \ \ \ \ | \bullet 3z\]

\[3z^{2} + 3 = (9 + 1)z\]

\[3z^{2} - 10z + 3 = 0\]

\[D = 100 - 36 = 64\]

\[z_{1} = \frac{10 - 8}{2 \bullet 3} = \frac{1}{3};\]

\[z_{2} = \frac{10 + 8}{2 \bullet 3} = 3;\]

\[1)\ \left\{ \begin{matrix} \frac{x}{y} = \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} - y^{2} = 8 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{1}{3}\text{y\ \ \ \ \ \ \ \ \ } \\ x^{2} - y^{2} = 8 \\ \end{matrix} \right.\ \]

\[\left( \frac{1}{3}y \right)^{2} - y^{2} = 8\]

\[\frac{1}{9}y^{2} - y^{2} = 8\]

\[- \frac{8}{9}y^{2} = 8\]

\[y^{2} = - 9\]

\[корней\ нет.\]

\[2)\ \left\{ \begin{matrix} \frac{x}{y} = 3\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - y^{2} = 8 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \\ x^{2} - y^{2} = 8 \\ \end{matrix} \right.\ \]

\[(3y)^{2} - y^{2} = 8\]

\[9y^{2} - y^{2} = 8\]

\[8y^{2} = 8\]

\[y^{2} = 1\]

\[y = \pm 1.\]

\[x_{1} = 3 \bullet ( - 1) = - 3;\]

\[x_{2} = 3 \bullet 1 = 3.\]

\[Ответ:\ \ ( - 3;\ - 1);\ \ (3;\ 1)\text{.\ }\]

\[3)\ \left\{ \begin{matrix} x^{2} = 13x + 4y \\ y^{2} = 4x + 13y \\ \end{matrix} \right.\ \ ( - )\]

\[x^{2} - y^{2} = 13x - 4x + 4y - 13y\]

\[x^{2} - y^{2} = 9x - 9y\]

\[(x - y)(x + y) = 9(x - y)\]

\[(x - y) \bullet (x + y - 9) = 0\]

\[1)\ \left\{ \begin{matrix} x - y = 0\ \ \ \ \ \ \ \ \ \ \\ y^{2} = 4x + 13y \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 13y - 4x = 0 \\ \end{matrix} \right.\ \]

\[y^{2} - 13y - 4y = 0\]

\[y^{2} - 17y = 0\]

\[y \bullet (y - 17) = 0\]

\[y_{1} = 0\ \ и\ \ y_{2} = 17;\]

\[x_{1} = 0\ \ и\ \ x_{2} = 17.\]

\[2)\ \left\{ \begin{matrix} x + y - 9 = 0\ \ \\ y^{2} = 4x + 13y \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} x = 9 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 13y - 4x = 0 \\ \end{matrix} \right.\ \]

\[y^{2} - 13y - 4(9 - y) = 0\]

\[y^{2} - 13y - 36 + 4y = 0\]

\[y^{2} - 9y - 36 = 0\]

\[D = 81 + 144 = 225\]

\[y_{1} = \frac{9 - 15}{2} = - 3;\]

\[y_{2} = \frac{9 + 15}{2} = 12;\]

\[x_{1} = 9 + 3 = 12;\]

\[x_{2} = 9 - 12 = - 3.\]

\[Ответ:\ \ (0;\ 0);\ \ (17;\ 17);\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ }(12;\ - 3);\ \ ( - 3;\ 12).\]

\[4)\ \left\{ \begin{matrix} 3x^{2} + y^{2} - 4x = 40 \\ 2x^{2} + y^{2} + 3x = 52 \\ \end{matrix} \right.\ \ ( - )\ \ \ \]

\[\left\{ \begin{matrix} y = \sqrt{40 - 3x^{2} + 4x} \\ 2x^{2} + y^{2} + 3x = 52\ \\ \end{matrix} \right.\ ;\]

\[3x^{2} - 2x^{2} + y^{2} - y^{2} - 4x - 3x = 40 - 52\]

\[x^{2} - 7x = - 12\]

\[x^{2} - 7x + 12 = 0\]

\[D = 49 - 48 = 1\]

\[x_{1} = \frac{7 - 1}{2} = 3;\ \]

\[x_{2} = \frac{7 + 1}{2} = 4;\]

\[y_{1} = \sqrt{40 - 3 \bullet 3^{2} + 4 \bullet 3} =\]

\[= \sqrt{40 - 27 + 12} = \sqrt{25} = \pm 5;\]

\[y_{2} = \sqrt{40 - 3 \bullet 4^{2} + 4 \bullet 4} =\]

\[= \sqrt{40 - 48 + 16} = \sqrt{8} = \pm 2\sqrt{2}.\]

\[Ответ:\ \ (3;\ - 5);\ \ (3;\ 5);\ \ \]

\[\text{\ \ \ \ \ }\left( 4;\ - 2\sqrt{2} \right);\ \ \left( 4;\ 2\sqrt{2} \right).\]