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МатематикаРешебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 10
Задание 10
\[\boxed{\mathbf{10.}}\]
\[\textbf{а)}\ \sqrt{x^{2} - 1} = \sqrt{- 2x}\]
\[\left\{ \begin{matrix} x^{2} - 1 = - 2x \\ - 2x \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + 2x - 1 = 0 \\ x \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - 1 = 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1} =\]
\[\text{=} - 1 + \sqrt{2} > 0 - не\ подходит;\]
\[x_{2} = - 1 - \sqrt{2}.\]
\[Ответ:x = - 1 - \sqrt{2}.\]
\[\textbf{б)}\ \sqrt{x^{2} + 3x} = \sqrt{x + 1}\]
\[\left\{ \begin{matrix} x^{2} + 3x = x + 1 \\ x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + 2x - 1 = 0 \\ x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - 1 = 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1} = - 1 + \sqrt{2};\]
\[x_{2} =\]
\[= - 1 - \sqrt{2} < - 1\ (не\ подходит).\]
\[Ответ:x = - 1 + \sqrt{2}.\]
\[\textbf{в)}\ \sqrt{x^{2} - 7} = \sqrt{- 2x - 6}\]
\[\left\{ \begin{matrix} x^{2} - 7 = - 2x - 6 \\ - 2x - 6 \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 7 + 2x + 6 = 0 \\ - 2x \geq 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + 2x - 1 = 0 \\ x \leq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - 1 = 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1} = - 1 + \sqrt{2} > - 3;\]
\[x_{2} = - 1 - \sqrt{2} > - 3.\]
\[Ответ:нет\ корней.\]
\[\textbf{г)}\ \sqrt{x^{2} + x} = \sqrt{1 - x}\]
\[\left\{ \begin{matrix} x^{2} + x = 1 - x \\ 1 - x \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + 2x - 1 = 0 \\ - x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + 2x - 1 = 0 \\ x \leq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + 2x - 1 = 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1} = - 1 + \sqrt{2} < 1;\]
\[x_{2} = - 1 - \sqrt{2} < 1.\]
\[Ответ:x = - 1 \pm \sqrt{2}.\]
