ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 696

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\[1)\ 4 \bullet 9^{x} - 13 \bullet 6^{x} +\]

\[+ 9 \bullet 4^{x} = 0\ \ \ \ \ |\ :4^{x};\]

\[4 \bullet \left( \frac{9}{4} \right)^{x} - 13 \bullet \left( \frac{6}{4} \right)^{x} + 9 = 0;\]

\[4 \bullet \left( \frac{3}{2} \right)^{2x} - 13 \bullet \left( \frac{3}{2} \right)^{x} + 9 = 0;\]

\[Пусть\ y = \left( \frac{3}{2} \right)^{x}:\]

\[4y^{2} - 13y + 9 = 0;\]

\[D = 13^{2} - 4 \bullet 4 \bullet 9 =\]

\[= 169 - 144 = 25\]

\[y_{1} = \frac{13 - 5}{2 \bullet 4} = \frac{8}{8} = 1;\]

\[y_{2} = \frac{13 + 5}{2 \bullet 4} = \frac{18}{8} = \frac{9}{4};\]

\[Первое\ значение:\]

\[\left( \frac{3}{2} \right)^{x} = 1;\]

\[\left( \frac{3}{2} \right)^{x} = \left( \frac{3}{2} \right)^{0};\ \]

\[x = 0;\]

\[Второе\ значение:\]

\[\left( \frac{3}{2} \right)^{x} = \frac{9}{4};\]

\[\left( \frac{3}{2} \right)^{x} = \left( \frac{3}{2} \right)^{2};\]

\[x = 2;\]

\[Ответ:\ \ x_{1} = 0;\ \ x_{2} = 2.\]

\[2)\ 16 \bullet 9^{x} - 25 \bullet 12^{x} +\]

\[+ 9 \bullet 16^{x} = 0\ \ \ \ \ |\ :16^{x};\]

\[16 \bullet \left( \frac{9}{16} \right)^{x} - 25 \bullet \left( \frac{12}{16} \right)^{x} +\]

\[+ 9 = 0;\]

\[16 \bullet \left( \frac{3}{4} \right)^{2x} - 25 \bullet \left( \frac{3}{4} \right)^{x} + 9 = 0;\]

\[Пусть\ y = \left( \frac{3}{4} \right)^{x}:\]

\[16y^{2} - 25y + 9 = 0;\]

\[D = 25^{2} - 4 \bullet 16 \bullet 9 =\]

\[= 625 - 576 = 49\]

\[y_{1} = \frac{25 - 7}{2 \bullet 16} = \frac{18}{32} = \frac{9}{16};\]

\[y_{2} = \frac{25 + 7}{2 \bullet 16} = \frac{32}{32} = 1;\]

\[Первое\ значение:\ \ \]

\[\left( \frac{3}{4} \right)^{x} = \frac{9}{16};\]

\[\left( \frac{3}{4} \right)^{x} = \left( \frac{3}{4} \right)^{2};\]

\[x = 2;\]

\[Второе\ значение:\]

\[\left( \frac{3}{4} \right)^{x} = 1;\]

\[\left( \frac{3}{4} \right)^{x} = \left( \frac{3}{4} \right)^{0};\]

\[x = 0;\]

\[Ответ:\ \ x_{1} = 2;\ \ x_{2} = 0.\]

\[3)\ \sqrt[x]{2} \bullet \sqrt[{2x}]{3} = 12;\]

\[\sqrt[{2x}]{2^{2}} \bullet \sqrt[{2x}]{3} = 12;\]

\[\sqrt[{2x}]{4 \bullet 3} = 12;\]

\[\sqrt[{2x}]{12} = 12;\]

\[12^{\frac{1}{2x}} = 12^{1};\]

\[\frac{1}{2x} = 1;\]

\[2x = 1;\]

\[x = 0,5;\]

\[Ответ:\ \ x = 0,5.\]

\[4)\ \sqrt[x]{5} \bullet 5^{x} = 25;\]

\[5^{\frac{1}{x}} \bullet 5^{x} = 25;\]

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

\[\frac{1}{x} + x = 2;\]

\[1 + x^{2} = 2x;\]

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

\[(x - 1)^{2} = 0;\]

\[x - 1 = 0;\]

\[x = 1;\]

\[Ответ:\ \ x = 1.\]