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МатематикаГДЗ по алгебре 9 класс Мерзляк Проверь себя №3
Проверь себя №3
\[\boxed{\mathbf{Задание}\mathbf{\ }\mathbf{№}\mathbf{3.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[\mathbf{№1.}\]
\[x^{2} > 4\]
\[x = \pm 2\]
\[\mathbf{Ответ:В).}\]
\[\mathbf{№2.}\]
\[x^{2} + 8x - 9 \geq 0\]
\[x_{1} + x_{2} = - 8,\ \ x_{1} = - 9\]
\[x_{1} \cdot x_{2} = - 9,\ \ x_{2} = 1\]
\[\mathbf{Ответ:Б).}\]
\[\mathbf{№3.}\]
\[3x^{2} + 5x - 8 < 0\]
\[D = 25 + 96 = 121\]
\[x_{1,2} = \frac{- 5 \pm 11}{6}\]
\[x = - \frac{8}{3} = - 2\frac{2}{3};\ \ \ \ x = 1\]
\[x = - 2;\ - 1;0.\]
\[Ответ:А).\]
\[\mathbf{№4.}\]
\[А)\ x^{2} - 14x + 49 > 0\]
\[(x - 7)^{2} > 0\]
\[x \neq 7\]
\[Б) - 3x^{2} + x + 2 \leq 0\]
\[D = 1 + 4 \cdot 3 \cdot 2 =\]
\[= 25 > 0 \Longrightarrow 2\ корня.\]
\[В)\ x^{2} - 3x + 4 > 0\]
\[D = 9 - 4 \cdot 4 < 0 \Longrightarrow нет\]
\[\ корней.\]
\[Так\ как\ a = 1 > 0,\ ветви\ \]
\[направлены\ вверх.\]
\[x - любое\ число.\]
\[Г) - x^{2} + 7x - 10 < 0\]
\[D = 49 - 40 =\]
\[= 9 > 0 \Longrightarrow 2\ корня.\]
\[\mathbf{Ответ:В).}\]
\[\mathbf{№5.}\]
\[f(x) = \frac{5}{\sqrt{8x - 4x^{2}}}\]
\[8x - 4x^{2} > 0\]
\[4x(2 - x) > 0\ \]
\[\ x = 0,\ \ x = 2\]
\[\mathbf{Ответ:Г).}\]
\[\mathbf{№6.}\]
\[А)\ x^{2} - 6x + 10 < 0\]
\[D = 9 - 10 = - 1 < 0 \Longrightarrow нет\]
\[\ корней.\]
\[Б) - 5x^{2} + 3x + 2 > 0\]
\[D = 9 + 40 = 49 > 0 \Longrightarrow 2\]
\[\ корня.\]
\[В) - 3x^{2} + 8x + 3 < 0\]
\[D = 64 + 36 = 100 > 0 \Longrightarrow 2\ \]
\[корня.\]
\[Г) - x^{2} - 10x > 0\]
\[- x(x + 10) > 0\]
\[2\ корня.\]
\[\mathbf{Ответ:А).}\]
\[\mathbf{№7.}\]
\[\left\{ \begin{matrix} y - x = 2\ \ \ \\ xy - y = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = 2 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x(2 + x) - 2 - x - 10 = 0 \\ \end{matrix} \right.\ \]
\[2x + x² - 2 - x - 10 = 0\]
\[x² + x - 12 = 0\]
\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 4\]
\[x_{1} \cdot x_{2} = - 12,\ \ x_{2} = 3\]
\[\left\{ \begin{matrix} x = - 4 \\ y = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = 3 \\ y = 5 \\ \end{matrix} \right.\ \]
\[- 4 \cdot ( - 2) + 3 \cdot 5 = 8 + 15 = 23\]
\[Ответ:А).\]
\[\mathbf{№8.}\]
\[x^{2} + y^{2} = 5 \Longrightarrow окружность.\]
\[xy = - 3 \Longrightarrow y =\]
\[= - \frac{3}{x} \Longrightarrow гипербола.\]
\[\mathbf{Ответ:\ \ В).}\]
\[\mathbf{№9.}\]
\[\mathbf{Ответ:В).}\]
\[\mathbf{№10.}\]
\[\left\{ \begin{matrix} x^{2} - y = 4 \\ x + y = 1\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x^{2} - y = 4 \\ y = 1 - x\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x^{2} - 1 + x - 4 = 0 \\ y = 1 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x² + x - 5 = 0\]
\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 3\]
\[x_{1} \cdot x_{2} = - 5,\ \ x_{2} = 2\]
\[\left\{ \begin{matrix} x = - 3 \\ y = 4\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = 2\ \ \\ y = - 1 \\ \end{matrix} \right.\ \]
\[Проще\ решить\ графически:\]
\[y = x^{2} - 4;\ \ \ \ \ \ y = 1 - x\]
\[Ответ:В).\]
\[\mathbf{№11.}\]
\[\left\{ \begin{matrix} x - y = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2xy - y^{2} = - 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = 5 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (5 + y)^{2} + 2y(5 + y) - y^{2} + 7 = 0 \\ \end{matrix} \right.\ \]
\[25 + 10y + y^{2} + 10y + 2y^{2} -\]
\[- y^{2} + 7 = 0\]
\[2y² + 20y + 32 = 0\ \ \ \ |\ :2\]
\[y^{2} + 10y + 16 = 0\]
