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МатематикаРешебник по алгебре 9 класс Мерзляк Проверь себя №2
Проверь себя №2
\[\boxed{\mathbf{Задание}\mathbf{\ 2.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[\mathbf{№1.}\]
\[f( - 3) = 18 - 1 = 17\]
\[\mathbf{Ответ:Г.}\]
\[\mathbf{№2.}\]
\[Ответ:В.\]
\[\mathbf{№3.}\]
\[\textbf{а)}\ y = \sqrt{6 + x},\ \ 6 + x \geq 0,\]
\[\ \ x \geq - 6\]
\[\textbf{б)}\ y = \frac{1}{\sqrt{6 - x}},\ \ 6 - x > 0,\]
\[\ \ x < 6\]
\[\textbf{в)}\ y = \frac{1}{\sqrt{6 + x}},\ \ 6 + x > 0,\]
\[\ \ x > - 6\]
\[\textbf{г)}\ y = \sqrt{6 - x},\ \ 6 - x \geq 0,\]
\[\ \ x \leq 6\]
\[\mathbf{№4.}\]
\[Ответ:В.\]
\[\mathbf{№5.}\]
\[\mathbf{Ответ:А.}\]
\[\mathbf{№6.}\]
\[\mathbf{Ответ:Г.}\]
\[\mathbf{№7.}\]
\[\mathbf{Ответ:Г.}\]
\[\mathbf{№8.}\]
\[y = 3 \cdot (x - 4)^{2} - 5\]
\[график\ функции\ x^{2}смещаем\ на\]
\[\ 4\ вправо\ и\ на\ 5\ вниз.\]
\[Ответ:В.\]
\[\mathbf{№9.}\]
\[\mathbf{Ответ:В.}\]
\[\mathbf{№10.}\]
\[y = 2x^{2} - 12x + 3\]
\[x_{0} = \frac{12}{2 \cdot 2} = 3\]
\[\mathbf{Ответ:В.}\]
\[\mathbf{№11.}\]
\[\mathbf{Ответ:В.}\]
\[\mathbf{№12.}\]
\[\mathbf{Ответ:Г.}\]
\[\mathbf{№13.}\]
\[\mathbf{Ответ:Б.}\]
\[\mathbf{№14.}\]
\[y = 2x^{2} + x - 6\]
\[2x^{2} + x - 6 = 0\]
\[D = 1 + 48 = 49\]
\[x_{1,2} = \frac{- 1 \pm 7}{4}\]
\[x_{1} = - 2,\ \ x_{2} = 1,5\]
\[Ответ:Г.\ \]
\[\mathbf{№15.}\]
\[y = x^{2} + bx + c\]
\[9 + 3b + c = 8\]
\[\left\{ \begin{matrix} 3b + c = - 1 \\ - \frac{b}{2} = 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} 3b + c = - 1 \\ b = - 6\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} - 18 + c = - 1 \\ b = - 6\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} c = 17 \\ b = - 6 \\ \end{matrix} \right.\ \]
\[\mathbf{Ответ:Б.}\]
\[\mathbf{№16.}\]
\[\mathbf{Ответ:В.}\]
\[\mathbf{№17.}\]
\[y = 3x^{2} - 6x + a,\]
\[\ \ вершина\ \ (x;4).\]
\[y_{0} = - \frac{D}{4 \cdot 3}\]
\[D = 36 - 12a = 12 \cdot (3 - a)\]
\[a - 3 = 4\]
\[a = 7\]
\[Ответ:В.\]
\[\mathbf{№18.}\]
\[m - n = 8,\ \ m = 8 + n\]
\[\mathbf{Ответ:А.}\]
\[\boxed{\mathbf{Задание}\mathbf{\ 2.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[\mathbf{№1.}\]
\[x^{4} + 7x^{2} - 18 = 0\]
\[x^{2} = y:\]
\[y^{2} + 7x - 18 = 0\]
\[D = 49 + 72 = 121\]
\[y_{1} = \frac{- 7 + 11}{2} = 2;\]
\[y_{2} = \frac{- 7 - 11}{2} = - 9 < 0.\]
\[x^{2} = 2\]
\[x = \pm \sqrt{2}.\]
\[Ответ:Б).\]
\[\mathbf{№2.}\]
\[\left( x^{2} - 4x \right)^{2} - 2\left( x^{2} - 4x \right) - 15 = 0\]
\[x^{2} - 4x = y:\]
\[y^{2} - 2y - 15 = 0\]
\[D_{1} = 1 + 15 = 16\]
\[y_{1} = 1 + 4 = 5;\]
\[y_{2} = 1 - 4 = - 3.\]
\[1)\ x^{2} - 4x = 5\]
\[x^{2} - 4x - 5 = 0\]
\[D_{1} = 4 + 5 = 9\]
\[x_{1} = 2 + 3 = 5;\]
\[x_{2} = 2 - 3 = - 1.\]
\[2)\ x^{2} - 4x = - 3\]
\[x^{2} - 4x + 3 = 0\]
\[D_{1} = 4 - 3 = 1\]
\[x_{1} = 2 + 1 = 3;\]
\[x_{2} = 2 - 1 = 1.\]
