ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1338

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

\[2x - 3 \geq 0\]

\[2x \geq 3\]

\[x \geq 1,5.\]

\[x \geq 1,5:\]

\[2x - 3 = 7\]

\[2x = 10\]

\[x = 5.\]

\[x < 1,5:\]

\[- (2x - 3) = 7\]

\[- 2x + 3 = 7\]

\[- 2x = 4\]

\[x = - 2.\]

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

\[2)\ |x + 6| = 2x\]

\[x + 6 \geq 0\]

\[x \geq - 6.\]

\[x \geq - 6:\]

\[x + 6 = 2x\]

\[- x = - 6\]

\[x = 6.\]

\[x < - 6:\]

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

\[- x - 6 = 2x\]

\[- 3x = 6\]

\[x = - 2.\]

\[Ответ:\ \ x = 6.\]

\[3)\ 2x - 7 = |x - 4|\]

\[x - 4 \geq 0\]

\[x \geq 4.\]

\[x \geq 4:\]

\[2x - 7 = x - 4\]

\[2x - x = - 4 + 7\]

\[x = 3.\]

\[x < 4:\]

\[2x - 7 = - (x - 4)\]

\[2x - 7 = - x + 4\]

\[3x = 11\]

\[x = \frac{11}{3} = 3\frac{2}{3}.\]

\[Ответ:\ \ x = 3\frac{2}{3}.\]