При каких значениях m система уравнений 2x-3y=8; x+y=-1; 3x-y=m имеет решение?

Ответ:

\[\left\{ \begin{matrix} 2x - 3y = 8 \\ x + y = - 1\ \ \\ 3x - y = m\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = - x - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x - 3 \cdot ( - x - 1) = 8 \\ 3x - ( - x - 1) = m\ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2x + 3x + 3 = 8 \\ 3x + x + 1 = m\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} 5x = 5\ \ \ \ \ \ \ \ \ \ \\ m = 4x + 1 \\ \end{matrix}\ \right.\ \]

\[\left\{ \begin{matrix} x = 1 \\ m = 5 \\ \end{matrix} \right.\ \]

\[Ответ:при\ m = 5.\]

\[\left\{ \begin{matrix} x - y = 7 \\ xy = - 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 7 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (7 + y) \cdot y = - 12 \\ \end{matrix} \right.\ \]

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

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

\[y_{1} + y_{2} = - 7;\ \ \ y_{1} \cdot y_{2} = 12\]

\[y_{1} = - 4 \rightarrow x_{1} = 3;\ \]

\[y_{2} = - 3 \rightarrow x_{2} = 4.\]

\[Ответ:(3;\ - 4);(4;\ - 3).\]

\[2x + y = 2 \rightarrow y = 2 - 2x \rightarrow прямая.\]

\[x^{2} - y = - 2 \rightarrow y = x^{2} + 2 \rightarrow парабола.\]

\[2 - 2x = x^{2} + 2\]

\[x^{2} + 2x = 0\]

\[x(x + 2) = 0\]

\[x_{1} = 0 \rightarrow y_{1} = 2;\]

\[x_{2} = - 2 \rightarrow y_{2} = 6.\]

\[Ответ:(0;2);( - 2;6).\]

\[c = 29\ см;a = b + 1\ (см).\]

\[\left\{ \begin{matrix} a = b + 1\ \ \ \ \ \ \ \\ a^{2} + b^{2} = c^{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = b + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (b + 1)^{2} + b^{2} = 841 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a = b + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (b + 1)^{2} + b^{2} = 841 \\ \end{matrix} \right.\ \]

\[b^{2} + 2b + 1 + b^{2} - 841 = 0\]

\[2b^{2} + 2b - 840 = 0\ \ \ \ |\ :2\]

\[b^{2} + b - 420 = 0\]

\[b_{1} + b_{2} = - 1;\ \ b_{1} \cdot b_{2} = - 420\]

\[b_{1} = - 21\ (не\ подходит).\]

\[b_{2} = 20\ (см) - один\ катет.\]

\[a = 20 + 1 = 21\ (см) - другой\ катет.\]

\[Ответ:21\ см\ и\ 20\ см.\]

\[x^{3} = \frac{4}{x}\]

\[y = x^{3};\ \ y = \frac{4}{x}.\]

\[Ответ:уравнение\ имеет\ 2\ корня.\]

\[\left\{ \begin{matrix} (x + 1)(y - 2) = 0 \\ y^{2} - xy - 6 = 0\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x + 1 = 0\]

\[x = - 1:\]

\[y^{2} + 1 - 6 = 0\]

\[y_{1} + y_{2} = - 1;y_{1} \cdot y_{2} = - 6\]

\[y_{1} = - 3;\ \ y_{2} = 2.\]

\[y - 2 = 0\]

\[y = 2:\]

\[4 - 2x - 6 = 0\]

\[- 2x = 2\]

\[x = - 1.\]

\[Ответ:( - 1;\ - 3);( - 1;2).\]

\[Так\ как\ парабола\ проходит\ через\ \]

\[начало\ координат,\ то\ уравнение\ \]

\[имеет\ вид:\]

\[y = ax^{2}.\]

\[Точка\ (2; - 1):\]

\[- 1 = a \cdot 2^{2}\]

\[a = - \frac{1}{4}\text{.\ }\]

\[Уравнение\ параболы:\]

\[y = - \frac{1}{4}x^{2}.\]

\[При\ y = - 9:\]

\[- \frac{1}{4}x^{2} = - 9\]

\[x^{2} = 36\]

\[x = \pm 6.\]

\[Парабола\ пересекает\ прямую\ y = - 9\]

\[в\ точках\ ( - 6;\ - 9);(6;\ - 9).\]