Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 339

\[\boxed{\mathbf{339}\mathbf{.}}\]

\[1)\ \frac{1}{2}\lg\left( x^{2} + x - 5 \right) =\]

\[= \lg(5x) + \lg\frac{1}{5x}\]

\[\lg\left( x^{2} + x - 5 \right)^{\frac{1}{2}} = \lg\left( 5x \bullet \frac{1}{5x} \right)\]

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

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

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

\[x^{2} + x - 6 = 0\]

\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{- 1 + 5}{2} = 2.\]

\[имеет\ смысл\ при:\]

\[5x > 0\]

\[x > 0.\]

\[Проверка:\]

\[Ответ:\ \ x = 2.\]

\[2)\ \frac{1}{2}\lg\left( x^{2} - 4x - 1 \right) =\]

\[= \lg(8x) - \lg(4x)\]

\[\lg\left( x^{2} - 4x - 1 \right)^{\frac{1}{2}} = \lg\frac{8x}{4x}\]

\[\lg\sqrt{x^{2} - 4x - 1} = \lg 2\]

\[\sqrt{x^{2} - 4x - 1} = 2\]

\[x^{2} - 4x - 1 = 4\]

\[x^{2} - 4x - 5 = 0\]

\[D = 4^{2} + 4 \bullet 5 = 16 + 20 = 36\]

\[x_{1} = \frac{4 - 6}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{4 + 6}{2} = 5.\]

\[имеет\ смысл\ при:\]

\[8x > 0\]

\[x > 0.\]

\[4x > 0\]

\[x > 0.\]

\[Проверка:\]

\[= \lg(2 \bullet 2) - \lg 4 =\]

\[= \lg 4 - \lg 4 = 0\]

\[Ответ:\ \ x = 5.\]