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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 143
Задание 143
\[\boxed{\text{143\ (143).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ 3x^{2} - 6x + b = 0\]
\[Квадратное\ уравнение\ имеет\ \]
\[два\ различных\ корня\ \]
\[при\ D > 0.\]
\[D = 36 - 12b\]
\[36 - 12b > 0\]
\[36 > 12b\]
\[b < 3\]
\[Ответ:имеет\ два\ различных\ \]
\[корня\ при\ b \in ( - \infty;3).\]
\[2)\ x^{2} - x - 2b = 0\ \]
\[Квадратное\ уравнение\ не\ \]
\[имеет\ корней\ при\ D < 0.\]
\[D = 1 + 8b\]
\[1 + 8b < 0\]
\[8b < - 1\]
\[b < - \frac{1}{8}\]
\[Ответ:уравнение\ не\ имеет\ \]
\[корней\ при\ b \in \left( - \infty;\ - \frac{1}{8} \right).\]
\[\boxed{\mathbf{143.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ x - 3\sqrt{x} + 2 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} - 3t + 2 = 0 \\ \end{matrix} \right.\ \ \]
\[t_{1} + t_{2} = 3,\ \ t_{1} \cdot t_{2} = 2,\ \ \]
\[t_{1} = 2,\ \ t_{2} = 1\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = 2 \\ t = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = 2 \\ \sqrt{x} = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 4 \\ x = 1 \\ \end{matrix} \right.\ \]
\[Ответ:x = 4;x = 1.\]
\[2)\ x - \sqrt{x} - 12 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} - t - 12 = 0 \\ \end{matrix} \right.\ \ \]
\[t_{1} + t_{2} = 1,\ \ t_{1} \cdot t_{2} = - 12,\ \]
\[\ t_{1} = 4,\ \ t_{2} = - 3\]
\[Ответ:x = 16.\]
\[3)\ 3x - 10\sqrt{x} + 3 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3t^{2} - 10t + 3 = 0 \\ \end{matrix} \right.\ \text{\ \ }\]
\[t_{1} + t_{2} = \frac{10}{3},\ \ t_{1} \cdot t_{2} = 1,\ \]
\[\ t_{1} = \frac{1}{3},\ \ t_{2} = 3\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = 3 \\ t = \frac{1}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = 3 \\ \sqrt{x} = \frac{1}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 9 \\ x = \frac{1}{9} \\ \end{matrix} \right.\ \]
\[Ответ:x = 9;x = \frac{1}{9}.\]
\[4)\ 8\sqrt{x} + x + 7 = 0\]
\[\ \left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} + 8t + 7 = 0 \\ \end{matrix} \right.\ \text{\ \ }\]
\[t_{1} + t_{2} = - 8,\ \ t_{1} \cdot t_{2} = 7,\ \]
\[\ t_{1} = - 7,\ \ t_{2} = - 1\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = - 7 \\ t = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = - 7 \\ \sqrt{x} = - 1 \\ \end{matrix} \right.\ \]
\[Ответ:нет\ корней.\]
\[5)\ 6\sqrt{x} - 27 + x = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} + 6t - 27 = 0 \\ \end{matrix} \right.\ \]
\[\ t_{1} + t_{2} = - 6,\ \ t_{1} \cdot t_{2} = - 27,\]
\[\text{\ \ }t_{1} = - 9,\ \ t_{2} = 3\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = - 9 \\ t = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = - 9 \\ \sqrt{x} = 3 \\ \end{matrix} \right.\ ,\ \ \]
\[x = 9\]
\[Ответ:x = 9.\]
\[6)\ 8x - 10\sqrt{x} + 3 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8t^{2} - 10t + 3 = 0 \\ \end{matrix} \right.\ \]
\[D = 100 - 96 = 4,\ \ \]
\[t_{1,2} = \frac{10 \pm 2}{16}\]
\[Ответ:x = \frac{1}{4};x = \frac{9}{16}.\]
