ГДЗ по алгебре 8 класс Мерзляк Задание 411

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Пояснение.

Решение.

\[Подкоренное\ выражение\ может\]

\[\ принимать\ только\ \]

\[неотрицательные\]

\[значения.\]

\[1)\ \sqrt{x} = - x\]

\[\left\{ \begin{matrix} x = x^{2} \\ x \geq 0\ \ \\ \end{matrix} \right.\ \]

\[x = 0\]

\[Ответ:x = 0.\]

\[2)\ \sqrt{x} + \sqrt{x - 1} = 0\]

\[\sqrt{x} = - \sqrt{x - 1}\]

\[\left\{ \begin{matrix} x = - x + 1 \\ \left\{ \begin{matrix} x \geq 0\ \ \ \ \ \ \\ x - 1 \geq 0 \\ \end{matrix}\text{\ \ } \right.\ \\ \end{matrix} \right.\ \]

\[2x = 1\]

\[\left\{ \begin{matrix} x = 0,5 \\ x \geq 0\ \ \ \ \\ x \geq 1\ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:нет\ корней.\]

\[3)\ \sqrt{x^{2} - x} + \sqrt{x - 1} = 0\]

\[\sqrt{x^{2} - x} = - \sqrt{x - 1}\]

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

\[\left\{ \begin{matrix} x^{2} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \cdot (x - 1) \geq 0 \\ x - 1 \geq 0\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1\ \ \ \ \\ x = - 1\ \\ x \geq 0\ \ \ \ \\ x \geq 1\ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = 1.\]

\[4)\ \sqrt{x^{2} + 2x} + \sqrt{x^{2} - 4} = 0\]

\[\sqrt{x^{2} + 2x} = - \sqrt{x^{2} - 4}\]

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

\[\left\{ \begin{matrix} 2x^{2} + 2x = 4 \\ x \cdot (x + 2) \geq 0 \\ x^{2} \geq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} + x - 2 = 0 \\ x \cdot (x + 2) \geq 0 \\ x^{2} \geq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(x + 2) \cdot (x - 1) = 0\]

\[\left\{ \begin{matrix} x = - 2 \\ x = 1\ \ \ \\ x \leq - 2 \\ x \geq 2\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ x = - 2.\]

\[5)\ (x - 1) \cdot \sqrt{x + 1} = 0\]

\[\left\{ \begin{matrix} x - 1 = 0 \\ \sqrt{x + 1} = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1\ \ \ \ \ \ \\ x + 1 = 0 \\ x + 1 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1\ \ \\ x = - 1 \\ x \geq - 1 \\ \end{matrix} \right.\ \]

\[Ответ:x = - 1;x = 1.\]

\[6)\ (x + 1) \cdot \sqrt{x - 1} = 0\]

\[\left\{ \begin{matrix} x + 1 = 0 \\ \sqrt{x - 1} = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = - 1\ \ \ \\ x - 1 = 0 \\ x - 1 \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = - 1 \\ x = 1\ \ \ \\ x \geq 1\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = 1.\ \]