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МатематикаРешебник по алгебре 8 класс Мерзляк ФГОС Задание 781
Задание 781
\[\boxed{\mathbf{781\ (781).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ x - 3\sqrt{x} + 2 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} - 3t + 2 = 0 \\ \end{matrix} \right.\ \ \]
\[t_{1} + t_{2} = 3,\ \ t_{1} \cdot t_{2} = 2,\ \ \]
\[t_{1} = 2,\ \ t_{2} = 1\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = 2 \\ t = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = 2 \\ \sqrt{x} = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 4 \\ x = 1 \\ \end{matrix} \right.\ \]
\[Ответ:x = 4;x = 1.\]
\[2)\ x - \sqrt{x} - 12 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} - t - 12 = 0 \\ \end{matrix} \right.\ \ \]
\[t_{1} + t_{2} = 1,\ \ t_{1} \cdot t_{2} = - 12,\ \]
\[\ t_{1} = 4,\ \ t_{2} = - 3\]
\[Ответ:x = 16.\]
\[3)\ 3x - 10\sqrt{x} + 3 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3t^{2} - 10t + 3 = 0 \\ \end{matrix} \right.\ \text{\ \ }\]
\[t_{1} + t_{2} = \frac{10}{3},\ \ t_{1} \cdot t_{2} = 1,\ \]
\[\ t_{1} = \frac{1}{3},\ \ t_{2} = 3\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = 3 \\ t = \frac{1}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = 3 \\ \sqrt{x} = \frac{1}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 9 \\ x = \frac{1}{9} \\ \end{matrix} \right.\ \]
\[Ответ:x = 9;x = \frac{1}{9}.\]
\[4)\ 8\sqrt{x} + x + 7 = 0\]
\[\ \left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} + 8t + 7 = 0 \\ \end{matrix} \right.\ \text{\ \ }\]
\[t_{1} + t_{2} = - 8,\ \ t_{1} \cdot t_{2} = 7,\ \]
\[\ t_{1} = - 7,\ \ t_{2} = - 1\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = - 7 \\ t = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = - 7 \\ \sqrt{x} = - 1 \\ \end{matrix} \right.\ \]
\[Ответ:нет\ корней.\]
\[5)\ 6\sqrt{x} - 27 + x = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t^{2} + 6t - 27 = 0 \\ \end{matrix} \right.\ \]
\[\ t_{1} + t_{2} = - 6,\ \ t_{1} \cdot t_{2} = - 27,\]
\[\text{\ \ }t_{1} = - 9,\ \ t_{2} = 3\]
\[\left\{ \begin{matrix} \sqrt{x} = t \\ t = - 9 \\ t = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} \sqrt{x} = - 9 \\ \sqrt{x} = 3 \\ \end{matrix} \right.\ ,\ \ \]
\[x = 9\]
\[Ответ:x = 9.\]
\[6)\ 8x - 10\sqrt{x} + 3 = 0\]
\[\left\{ \begin{matrix} \sqrt{x} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8t^{2} - 10t + 3 = 0 \\ \end{matrix} \right.\ \]
\[D = 100 - 96 = 4,\ \ \]
\[t_{1,2} = \frac{10 \pm 2}{16}\]
\[Ответ:x = \frac{1}{4};x = \frac{9}{16}.\]
\[\boxed{\mathbf{7}\mathbf{8}\mathbf{1}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[x_{1}² + x_{2}² = 9\]
\[\left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} = 9\]
\[x^{2} + (a - 1)x - 2a = 0\]
\[x_{1} + x_{2} = 1 - a\]
\[x_{1}x_{2} = - 2a\]
\[(1 - a)^{2} + 4a = 9\]
\[1 - 2a + a^{2} + 4a - 9 = 0\]
\[a^{2} + 2a - 8 = 0\]
\[a = - 4 \Longrightarrow не\ подходит.\]
\[a = 2\]
\[Ответ:при\ a = 2.\]
