Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 11

\[\boxed{\mathbf{11.}}\]

\[\textbf{а)}\ \sqrt{x - 4}\sqrt{x + 4} = \sqrt{6}\]

\[1)\ x - 4 \geq 0\]

\[x > 4.\]

\[2)\ x + 4 \geq 0\]

\[x \geq - 4.\]

\[M = \lbrack 4; + \infty).\]

\[3)\ (x - 4)(x + 4) = 6\]

\[x^{2} - 16 = 6\]

\[x^{2} = 22\]

\[x = - \sqrt{22} < 4;\]

\[x = \sqrt{22} > 4.\]

\[Ответ:x = \sqrt{22}.\]

\[\textbf{б)}\ \sqrt{x - 3}\sqrt{x + 3} = \sqrt{5}\]

\[1)\ x - 3 \geq 0\]

\[x \geq 3.\]

\[2)\ x + 3 \geq 0\]

\[x \geq - 3.\]

\[M = \lbrack 3; + \infty).\]

\[3)\ (x - 3)(x + 3) = 5\]

\[x^{2} - 9 = 5\]

\[x^{2} = 14\]

\[x = - \sqrt{14} < 3;\]

\[x = \sqrt{14} > 3.\]

\[Ответ:x = \sqrt{14}.\]

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

\[1)\ x - 5 \geq 0\]

\[x \geq 5.\]

\[2)\ x + 5 \geq 0\]

\[x \geq - 5.\]

\[M = \lbrack 5; + \infty).\]

\[3)\ (x - 5)(x + 5) = 2\]

\[x^{2} - 25 = 2\]

\[x^{2} = 27\]

\[x = - 3\sqrt{3} < 5;\]

\[x = 3\sqrt{3} > 5.\]

\[Ответ:x = 3\sqrt{3}.\]

\[\textbf{г)}\ \sqrt{x - 2}\sqrt{x + 2} = \sqrt{8}\]

\[1)\ x - 2 \geq 0\]

\[x \geq 2.\]

\[2)\ x + 2 \geq 0\]

\[x \geq - 2.\]

\[M = \lbrack 2; + \infty).\]

\[3)\ (x - 2)(x + 2) = 8\]

\[x^{2} - 4 = 8\]

\[x^{2} = 12\]

\[x = 2\sqrt{3} > 2;\]

\[x = - 2\sqrt{3} < 2.\]

\[Ответ:x = 2\sqrt{3}.\]