ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 746

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\[1)\ \frac{{0,2}^{x + 0,5}}{\sqrt{5}} = 5 \bullet {0,04}^{x}\ \]

\[\left( \frac{1}{5} \right)^{x + 0,5}\ :5^{\frac{1}{2}} = 5 \bullet \left( \frac{4}{100} \right)^{x}\ \]

\[\left( \frac{1}{5} \right)^{x + 0,5}\ :\left( \frac{1}{5} \right)^{- 0,5} = 5 \bullet \left( \frac{1}{25} \right)^{x}\ \]

\[\left( \frac{1}{5} \right)^{x + 0,5 - ( - 0,5)} = \left( \frac{1}{5} \right)^{- 1} \bullet \left( \frac{1}{5} \right)^{2x}\ \]

\[\left( \frac{1}{5} \right)^{x + 1} = \left( \frac{1}{5} \right)^{2x - 1}\ \]

\[x + 1 = 2x - 1\ \]

\[- x = - 2\ \]

\[x = 2\ \]

\[Ответ:\ \ x = 2.\]

\[2)\ 4 \bullet 3^{x} - 9 \bullet 2^{x} =\]

\[= 5 \bullet 3^{\frac{x}{2}} \bullet 2^{\frac{x}{2}}\ \ \ \ \ |\ :2^{x}\ \]

\[4 \bullet \frac{3^{x}}{2^{x}} - 9 = 5 \bullet \frac{3^{\frac{x}{2}}}{2^{\frac{x}{2}}}\ \]

\[4 \bullet \left( \frac{3}{2} \right)^{x} - 5 \bullet \left( \frac{3}{2} \right)^{\frac{x}{2}} - 9 = 0\ \]

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

\[4y^{2} - 5y - 9 = 0\ \]

\[D = 5^{2} + 4 \bullet 4 \bullet 9 =\]

\[= 25 + 144 = 169\]

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

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

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

\[нет\ корней.\ \]

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

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

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

\[x = 4\ \]

\[Ответ:\ \ x = 4.\]

\(3)\ 2 \bullet 4^{x} - 3 \bullet 10^{x} - 5 \bullet 25^{x}\)=

\[= 0\ \ \ \ \ |\ :25^{x}\ \]

\[2 \bullet \frac{4^{x}}{25^{x}} - 3 \bullet \frac{10^{x}}{25^{x}} - 5 = 0\ \]

\[2 \bullet \left( \frac{2}{5} \right)^{2x} - 3 \bullet \left( \frac{2}{5} \right)^{x} - 5 = 0\ \]

\[Пусть\ y = \left( \frac{2}{5} \right)^{x}:\]

\[2y^{2} - 3y - 5 = 0\ \]

\[D = 3^{2} + 4 \bullet 2 \bullet 5 =\]

\[= 9 + 40 = 49\]

\[y_{1} = \frac{3 - 7}{2 \bullet 2} = - \frac{4}{4} = - 1;\ \]

\[y_{1} = \frac{3 + 7}{2 \bullet 2} = \frac{10}{4} = \frac{5}{2}\text{\ .}\]

\[1)\ \left( \frac{2}{5} \right)^{x} = - 1\]

\[нет\ корней.\ \]

\[2)\ \left( \frac{2}{5} \right)^{x} = \frac{5}{2}\ \]

\[\left( \frac{2}{5} \right)^{x} = \left( \frac{2}{5} \right)^{- 1}\ \]

\[x = - 1.\ \]

\[Ответ:\ \ x = - 1.\]

\[4)\ 4 \bullet 9^{x} + 12^{x} -\]

\[- 3 \bullet 16^{x} = 0\ \ \ \ \ |\ :16^{x}\ \]

\[4 \bullet \frac{9^{x}}{16^{x}} + \frac{12^{x}}{16^{x}} - 3 = 0\ \]

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

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

\[4y^{2} + y - 3 = 0\ \]

\[D = 1^{2} + 4 \bullet 4 \bullet 3 =\]

\[= 1 + 48 = 49\]

\[y_{1} = \frac{- 1 - 7}{2 \bullet 4} = - \frac{8}{8} = - 1;\ \]

\[y_{2} = \frac{- 1 + 7}{2 \bullet 4} = \frac{6}{8} = \frac{3}{4}.\]

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

\[нет\ корней.\ \]

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

\[x = 1.\ \]

\[Ответ:\ \ x = 1.\]