\[y_{1} + y_{2} = - 10,\ \ y_{1} = - 8\]
\[y_{1} \cdot y_{2} = 16,\ \ y_{2} = - 2\]
\[\left\{ \begin{matrix} y = - 8 \\ x = - 3 \\ \end{matrix} \right.\ \ \ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} y = - 2 \\ x = 3\ \ \ \\ \end{matrix} \right.\ \]
\[- 8 + ( - 3) = - 11;\ \ \]
\[- 2 + 3 = 1 \Longrightarrow максимальное\]
\[\ значение.\]
\[\mathbf{Ответ:А).}\]
\[\mathbf{№12.}\]
\[\left\{ \begin{matrix} \frac{2}{x} + \frac{1}{y} = 4 \\ \frac{1}{x} - \frac{3}{y} = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\frac{1}{x} = t;\ \ \ \frac{1}{y} = k\]
\[\left\{ \begin{matrix} 2t + k = 4\ \ \ | \cdot 3 \\ t - 3k = 9\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 6t + 3k = 12 \\ t - 3k = 9\ \ \ \ \ \\ \end{matrix} \right.\ \ ( + )\]
\[7t = 21\]
\[t = 3 \Longrightarrow \frac{1}{x} = 3 \Longrightarrow x = \frac{1}{3}\text{\ \ \ \ }\]
\[k = 4 - 2t = 4 - 2 \cdot 3 =\]
\[= - 2 \Longrightarrow \frac{1}{y} = - 2 \Longrightarrow y = - \frac{1}{2}\]
\[x - y = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}\]
\[Ответ:Г).\]
\[\mathbf{№13.}\]
\[\left\{ \begin{matrix} 2x - xy = 5 \\ y + xy = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x - xy = 5 \\ y = 6 - xy\ \ \\ \end{matrix} \right.\ \ \ \ | +\]
\[\left\{ \begin{matrix} 2x + y = 11 \\ y = 6 - xy\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} y = 11 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 11 - 2x - 6 + 11x - 2x^{2} = 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = 11 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \\ - 2x^{2} + 9x + 5 = 0 \\ \end{matrix} \right.\ \]
\[- 2x^{2} + 9x + 5 = 0\]
\[D = 81 + 40 = 121\]
\[x_{1,2} = \frac{- 9 \pm 11}{- 4}\]
\[\left\{ \begin{matrix} x = 5 \\ y = 1 \\ \end{matrix} \right.\ \ \ \ \ \ \ или\ \ \ \ \ \left\{ \begin{matrix} x = - 0,5 \\ y = 12\ \ \ \ \\ \end{matrix} \right.\ \]
\[|5 \cdot 1 + 0,5 \cdot 12| = 11\]
\[Ответ:Б).\]
\[\mathbf{№14.}\]
\[3x² - bx + 3 = 0\]
\[D = b^{2} - 36\]
\[Не\ имеет\ корней\ при\ D < 0:\]
\[b^{2} - 36 < 0\]
\[(b - 6)(b + 6) < 0\]
\[- 6 < b < 6\]
\[Ответ:А).\]
\[\mathbf{№15.}\]
\[\left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ x - y = a\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ x = a + y\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[(a + y)^{2} + y^{2} - 25 = 0\]
\[a^{2} + 2ay + y^{2} + y^{2} - 25 = 0\]
\[2y^{2} + 2ay + a^{2} - 25 = 0\]
\[D = 4a^{2} - 8a^{2} + 200 =\]
\[= - {4a}^{2} + 200\]
\[Имеет\ единственное\ решение,\ \]
\[если\ D = 0:\]
\[- 4a^{2} + 200 = 0\]
\[a^{2} = 50\]
\[a = \pm 5\sqrt{2}\]
\[\mathbf{Ответ:Г).}\]
\[\mathbf{№16.}\]
\[ax^{2} - 4x + a \geq 0\]
\[D = 16 - 4a²\]
\[a^{2} = 4\]
\[a = \pm 2\]
\[При\ a = 2:\ \ 2x^{2} - 4x +\]
\[+ 2 \geq 0 \Longrightarrow x - любое\ число;\]
\[при\ a = - 2 \Longrightarrow - 2x^{2} - 4x -\]
\[- 2 \geq 0 \Longrightarrow x = - 1\]
\[Ответ:В).\]
\[\mathbf{№17.}\]
\[ax² - 2x + a < 0\]
\[D = 4 - 4a^{2}\]
\[Неравенство\ не\ имеет\]
\[\ решения,\ если\ a > 0;\ \ D \leq 0:\]
\[4 - 4a^{2} < 0\]
\[4a^{2} > 4\]
\[a^{2} > 1;\ \ a = \pm 1\]
\[\mathbf{Ответ:\ \ Б).}\]
\[\mathbf{№18.}\]
\[\left\{ \begin{matrix} 2x - y = a \\ y = x^{2} - 8 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\ \left\{ \begin{matrix} 2x - x^{2} + 8 - a = 0 \\ y = x^{2} - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[- x^{2} + 2x + 8 - a = 0\]
\[D = 4 + 4 \cdot (8 - a) = 4 + 32 -\]
\[- 4a = 36 - 4a\]
\[36 - 4a = 0\]
\[a = 9\]
\[\mathbf{Ответ:Б).}\]
