Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 337

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\[1)\log_{2}(x - 5) + \log_{2}(x + 2) = 3\]

\[\log_{2}\left( (x - 5)(x + 2) \right) = \log_{2}2^{3}\]

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

\[x^{2} + 2x - 5x - 10 = 8\]

\[x^{2} - 3x - 18 = 0\]

\[D = 3^{2} + 4 \bullet 18 = 9 + 72 = 81\]

\[x_{1} = \frac{3 - 9}{2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{3 + 9}{2} = 6.\]

\[имеет\ смысл\ при:\]

\[x - 5 > 0\ \]

\[x > 5;\]

\[x + 2 > 0\]

\[x > - 2.\]

\[Ответ:\ \ x = 6.\]

\[2)\log_{3}(x - 2) + \log_{3}(x + 6) = 2\]

\[\log_{3}\left( (x - 2)(x + 6) \right) = \log_{3}3^{2}\]

\[(x - 2)(x + 6) = 3^{2}\]

\[x^{2} + 6x - 2x - 12 = 9\]

\[x^{2} + 4x - 21 = 0\]

\[D = 4^{2} + 4 \bullet 21 = 16 + 84 =\]

\[= 100\]

\[x_{1} = \frac{- 4 - 10}{2} = - 7;\text{\ \ }\]

\[x_{2} = \frac{- 4 + 10}{2} = 3.\]

\[имеет\ смысл\ при:\]

\[x - 2 > 0\ \]

\[x > 2;\]

\[x + 6 > 0\ \]

\[x > - 6.\]

\[Ответ:\ \ x = 3.\]

\[3)\lg\left( x + \sqrt{3} \right) + \lg\left( x - \sqrt{3} \right) = 0\]

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

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

\[x^{2} - 3 = 1\]

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

\[x = \pm 2.\]

\[имеет\ смысл\ при:\]

\[x + \sqrt{3} > 0\ \]

\[x > - \sqrt{3}.\]

\[x - \sqrt{3} > 0\ \]

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

\[3 < 4 \Longrightarrow \ \sqrt{3} < 2.\]

\[Ответ:\ \ x = 2.\]

\[4)\lg(x - 1) + \lg(x + 1) = 0\]

\[\lg\left( (x - 1)(x + 1) \right) = \lg 1\]

\[(x - 1)(x + 1) = 1\]

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

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

\[x = \pm \sqrt{2}.\]

\[имеет\ смысл\ при:\]

\[x - 1 > 0\ \]

\[x > 1.\]

\[x + 1 > 0\ \]

\[x > - 1.\]

\[2 > 1 \Longrightarrow \sqrt{2} > 1.\]

\[Ответ:\ \ x = \sqrt{2}.\]