Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1339

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\[1)\ |6 - 2x| = 3x + 1\]

\[6 - 2x \geq 0\]

\[2x \leq 6\]

\[x \leq 3.\]

\[x \leq 3:\]

\[6 - 2x = 3x + 1\]

\[- 5x = - 5\]

\[x = 1.\]

\[x > 3:\]

\[- (6 - 2x) = 3x + 1\]

\[- 6 + 2x = 3x + 1\]

\[- x = 7\]

\[x = - 7.\]

\[Ответ:\ \ x = 1.\]

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

\[x - 2 \geq 0\]

\[x \geq 2;\]

\[x \geq 0.\]

\[x \geq 2:\]

\[2(x - 2) = x - 1\]

\[2x - 4 = x - 1\]

\[2x - x = - 1 + 4\]

\[x = 3.\]

\[0 \leq x < 2:\]

\[- 2(x - 2) = x - 1\]

\[- 2x + 4 = x - 1\]

\[- 3x = - 5\]

\[x = \frac{5}{3} = 1\frac{2}{3}.\]

\[x < 0:\]

\[- 2(x - 2) = - x - 1\]

\[- 2x + 4 = - x - 1\]

\[- x = - 5\]

\[x = 5.\]

\[Ответ:\ \ x_{1} = 3;\ \ x_{2} = 1\frac{2}{3}.\]