Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 41

\[\boxed{\mathbf{41.}}\]

\[= \sqrt{4x} + \sqrt[4]{4x + 1} + \sqrt[6]{4x + 2}\]

\[\left\{ \begin{matrix} x^{2} + 4 = 4x\ \ \ \ \ \ \ \ \\ x^{2} + 5 = 4x + 1 \\ x^{2} + 6 = 4x + 2 \\ 4x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \\ 4x + 2 \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 4x + 4 = 0 \\ x^{2} - 4x + 4 = 0 \\ x^{2} - 4x + 4 = 0 \\ x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq - \frac{1}{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x \geq - \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

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

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

\[Ответ:x = 2.\]

\[= \sqrt{2x} + \sqrt[4]{4x + 1} + \sqrt[6]{6x + 4}\]

\[\left\{ \begin{matrix} x + 1 = 2x\ \ \ \ \ \ \ \ \ \\ 2x + 3 = 4x + 1 \\ 3x + 7 = 6x + 4 \\ 2x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \\ 6x + 4 \geq 0\ \ \ \ \ \ \ \ \ \ \\ x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ 2x + 3 \geq 0\ \ \ \ \ \ \ \ \ \ \\ 3x + 7 \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1\ \ \ \\ 2x = 2\ \ \\ 3x = 3\ \ \\ x \geq 0\ \ \ \ \\ x \geq - \frac{1}{4} \\ x \geq - \frac{2}{3} \\ x \geq - 1 \\ x \geq - \frac{3}{2} \\ x \geq - \frac{7}{3} \\ \end{matrix} \right.\ \]

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

\[Ответ:x = 1.\]