ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 358

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\[1)\ y = \log_{5}\left( x^{2} - 4x + 3 \right)\]

\[x^{2} - 4x + 3 > 0\]

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2} = 1;\text{\ \ }x_{2} = \frac{4 + 2}{2} = 3.\]

\[(x - 1)(x - 3) > 0\]

\[x < 1;\text{\ \ }x > 3.\]

\[Ответ:\ \ x < 1;\ \ x > 3.\]

\[2)\ y = \log_{6}\frac{3x + 2}{1 - x}\]

\[\frac{3x + 2}{1 - x} > 0\]

\[(3x + 2)(1 - x) > 0\]

\[(3x + 2)(x - 1) < 0\]

\[- \frac{2}{3} < x < 1\]

\[Ответ:\ \ - \frac{2}{3} < x < 1.\]

\[3)\ y = \sqrt{\lg x + \lg(x + 2)}\]

\[1)\ x > 0;\]

\[x + 2 > 0 \Longrightarrow x > - 2.\]

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

\[\lg\left( x(x + 2) \right) \geq \lg 1\]

\[x(x + 2) \geq 1\]

\[x^{2} + 2x - 1 \geq 0\]

\[D = 2^{2} + 4 = 4 + 4 = 8\]

\[x = \frac{- 2 \pm \sqrt{8}}{2} = \frac{- 2 \pm 2\sqrt{2}}{2} =\]

\[= - 1 \pm \sqrt{2};\]

\[x \leq - 1 - \sqrt{2}\text{\ \ }и\ \ x \geq - 1 + \sqrt{2}\]

\[Ответ:\ \ x \geq \sqrt{2} - 1.\]

\[4)\ y = \sqrt{\lg(x - 1) + \lg(x + 1)}\]

\[1)\ x - 1 > 0 \Longrightarrow \ x > 1;\]

\[x + 1 > 0 \Longrightarrow x \geq - 1.\]

\[2)\ \lg(x - 1) + \lg(x + 1) \geq 0\]

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

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

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

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

\[x \leq - \sqrt{2};\text{\ \ }x \geq \sqrt{2}.\]

\[Ответ:\ \ x \geq \sqrt{2}.\]