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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 696
Задание 696
\[\boxed{\mathbf{696}.}\]
\[1)\ 4 \bullet 9^{x} - 13 \bullet 6^{x} +\]
\[+ 9 \bullet 4^{x} = 0\ \ \ \ \ |\ :4^{x};\]
\[4 \bullet \left( \frac{9}{4} \right)^{x} - 13 \bullet \left( \frac{6}{4} \right)^{x} + 9 = 0;\]
\[4 \bullet \left( \frac{3}{2} \right)^{2x} - 13 \bullet \left( \frac{3}{2} \right)^{x} + 9 = 0;\]
\[Пусть\ y = \left( \frac{3}{2} \right)^{x}:\]
\[4y^{2} - 13y + 9 = 0;\]
\[D = 13^{2} - 4 \bullet 4 \bullet 9 =\]
\[= 169 - 144 = 25\]
\[y_{1} = \frac{13 - 5}{2 \bullet 4} = \frac{8}{8} = 1;\]
\[y_{2} = \frac{13 + 5}{2 \bullet 4} = \frac{18}{8} = \frac{9}{4};\]
\[Первое\ значение:\]
\[\left( \frac{3}{2} \right)^{x} = 1;\]
\[\left( \frac{3}{2} \right)^{x} = \left( \frac{3}{2} \right)^{0};\ \]
\[x = 0;\]
\[Второе\ значение:\]
\[\left( \frac{3}{2} \right)^{x} = \frac{9}{4};\]
\[\left( \frac{3}{2} \right)^{x} = \left( \frac{3}{2} \right)^{2};\]
\[x = 2;\]
\[Ответ:\ \ x_{1} = 0;\ \ x_{2} = 2.\]
\[2)\ 16 \bullet 9^{x} - 25 \bullet 12^{x} +\]
\[+ 9 \bullet 16^{x} = 0\ \ \ \ \ |\ :16^{x};\]
\[16 \bullet \left( \frac{9}{16} \right)^{x} - 25 \bullet \left( \frac{12}{16} \right)^{x} +\]
\[+ 9 = 0;\]
\[16 \bullet \left( \frac{3}{4} \right)^{2x} - 25 \bullet \left( \frac{3}{4} \right)^{x} + 9 = 0;\]
\[Пусть\ y = \left( \frac{3}{4} \right)^{x}:\]
\[16y^{2} - 25y + 9 = 0;\]
\[D = 25^{2} - 4 \bullet 16 \bullet 9 =\]
\[= 625 - 576 = 49\]
\[y_{1} = \frac{25 - 7}{2 \bullet 16} = \frac{18}{32} = \frac{9}{16};\]
\[y_{2} = \frac{25 + 7}{2 \bullet 16} = \frac{32}{32} = 1;\]
\[Первое\ значение:\ \ \]
\[\left( \frac{3}{4} \right)^{x} = \frac{9}{16};\]
\[\left( \frac{3}{4} \right)^{x} = \left( \frac{3}{4} \right)^{2};\]
\[x = 2;\]
\[Второе\ значение:\]
\[\left( \frac{3}{4} \right)^{x} = 1;\]
\[\left( \frac{3}{4} \right)^{x} = \left( \frac{3}{4} \right)^{0};\]
\[x = 0;\]
\[Ответ:\ \ x_{1} = 2;\ \ x_{2} = 0.\]
\[3)\ \sqrt[x]{2} \bullet \sqrt[{2x}]{3} = 12;\]
\[\sqrt[{2x}]{2^{2}} \bullet \sqrt[{2x}]{3} = 12;\]
\[\sqrt[{2x}]{4 \bullet 3} = 12;\]
\[\sqrt[{2x}]{12} = 12;\]
\[12^{\frac{1}{2x}} = 12^{1};\]
\[\frac{1}{2x} = 1;\]
\[2x = 1;\]
\[x = 0,5;\]
\[Ответ:\ \ x = 0,5.\]
\[4)\ \sqrt[x]{5} \bullet 5^{x} = 25;\]
\[5^{\frac{1}{x}} \bullet 5^{x} = 25;\]
\[5^{\frac{1}{x} + x} = 5^{2};\]
\[\frac{1}{x} + x = 2;\]
\[1 + x^{2} = 2x;\]
\[x^{2} - 2x + 1 = 0;\]
\[(x - 1)^{2} = 0;\]
\[x - 1 = 0;\]
\[x = 1;\]
\[Ответ:\ \ x = 1.\]
