Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1596

\[\boxed{\mathbf{1596}\mathbf{.}}\]

\[\left| 2\sqrt{x} + 1 - x \right| + \left| x - 2\sqrt{x} + 2 \right| = 7\]

\[\left| 2\sqrt{x} + 1 - x \right| + \left| 3 - \left( 2\sqrt{x} + 1 - x \right) \right| = 7\]

\[y = 2\sqrt{x} + 1 - x:\]

\[|y| + |3 - y| = 7.\]

\[3 - y \geq 0\]

\[y \leq 3;\]

\[y \geq 0.\]

\[0 \leq y \leq 3:\]

\[y + 3 - y = 7\]

\[0y = 4\]

\[корней\ нет.\]

\[y > 3:\]

\[y - (3 - y) = 7\]

\[2y = 10\]

\[y = 5.\]

\[y < 0:\]

\[- y + 3 - y = 7\]

\[- 2y = 4\]

\[y = - 2.\]

\[1)\ 2\sqrt{x} + 1 - x = 5\]

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

\[D = 4 - 16 = - 12 < 0\]

\[корней\ нет.\]

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

\[x - 2\sqrt{x} - 3 = 0\]

\[D = 4 + 12 = 16\]

\[\sqrt{x_{1}} = \frac{2 - 4}{2} = - 1;\]

\[\sqrt{x_{2}} = \frac{2 + 4}{2} = 3;\]

\[x_{1} = ( - 1)^{2} = 1;\ \]

\[x_{2} = 3^{2} = 9.\]

\[Проверка:\]

\[\left| 2\sqrt{1} + 1 - 1 \right| + \left| 1 - 2\sqrt{1} + 2 \right| =\]

\[= |2| + |1| = 3 \neq 7;\]

\[\left| 2\sqrt{9} + 1 - 9 \right| + \left| 9 - 2\sqrt{9} + 2 \right| =\]

\[= |6 - 8| + |11 - 6| = 2 + 5 = 7.\]

\[Ответ:\ \ x = 9.\]