При каких значениях c система уравнений 2x+3y=4; x-y=-3; x+2y=c имеет решение?

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

\[\left\{ \begin{matrix} 2y - 6 + 3y = 4 \\ y - 3 + 2y = c\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} 5y = 10\ \ \ \ \ \\ c = 3y - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 \\ c = 3 \\ \end{matrix} \right.\ \]

\[Ответ:при\ c = 3.\]

\[\left\{ \begin{matrix} xy = 8\ \ \ \ \ \\ x + y = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = 6 - y\ \ \ \ \ \ \ \ \ \ \\ (6 - y) \cdot y = 8 \\ \end{matrix} \right.\ \]

\[6y - y^{2} = 8\]

\[y^{2} - 6y + 8 = 0\]

\[D_{1} = 9 - 8 = 1\]

\[y_{1} = 3 + 1 = 4 \rightarrow x_{1} = 2;\]

\[y_{2} = 3 - 1 = 2 \rightarrow x_{2} = 4.\]

\[Ответ:(2;4);(4;2).\]

\[x^{2} - y = - 1 \rightarrow y = x^{2} + 1 \rightarrow парабола.\]

\[x + y = 1 \rightarrow y = 1 - x \rightarrow прямая.\]

\[x^{2} + 1 = 1 - x\]

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

\[x(x + 1) = 0\]

\[x_{1} = 0 \rightarrow y = 1;\]

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

\[Ответ:(0;1);( - 1;2).\]

\[P = 34\ см;\ \ d = 13\ см.\]

\[a + b = 34\ :2 = 17\ см.\]

\[\left\{ \begin{matrix} a + b = 17\ \ \ \ \ \ \\ a^{2} + b^{2} = 169 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 17 - b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (17 - b)^{2} + b^{2} = 169 \\ \end{matrix} \right.\ \]

\[289 - 34b + b^{2} + b^{2} = 169\]

\[2b^{2} - 34b + 120 = 0\ \ \ |\ :2\]

\[b^{2} - 17b + 60 = 0\]

\[b_{1} + b_{2} = 17;\ \ b_{1} \cdot b_{2} = 60\]

\[b_{1} = 12 \rightarrow a_{1} = 5;\text{\ \ }\]

\[b_{2} = 5 \rightarrow a_{2} = 12.\]

\[Ответ:длины\ сторон\ 5\ см\ и\ 12\ см.\]

\[0,5x^{2} = \frac{1}{x}\]

\[y = 0,5x^{2};\ \ y = \frac{1}{x}.\]

\[Ответ:уравнение\ имеет\ один\ корень.\]

\[\left\{ \begin{matrix} (x + 2)(y - 1) = 0 \\ x^{2} - xy - 12 = 0\ \ \\ \end{matrix} \right.\ \]

\[x + 2 = 0\]

\[x = - 2:\]

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

\[2y - 8 = 0\]

\[2y = 8\]

\[y = 4.\]

\[y - 1 = 0\]

\[y = 1:\]

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

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

\[x_{1} = 4;\ \ \ \ x_{2} = 3.\]

\[Ответ:( - 2;4);(4;1);(3;1).\]

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

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

\[имеет\ вид:\]

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

\[Точка\ (3;\ - 3):\]

\[- 3 = a \cdot 3^{2}\]

\[9a = - 3\]

\[a = - \frac{1}{3}.\]

\[Уравнение:\]

\[y = - \frac{1}{3}x^{2}.\]

\[При\ y = - 27:\]

\[- \frac{1}{3}x^{2} = - 27\]

\[x^{2} = 27 \cdot 3\]

\[x^{2} = 81\]

\[x = \pm 9.\]

\[Парабола\ пересекает\ прямую\ y = - 27\ \]

\[в\ точках\ ( - 9;\ - 27);(9;\ - 27).\]