Решебник по алгебре 9 класс Мерзляк Проверь себя №3

 

\[\boxed{\mathbf{Задание}\mathbf{\ }\mathbf{№}\mathbf{3.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\mathbf{№1.}\]

\[x^{2} > 4\]

\[x = \pm 2\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№2.}\]

\[x^{2} + 8x - 9 \geq 0\]

\[x_{1} + x_{2} = - 8,\ \ x_{1} = - 9\]

\[x_{1} \cdot x_{2} = - 9,\ \ x_{2} = 1\]

\[\mathbf{Ответ:Б).}\]

\[\mathbf{№3.}\]

\[3x^{2} + 5x - 8 < 0\]

\[D = 25 + 96 = 121\]

\[x_{1,2} = \frac{- 5 \pm 11}{6}\]

\[x = - \frac{8}{3} = - 2\frac{2}{3};\ \ \ \ x = 1\]

\[x = - 2;\ - 1;0.\]

\[Ответ:А).\]

\[\mathbf{№4.}\]

\[А)\ x^{2} - 14x + 49 > 0\]

\[(x - 7)^{2} > 0\]

\[x \neq 7\]

\[Б) - 3x^{2} + x + 2 \leq 0\]

\[D = 1 + 4 \cdot 3 \cdot 2 =\]

\[= 25 > 0 \Longrightarrow 2\ корня.\]

\[В)\ x^{2} - 3x + 4 > 0\]

\[D = 9 - 4 \cdot 4 < 0 \Longrightarrow нет\]

\[\ корней.\]

\[Так\ как\ a = 1 > 0,\ ветви\ \]

\[направлены\ вверх.\]

\[x - любое\ число.\]

\[Г) - x^{2} + 7x - 10 < 0\]

\[D = 49 - 40 =\]

\[= 9 > 0 \Longrightarrow 2\ корня.\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№5.}\]

\[f(x) = \frac{5}{\sqrt{8x - 4x^{2}}}\]

\[8x - 4x^{2} > 0\]

\[4x(2 - x) > 0\ \]

\[\ x = 0,\ \ x = 2\]

\[\mathbf{Ответ:Г).}\]

\[\mathbf{№6.}\]

\[А)\ x^{2} - 6x + 10 < 0\]

\[D = 9 - 10 = - 1 < 0 \Longrightarrow нет\]

\[\ корней.\]

\[Б) - 5x^{2} + 3x + 2 > 0\]

\[D = 9 + 40 = 49 > 0 \Longrightarrow 2\]

\[\ корня.\]

\[В) - 3x^{2} + 8x + 3 < 0\]

\[D = 64 + 36 = 100 > 0 \Longrightarrow 2\ \]

\[корня.\]

\[Г) - x^{2} - 10x > 0\]

\[- x(x + 10) > 0\]

\[2\ корня.\]

\[\mathbf{Ответ:А).}\]

\[\mathbf{№7.}\]

\[\left\{ \begin{matrix} y - x = 2\ \ \ \\ xy - y = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x(2 + x) - 2 - x - 10 = 0 \\ \end{matrix} \right.\ \]

\[2x + x² - 2 - x - 10 = 0\]

\[x² + x - 12 = 0\]

\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 4\]

\[x_{1} \cdot x_{2} = - 12,\ \ x_{2} = 3\]

\[\left\{ \begin{matrix} x = - 4 \\ y = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = 3 \\ y = 5 \\ \end{matrix} \right.\ \]

\[- 4 \cdot ( - 2) + 3 \cdot 5 = 8 + 15 = 23\]

\[Ответ:А).\]

\[\mathbf{№8.}\]

\[x^{2} + y^{2} = 5 \Longrightarrow окружность.\]

\[xy = - 3 \Longrightarrow y =\]

\[= - \frac{3}{x} \Longrightarrow гипербола.\]

\[\mathbf{Ответ:\ \ В).}\]

\[\mathbf{№9.}\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№10.}\]

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

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

\[x² + x - 5 = 0\]

\[x_{1} + x_{2} = - 1,\ \ x_{1} = - 3\]

\[x_{1} \cdot x_{2} = - 5,\ \ x_{2} = 2\]

\[\left\{ \begin{matrix} x = - 3 \\ y = 4\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = 2\ \ \\ y = - 1 \\ \end{matrix} \right.\ \]

\[Проще\ решить\ графически:\]

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

\[Ответ:В).\]

\[\mathbf{№11.}\]

\[\left\{ \begin{matrix} x - y = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2xy - y^{2} = - 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 5 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (5 + y)^{2} + 2y(5 + y) - y^{2} + 7 = 0 \\ \end{matrix} \right.\ \]

\[25 + 10y + y^{2} + 10y + 2y^{2} -\]

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

\[2y² + 20y + 32 = 0\ \ \ \ |\ :2\]

\[y^{2} + 10y + 16 = 0\]

\[y_{1} + y_{2} = - 10,\ \ y_{1} = - 8\]

\[y_{1} \cdot y_{2} = 16,\ \ y_{2} = - 2\]