\[Ответ:А).\]
\[\mathbf{№3.}\]
\[x - \sqrt{x} - 12 = 0\]
\[\sqrt{x} = y \geq 0:\]
\[y^{2} - y - 12 = 0\]
\[y_{1} + y_{2} = 1;\ \ y_{1} \cdot y_{2} = - 12\]
\[y_{1} = 4;\ \ \ \ y_{2} = - 3 < 0.\]
\[\sqrt{x} = 4\]
\[x = 16.\]
\[Ответ:\ \ В).\]
\[\mathbf{№4.}\]
\[\left\{ \begin{matrix} y - x = 2\ \ \ \ \ \\ xy - y = 10 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = x + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x(x + 2) - (x + 2) = 10 \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - x - 2 - 10 = 0\]
\[x^{2} + x - 12 = 0\]
\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 12\]
\[x_{1} = - 4;\ \ \ \ \ x_{2} = 3;\]
\[y_{1} = - 4 + 2 = - 2;\]
\[y_{2} = 3 + 2 = 5\]
\[x_{1}y_{1} = - 4 \cdot ( - 2) = 8;\]
\[x_{2}y_{2} = 3 \cdot 5 = 15;\]
\[Сумма:\]
\[8 + 15 = 23.\]
\[Ответ:А).\]
\[\mathbf{№5.}\]
\[Ответ:В).\]
\[\mathbf{№6.}\]
\[\left\{ \begin{matrix} x^{2} - y = 4 \\ x + y = 1\ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = x^{2} - 4 \\ y = x + 1\ \ \\ \end{matrix} \right.\ \]
\[x^{2} - 4 = x + 1\]
\[x^{2} - x - 5 = 0\]
\[D = 1 + 5 = 6 > 0\]
\[два\ корня.\]
\[Ответ:В).\]
\[\mathbf{№7.}\]
\[\left\{ \begin{matrix} x - y = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2xy - y^{2} = - 7 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = x - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2x(x - 5) - (x - 5)^{2} = - 7 \\ \end{matrix} \right.\ \]
\[x^{2} + 2x^{2} - 10x - x^{2} + 10x - 25 + 7 = 0\]
\[2x^{2} - 18 = 0\]
\[2x^{2} = 18\]
\[x^{2} = 9\]
\[x = \pm 3.\]
\[y = 3 - 5 = - 2;\]
\[y = - 3 - 5 = - 8.\]
\[Наибольшее:3 - 2 = 1.\]
\[Ответ:А).\]
\[\mathbf{№8.}\]
\[\left\{ \begin{matrix} \frac{2}{x} + \frac{1}{y} = 4\ \ | \cdot 3 \\ \frac{1}{x} - \frac{3}{y} = 9\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \frac{6}{x} + \frac{3}{y} = 12 \\ \frac{1}{x} - \frac{3}{y} = 9\ \ \\ \end{matrix} \right.\ ( + )\]
\[\frac{7}{x} = 21\]
\[21x = 7\]
\[x = \frac{7}{21} = \frac{1}{3};\]
\[\frac{1}{y} = 4 - \frac{2}{x} = 4 - 2 \cdot 3 = - 2\]
\[y = - \frac{1}{2}.\]
\[\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}.\]
\[Ответ:Г).\]
\[\mathbf{№9.}\]
\[\left\{ \begin{matrix} 2x - xy = 5 \\ y + xy = 6\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x(2 - y) = 5 \\ y(1 + x) = 6 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} y = \frac{6}{1 + x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x\left( 2 - \frac{6}{1 + x} \right) = 5 \\ \end{matrix} \right.\ \]
\[\frac{x(2 + 2x - 6)}{1 + x} = 5\]
\[2x^{2} - 4x = 5(1 + x)\]
\[2x^{2} - 4x = 5 + 5x\]
\[2x^{2} - 9x - 5 = 0\]
\[D = 81 + 40 = 121\]
\[x_{1} = \frac{9 + 11}{4} = 5;\]
\[x_{2} = \frac{9 - 11}{4} = - \frac{2}{4} = - 0,5.\]
\[y_{1} = \frac{6}{1 + 5} = 1;\]
\[y_{2} = \frac{6}{1 - 0,5} = \frac{60}{5} = 12.\]
