Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 618

\[\boxed{\mathbf{618}.}\]

\[1)\ \sqrt{x + \sqrt{6x - 9}} +\]

\[+ \sqrt{x - \sqrt{6x - 9}} = \sqrt{6};\]

\[x + \sqrt{6x - 9} +\]

\[+ 02\sqrt{\left( x + \sqrt{6x - 9} \right)\left( x - \sqrt{6x - 9} \right)} +\]

\[+ x - \sqrt{6x - 9} = 6;\]

\[2\sqrt{x^{2} - (6x - 9)} = 6 - 2x;\]

\[\sqrt{x^{2} - 6x + 9} = 3 - x;\]

\[x^{2} - 6x + 9 = 9 - 6x + x^{2};\]

\[0x = 0 - при\ любом\ x.\]

\[Выражение\ имеет\ смысл\ при:\]

\[6x - 9 \geq 0;\]

\[6x \geq 9;\]

\[x \geq 1,5;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x - \sqrt{6x - 9} \geq 0;\]

\[x \geq \sqrt{6x - 9};\]

\[x^{2} \geq 6x - 9;\]

\[x^{2} - 6x + 9 \geq 0;\]

\[(x - 3)^{2} \geq 0 - при\ любом\ x.\]

\[Уравнение\ имеет\ решения\]

\[\ при:\]

\[6 - 2x \geq 0;\]

\[3 - x \geq 0;\]

\[x \leq 3;\]

\[Ответ:\ \ 1,5 \leq x \leq 3.\]

\[2)\ \sqrt{x + \sqrt{x + 11}} +\]

\[+ \sqrt{x - \sqrt{x + 11}} = 4;\]

\[x + \sqrt{x + 11} +\]

\[+ 2\sqrt{\left( x + \sqrt{x + 11} \right)\left( x - \sqrt{x + 11} \right)} +\]

\[+ x - \sqrt{x + 11} = 16;\]

\[2\sqrt{x^{2} - (x + 11)} = 16 - 2x;\]

\[\sqrt{x^{2} - x - 11} = 8 - x;\]

\[x^{2} - x - 11 = 64 - 16x + x^{2};\]

\[15x = 75;\]

\[x = 5;\]

\[Выполним\ проверку:\]

\[\sqrt{5 + \sqrt{5 + 11}} + \sqrt{5 - \sqrt{5 + 11}} =\]

\[= \sqrt{5 + \sqrt{16}} + \sqrt{5 - \sqrt{16}} =\]

\[= \sqrt{9} + \sqrt{1} = 4;\]

\[Ответ:\ \ x = 5.\]