Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 9

\[\boxed{\mathbf{9.}}\]

\[\textbf{а)}\ 2^{2x} = 2^{x - 9}\]

\[2x = x - 9\]

\[x = - 9.\]

\[\textbf{б)}\ 4^{2x - 7} = 4^{x - 1}\]

\[2x - 7 = x - 1\]

\[x = 6.\]

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

\[3x - 4 = x + 2\]

\[2x = 6\]

\[x = 3.\]

\[\textbf{г)}\ 3^{3x - 1} = 3^{7x - 2}\]

\[3x - 1 = 7x - 2\]

\[4x = 1\]

\[x = \frac{1}{4}.\]

\[\textbf{д)}\ 25^{x + 1} = 5^{x^{2} + 3x}\]

\[\left( 5^{2} \right)^{x + 1} = 5^{x^{2} + 3x}\]

\[2(x + 1) = x^{2} + 3x\]

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

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

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

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

\[\textbf{е)}\ 16^{x - 1} = 4^{x^{2} - x}\]

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

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

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

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

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

\[x_{1} = 1;\ \ \ x_{2} = 2.\]