\[\left| 5 \cdot 1 - ( - 0,5) \cdot 12 \right| = |5 + 6| = 11.\]
\[Ответ:Б).\]
\[\mathbf{№10.}\]
\[2x - y = a\]
\[y = 2x - a;\]
\[2x - a = x^{2} - 8\]
\[x^{2} - 2x - (8 - a) = 0\]
\[D = 0:\]
\[D_{1} = 1 + 8 - a = 9 - a\]
\[9 - a = 0\]
\[a = 9.\]
\[Ответ:Б).\]
\[\mathbf{№11.}\]
\[Пусть\ \text{x\ }км - расстояние,\ \]
\[которое\ катер\ проплыл\ по\]
\[\ реке;тогда\ (x + 5)\ км\]
\[- расстояние\ по\ озеру.\]
\[Скорость\ катера\ против\]
\[\ течения\ равна:\]
\[10 - 2 = 8\ \frac{км}{ч}.\]
\[\frac{x + 5}{10}\ ч - время\ катера\ по\]
\[\ озеру;\]
\[\frac{x}{8}\ ч - время\ катера\ против\ \]
\[течения.\]
\[15\ мин = \frac{1}{4}\ ч.\]
\[Составим\ уравнение:\]
\[\frac{x + 5}{10} - \frac{x}{8} = \frac{1}{4}.\]
\[\mathbf{Ответ:Б).}\]
\[\mathbf{№12.}\]
\[x\ деталей - первый\ рабочий;\]
\[\text{y\ }деталей - второй\ рабочий.\]
\[3x + 4y = 44\ \ детали -\]
\[работая\ вместе;\]
\[x = 2y - 2,\ так\ как\ первый\ \]
\[делал\ за\ час\ меньше,\ чем\ \]
\[второй\ за\ 2\ часа.\]
\[Получаем\ систему:\]
\[\left\{ \begin{matrix} 3x + 4y = 44 \\ x = 2y - 2\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\mathbf{Ответ:Г).}\]
\[\mathbf{№13.}\]
\[x\ ч - время\ работы\ первого\ \]
\[тракториста;\]
\[\text{y\ }ч - время\ работы\ второго\ \]
\[тракториста.\]
\[\frac{1}{x};\ \frac{1}{y} - производительность\ \]
\[трактористов.\]
\[2ч\ 40\ мин = 2\frac{2}{3}\ ч = \frac{8}{3}\ ч.\]
\[\frac{1}{x} + \frac{1}{y} = 1\ :\frac{8}{3} - вспашут,\]
\[\ работая\ вместе.\]
\[\frac{1}{x} + 2 \cdot \frac{1}{y} = \frac{1}{2} - вспашут,\]
\[\ работая\ по\ отдельности.\]
\[Получаем\ систему:\]
\[\left\{ \begin{matrix} \frac{1}{x} + \frac{1}{y} = \frac{3}{8}\text{\ \ \ \ \ } \\ \frac{1}{x} + \frac{2}{y} = \frac{1}{2}\text{\ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\mathbf{Ответ:В).}\]
\[\mathbf{№14.}\]
\[3 - 2(x - 3) \leq 18 - 5x\]
\[3 - 2x + 6 + 5x \leq 18\]
\[3x \leq 9\]
\[x \leq 3.\]
\[Ответ:А).\]
\[\mathbf{№15.}\]
\[\sqrt{14 - 3x}\]
\[14 - 3x \geq 0\]
\[- 3x \geq - 14\]
\[x \leq \frac{14}{3} \leq 4\frac{2}{3}\]
\[x = 1;2;3;4.\]
\[1 \cdot 2 \cdot 3 \cdot 4 = 24.\]
\[Ответ:\ \ Г).\]
\[\mathbf{№16.}\]
\[\left\{ \begin{matrix} x - \frac{x - 2}{3} \geq \frac{x - 3}{4} - \frac{x - 1}{2}\ \ | \cdot 12 \\ 1 - 0,5x > x - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 12x - 4x + 8 \geq 3x - 9 - 6x + 6 \\ - 1,5x > - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 11x \geq - 11 \\ x < \frac{50}{15}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq - 1 \\ x < 3\frac{1}{3} \\ \end{matrix} \right.\ \]
\[- 1 \leq x < 3\frac{1}{3}\]
\[x = - 1;0;1;2;3.\]
\[Ответ:В).\]
\[\mathbf{№17.}\]
\[\left\{ \begin{matrix} y > x^{2}\text{\ \ \ \ \ \ \ \ } \\ y \leq 2x + 4 \\ \end{matrix} \right.\ \]
\[Ответ:\ \ Г).\]
\[\mathbf{№18.}\]
\[x + 2y > 4\]
\[2y > 4 - x\]
\[y > 2 - 0,5x\]
\[Ответ:\ \ Г).\]
