ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 14

\[\boxed{\mathbf{14.}}\]

\[\textbf{а)}\lg\left( x^{2} - x - 6 \right) + 4^{x} + 16 =\]

\[= 17 \cdot 2^{x} + \lg\left( x^{2} - x - 6 \right)\]

\[\left( 2^{x} \right)^{2} - 17 \cdot 2^{x} + 16 = 0\]

\[\left\{ \begin{matrix} \left( 2^{x} \right)^{2} - 17 \cdot 2^{x} + 16 = 0 \\ x^{2} - x - 6 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - x - 6 = 0\]

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3;\ \ x_{2} = - 2;\]

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

\[x < - 2;\ \ x > 3.\]

\[\left\{ \begin{matrix} \left( 2^{x} \right)^{2} - 17 \cdot 2^{x} + 16 = 0 \\ x < - 2;\ \ \ \ x > 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left( 2^{x} \right)^{2} - 17 \cdot 2^{x} + 16 = 0\]

\[2^{x} = t:\]

\[t^{2} - 17t + 16 = 0\]

\[t_{1} + t_{2} = 17;\ \ t_{1} \cdot t_{2} = 16\]

\[t_{1} = 1;\ \ \ t_{2} = 16.\]

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

\[2^{x} = 2^{0}\]

\[x = 0 - не\ корень.\]

\[2)\ 2^{x} = 16\]

\[2^{x} = 2^{4}\]

\[x = 4 > 3 - корень.\]

\[Ответ:x = 4.\]

\[\textbf{б)}\ \sqrt{x^{2} - 9} + \lg\left( x^{2} + 3x \right) =\]

\[\text{=}1 + \sqrt{x^{2} - 9}\]

\[\sqrt{x^{2} - 9} + \lg\left( x^{2} + 3x \right) - \sqrt{x^{2} - 9} = \lg 10\]

\[\lg\left( x^{2} + 3x \right) = \lg 10\]

\[\left\{ \begin{matrix} x^{2} + 3x = 10 \\ x^{2} - 9 \geq 0\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} + 3x - 10 = 0\ \ \ \\ (x + 3)(x - 3) \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} + 3x - 10 = 0 \\ x \leq - 3;\ \ \ \ \ \ \ x \geq 3 \\ \end{matrix} \right.\ \]

\[x^{2} + 3x - 10 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 10\]

\[x_{1} = - 5 < - 3 - корень;\ \ \]

\[x_{2} = 2 - не\ корень.\]

\[Ответ:x = - 5.\]

\[\textbf{в)}\ \sqrt{- x^{2} + 4x - 3,5} + 9^{x} + 243 =\]

\[= 36 \cdot 3^{x} + \sqrt{- x^{2} + 4x - 3,5}\]

\[9^{x} - 36 \cdot 3^{x} + 243 = 0\]

\[\left\{ \begin{matrix} \left( 3^{x} \right)^{2} - 36 \cdot 3^{x} + 243 = 0 \\ - x^{2} + 4x - 3,5 \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- x^{2} + 4x - 3,5 \geq 0\]

\[x^{2} - 4x + 3,5 \leq 0\ \ \ \]

\[D_{1} = 4 - 3,5 = 0,5\]

\[x_{1} = 2 - \sqrt{0,5};\]

\[x_{2} = 2 + \sqrt{0,5}.\]

\[\left\{ \begin{matrix} \left( 3^{x} \right)^{2} - 36 \cdot 3^{x} + 243 = 0 \\ 2 - \sqrt{0,5} < x < 2 + \sqrt{0,5}\ \\ \end{matrix} \right.\ \]

\[3^{x} = t:\]

\[t^{2} - 36t + 243 = 0\]

\[D_{1} = 324 - 243 = 81\]

\[t_{1} = 18 + 9 = 27;\]

\[t_{2} = 18 - 9 = 9.\]

\[1)\ 3^{x} = 27\]

\[3^{x} = 3^{3}\]

\[x = 3 - не\ корень.\]

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

\[3^{x} = 3^{2}\]

\[x = 2 - корень.\ \]

\[Ответ:x = 2.\]

\[\textbf{г)}\lg\left( x^{2} + 21x \right) + tg\frac{\text{πx}}{2} =\]

\[= 2 + tg\frac{\text{πx}}{2}\]

\[\lg\left( x^{2} + 21x \right) = 2\]

\[\left\{ \begin{matrix} \lg{(x^{2} + 21x)} - \lg 10^{2} = 0 \\ \frac{\text{πx}}{2} \neq \frac{\pi}{2} + \pi n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\frac{\text{πx}}{2} \neq \frac{\pi}{2} + \pi n\]

\[\text{πx} \neq \pi + 2\pi n\]

\[x \neq 1 + 2n.\]

\[x^{2} + 21x - 100 = 0\]

\[x_{1} + x_{2} = - 21;\ \ \]

\[x_{1} \cdot x_{2} = - 100\]

\[x_{1} = - 25\ (не\ подходит);\ \ \ \]

\[x_{2} = 4 - корень\ уравнения.\]

\[Ответ:x = 4.\]