При каких значениях a система уравнений 3x-2y=7; x+y=4; 2x-y=a имеет решение?

Ответ:

\[\left\{ \begin{matrix} 3x - 2y = 7 \\ x + y = 4\ \ \ \ \\ 2x - y = a\ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 \cdot (4 - y) - 2y = 7 \\ 2 \cdot (4 - y) - y = a\ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 12 - 3y - 2y = 7 \\ 8 - 2y - y = a\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} - 5y = - 5 \\ a = 8 - 3y \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 1 \\ a = 5 \\ \end{matrix} \right.\ \]

\[Ответ:при\ a = 5.\]

\[\left\{ \begin{matrix} x + y = 5 \\ xy = 6\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = 5 - x\ \ \ \ \ \ \\ x(5 - x) = 6 \\ \end{matrix}\text{\ \ } \right.\ \]

\[5x - x^{2} = 6\]

\[x^{2} - 5x + 6 = 0\]

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 2 \rightarrow y = 3;\]

\[x_{2} = 3 \rightarrow y = 2.\]

\[Ответ:(2;3);(3;2).\]

\[\ x^{2} - 4y = 5 \rightarrow 4y = x^{2} - 5 \rightarrow\]

\[\rightarrow y = \frac{1}{4}x^{2} - \frac{5}{4} \rightarrow парабола.\]

\[x + y = 4 \rightarrow y = 4 - x \rightarrow прямая.\]

\[\frac{1}{4}x^{2} - \frac{5}{4} = 4 - x\ \ \ \ | \cdot 4\]

\[x^{2} - 5 = 16 - 4x\]

\[x^{2} + 4x - 21 = 0\]

\[x_{1} + x_{2} = - 4;\ \ x_{1} \cdot x_{2} = - 21\]

\[x_{1} = - 7 \rightarrow y_{1} = 11;\]

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

\[Ответ:( - 7;11);(3;1).\]

\[P = 28\ см;d = 10\ см.\]

\[a + b = 28\ :2 = 14\ см.\]

\[\left\{ \begin{matrix} a + b = 14\ \ \ \ \ \ \ \\ a^{2} + b^{2} = 100 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a = 14 - b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (14 - b)^{2} + b^{2} = 100 \\ \end{matrix} \right.\ \]

\[196 - 28b + b^{2} + b^{2} - 100 = 0\]

\[2b^{2} - 28b + 96 = 0\ \ \ \ |\ :2\]

\[b^{2} - 14b + 48 = 0\]

\[D_{1} = 49 - 48 = 1\]

\[b_{1} = 7 + 1 = 8 \rightarrow a_{1} = 6;\]

\[b_{2} = 7 - 1 = 6 \rightarrow a_{2} = 8.\]

\[Ответ:стороны\ равны\ 6\ см\ и\ 8\ см.\]

\[\frac{1}{x} = |x|\]

\[y = \frac{1}{x};\ \ y = |x|.\]

\[Ответ:уравнение\ имеет\ 1\ корень.\]

\[\left\{ \begin{matrix} (x - 3)(y + 1) = 0 \\ x^{2} - xy - 12 = 0\ \ \\ \end{matrix} \right.\ \]

\[x - 3 = 0\]

\[x = 3:\]

\[3^{2} - 3y - 12 = 0\]

\[9 - 3y - 12 = 0\]

\[- 3y = 3\]

\[y = - 1.\]

\[y + 1 = 0\]

\[y = - 1:\]

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

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

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

\[Ответ:(3;\ - 1);( - 4;\ - 1).\]

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

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

\[имеет\ вид:\]

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

\[Точка\ ( - 3;3):\]

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

\[9a = 3\]

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

\[Уравнение\ параболы:\]

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

\[При\ y = 27:\]

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

\[x^{2} = 27 \cdot 3 = 81\]

\[x = \pm 9.\]

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

\[в\ точках\ (9;27)\ и\ ( - 9;27).\]