ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 221

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\[1)\ 2^{|x - 2|} = 2^{|x + 4|}\]

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

\[\sqrt{(x - 2)^{2}} = \sqrt{(x + 4)^{2}}\]

\[(x - 2)^{2} = (x + 4)^{2}\]

\[x^{2} - 4x + 4 = x^{2} + 8x + 16\]

\[- 12x = 12\]

\[x = - 1\]

\[Ответ:\ \ x = - 1.\]

\[2)\ {1,5}^{|5 - x|} = {1,5}^{|x - 1|}\]

\[|5 - x| = |x - 1|\]

\[\sqrt{(5 - x)^{2}} = \sqrt{(x - 1)^{2}}\]

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

\[25 - 10x + x^{2} = x^{2} - 2x + 1\]

\[- 8x = - 24\]

\[x = 3\]

\[Ответ:\ \ x = 3.\]

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

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

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

\[(x + 1)^{2} = 4 - 4\sqrt{x^{2}} + x^{2}\]

\[x^{2} + 2x + 1 = 4 - 4\sqrt{x^{2}} + x^{2}\]

\[2x - 3 = - 4\sqrt{x^{2}}\]

\[4x^{2} - 12x + 9 = 16x^{2}\]

\[12x^{2} + 12x - 9 = 0\]

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

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

\[= 16 + 48 = 64\]

\[x_{1} = \frac{- 4 - 8}{2 \bullet 4} = - \frac{12}{8} = - \frac{3}{2} =\]

\[= - 1,5;\ \]

\[x_{2} = \frac{- 4 + 8}{2 \bullet 4} = \frac{4}{8} = \frac{1}{2} = 0,5.\]

\[Уравнение\ имеет\ решения\ при:\]

\[2x - 3 \leq 0\]

\[2x \leq 3\]

\[x \leq 1,5\]

\[Ответ:\ \ x_{1} = - 1,5;\ \ \ x_{2} = 0,5.\]

\[4)\ 3^{|x|} = 3^{|2 - x| - 1}\]

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

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

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

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

\[4 - 4x + x^{2} = x^{2} + 2\sqrt{x^{2}} + 1\]

\[3 - 4x = 2\sqrt{x^{2}}\]

\[9 - 24x + 16x^{2} = 4x^{2}\]

\[12x^{2} - 24x + 9 = 0\]

\[4x^{2} - 8x + 3 = 0\]

\[D = 8^{2} - 4 \bullet 4 \bullet 3 =\]

\[= 64 - 48 = 16\]

\[x_{1} = \frac{8 - 4}{2 \bullet 4} = \frac{4}{8} = \frac{1}{2} = 0,5;\ \]

\[x_{2} = \frac{8 + 4}{2 \bullet 4} = \frac{12}{8} = \frac{3}{2} = 1,5.\]

\[Уравнение\ имеет\ решени\ при:\]

\[3 - 4x \geq 0\]

\[4x \leq 3\]

\[x \leq 0,75\]

\[Ответ:\ \ x = 0,5.\]