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

\[- 8 + ( - 3) = - 11;\ \ \]

\[- 2 + 3 = 1 \Longrightarrow максимальное\]

\[\ значение.\]

\[\mathbf{Ответ:А).}\]

\[\mathbf{№12.}\]

\[\left\{ \begin{matrix} \frac{2}{x} + \frac{1}{y} = 4 \\ \frac{1}{x} - \frac{3}{y} = 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\frac{1}{x} = t;\ \ \ \frac{1}{y} = k\]

\[\left\{ \begin{matrix} 2t + k = 4\ \ \ | \cdot 3 \\ t - 3k = 9\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 6t + 3k = 12 \\ t - 3k = 9\ \ \ \ \ \\ \end{matrix} \right.\ \ ( + )\]

\[7t = 21\]

\[t = 3 \Longrightarrow \frac{1}{x} = 3 \Longrightarrow x = \frac{1}{3}\text{\ \ \ \ }\]

\[k = 4 - 2t = 4 - 2 \cdot 3 =\]

\[= - 2 \Longrightarrow \frac{1}{y} = - 2 \Longrightarrow y = - \frac{1}{2}\]

\[x - y = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}\]

\[Ответ:Г).\]

\[\mathbf{№13.}\]

\[\left\{ \begin{matrix} 2x - xy = 5 \\ y + xy = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x - xy = 5 \\ y = 6 - xy\ \ \\ \end{matrix} \right.\ \ \ \ | +\]

\[\left\{ \begin{matrix} 2x + y = 11 \\ y = 6 - xy\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} y = 11 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 11 - 2x - 6 + 11x - 2x^{2} = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 11 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \\ - 2x^{2} + 9x + 5 = 0 \\ \end{matrix} \right.\ \]

\[- 2x^{2} + 9x + 5 = 0\]

\[D = 81 + 40 = 121\]

\[x_{1,2} = \frac{- 9 \pm 11}{- 4}\]

\[\left\{ \begin{matrix} x = 5 \\ y = 1 \\ \end{matrix} \right.\ \ \ \ \ \ \ или\ \ \ \ \ \left\{ \begin{matrix} x = - 0,5 \\ y = 12\ \ \ \ \\ \end{matrix} \right.\ \]

\[|5 \cdot 1 + 0,5 \cdot 12| = 11\]

\[Ответ:Б).\]

\[\mathbf{№14.}\]

\[3x² - bx + 3 = 0\]

\[D = b^{2} - 36\]

\[Не\ имеет\ корней\ при\ D < 0:\]

\[b^{2} - 36 < 0\]

\[(b - 6)(b + 6) < 0\]

\[- 6 < b < 6\]

\[Ответ:А).\]

\[\mathbf{№15.}\]

\[\left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ x - y = a\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} + y^{2} = 25 \\ x = a + y\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(a + y)^{2} + y^{2} - 25 = 0\]

\[a^{2} + 2ay + y^{2} + y^{2} - 25 = 0\]

\[2y^{2} + 2ay + a^{2} - 25 = 0\]

\[D = 4a^{2} - 8a^{2} + 200 =\]

\[= - {4a}^{2} + 200\]

\[Имеет\ единственное\ решение,\ \]

\[если\ D = 0:\]

\[- 4a^{2} + 200 = 0\]

\[a^{2} = 50\]

\[a = \pm 5\sqrt{2}\]

\[\mathbf{Ответ:Г).}\]

\[\mathbf{№16.}\]

\[ax^{2} - 4x + a \geq 0\]

\[D = 16 - 4a²\]

\[a^{2} = 4\]

\[a = \pm 2\]

\[При\ a = 2:\ \ 2x^{2} - 4x +\]

\[+ 2 \geq 0 \Longrightarrow x - любое\ число;\]

\[при\ a = - 2 \Longrightarrow - 2x^{2} - 4x -\]

\[- 2 \geq 0 \Longrightarrow x = - 1\]

\[Ответ:В).\]

\[\mathbf{№17.}\]

\[ax² - 2x + a < 0\]

\[D = 4 - 4a^{2}\]

\[Неравенство\ не\ имеет\]

\[\ решения,\ если\ a > 0;\ \ D \leq 0:\]

\[4 - 4a^{2} < 0\]

\[4a^{2} > 4\]

\[a^{2} > 1;\ \ a = \pm 1\]

\[\mathbf{Ответ:\ \ Б).}\]

\[\mathbf{№18.}\]

