ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 226

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

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

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

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

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

\[= 169 - 144 = 25\]

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

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

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

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

\[x = 0.\]

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

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

\[x = 2.\]

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

\[16 \bullet \left( \frac{9}{16} \right)^{x} - 25 \bullet \left( \frac{12}{16} \right)^{x} + 9 = 0\]

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

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

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

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

\[= 625 - 576 = 49\]

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

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

\[1)\ \left( \frac{3}{4} \right)^{x} = \frac{9}{16}\]

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

\[x = 2.\]

\[2)\ \left( \frac{3}{4} \right)^{x} = 1\]

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

\[x = 0.\]

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

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

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

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

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

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

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

\[2x = 1\ \]

\[x = 0,5.\]

\[Ответ:\ \ x = 0,5.\]

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

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

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

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

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

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

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

\[x - 1 = 0\ \]

\[x = 1.\]

\[Ответ:\ \ x = 1.\]