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

\[\ \left\{ \begin{matrix} 2x - x^{2} + 8 - a = 0 \\ y = x^{2} - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- x^{2} + 2x + 8 - a = 0\]

\[D = 4 + 4 \cdot (8 - a) = 4 + 32 -\]

\[- 4a = 36 - 4a\]

\[36 - 4a = 0\]

\[a = 9\]

\[\mathbf{Ответ:Б).}\]

\[\boxed{\mathbf{Задание}\mathbf{\ 3.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\mathbf{№1.}\]

\[f(x) = 2x^{2} - 1;\ \ x = - 3;\]

\[f( - 3) = 18 - 1 = 17\]

\[\mathbf{Ответ:Г}\mathbf{)}\mathbf{.}\]

\[\mathbf{№}\mathbf{2}\mathbf{.}\]

\[\textbf{а)}\ y = \sqrt{6 + x},\ \ 6 + x \geq 0,\]

\[\ \ x \geq - 6\]

\[\textbf{б)}\ y = \frac{1}{\sqrt{6 - x}},\ \ 6 - x > 0,\]

\[\ \ x < 6\]

\[\textbf{в)}\ y = \frac{1}{\sqrt{6 + x}},\ \ 6 + x > 0,\]

\[\ \ x > - 6\]

\[\textbf{г)}\ y = \sqrt{6 - x},\ \ 6 - x \geq 0,\]

\[\ \ x \leq 6\]

\[Ответ:Б).\]

\[\mathbf{№}\mathbf{3}\mathbf{.}\]

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

\[Ответ:Г).\]

\[\mathbf{№}\mathbf{4}\mathbf{.}\]

\[\mathbf{Ответ:Г)}\mathbf{.}\]

\[\mathbf{№}\mathbf{5}\mathbf{.}\]

\[y = 3 \cdot (x - 4)^{2} - 5\]

\[график\ функции\ x^{2}смещаем\ на\]

\[\ 4\ вправо\ и\ на\ 5\ вниз.\]

\[Ответ:В).\]

\[\mathbf{№}\mathbf{6}\mathbf{.}\]

\[Ответ:В)\]

\[\mathbf{№7.}\]

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

\[x_{0} = \frac{12}{2 \cdot 2} = 3\]

\[\mathbf{Ответ:В.}\]

\[\mathbf{№8.}\]

\[Ответ:Б).\]

\[\mathbf{№9.}\]

\[\mathbf{Ответ:Г)}\mathbf{.}\]

\[\mathbf{№10.}\]

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

\[x_{0} = \frac{12}{2 \cdot 2} = 3\]

\[y = x^{2} + bx + c\]

\[9 + 3b + c = 8\]

\[\left\{ \begin{matrix} 3b + c = - 1 \\ - \frac{b}{2} = 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} 3b + c = - 1 \\ b = - 6\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} - 18 + c = - 1 \\ b = - 6\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} c = 17 \\ b = - 6 \\ \end{matrix} \right.\ \]

\[\mathbf{Ответ:Б).}\]

\[\mathbf{№11.}\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№12.}\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№13.}\]

\[x^{2} + 8x - 9 \geq 0\]

\[D_{1} = 16 + 9 = 25\]

\[x_{1} = - 4 + 5 = 1;\]

\[x_{2} = - 4 - 5 = - 9.\]

\[(x + 9)(x - 1) \geq 0\]

\[x \leq - 9;\ \ \ x \geq 1.\]

\[Ответ:Б).\]

\[\mathbf{№14.}\]

\[3x^{2} + 5x - 8 < 0\]

\[D = 25 + 96 = 121\]

\[x_{1} = \frac{- 5 + 11}{6} = 1;\]

\[x_{2} = \frac{- 5 - 11}{6} = - \frac{16}{6} = - \frac{8}{3} = - 2\frac{2}{3};\]

\[3(x - 1)\left( x + 2\frac{2}{3} \right) < 0\]

\[- 2\frac{2}{3} < x < 1\]

\[x = - 2;\ - 1;0.\]

\[Ответ:А).\ \]

\[\mathbf{№15.}\]

\[А)\ x^{2} - 14x + 49 > 0\]

\[(x - 7)^{2} > 0\]

\[x \neq 0.\]

\[Б) - 3x^{2} + x + 2 \leq 0\]

\[3x^{2} - x - 2 \geq 0\]

\[D = 1 + 24 = 25\]

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

\[x_{2} = \frac{1 - 5}{6} = - \frac{4}{6} = - \frac{2}{3};\]

\[3\left( x + \frac{2}{3} \right)(x - 1) \geq 0\]

\[x \leq - \frac{2}{3};\ \ \ x \geq 1.\]

\[В)\ x^{2} - 3x + 4 > 0\]

\[D = 9 - 16 < 0\]

\[x - любое\ число.\]

\[Ответ:В).\]

\[\mathbf{№16.}\]

\[А)\ x^{2} - 6x + 10 < 0\]

\[D_{1} = 9 - 10 < 0;\ \ a > 0\]

\[нет\ решений.\]

\[Ответ:А).\]

\[\mathbf{№17.}\]

\[3x^{2} - bx + 3 = 0\]

\[D < 0:\]

\[D = b^{2} - 36\]

\[b^{2} - 36 < 0\]

\[(b + 6)(b - 6) < 0\]

\[- 6 < b < 6.\]

\[Ответ:А).\]

\[\mathbf{№18.}\]

\[ax^{2} - 4x + a \geq 0;\ \ D = 0\]

\[D_{1} = 4 - a^{2}\]

\[4 - a^{2} = 0\]

\[a^{2} = 4\]

\[a = \pm 2\]

\[Ответ:\ А).\]