Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 589

\[\boxed{\mathbf{589}.}\]

\[1)\ 2x - 1 \geq 2\ \ и\ \ 2(x - 1) \geq 1;\]

\[Преобразуем\ второе\]

\[\ неравенство:\]

\[2(x - 1) \geq 1;\]

\[2x - 2 \geq 1;\]

\[2x - 1 \geq 2;\]

\[Ответ:\ \ равносильны.\]

\[2)\ (x - 1)(x + 2) < 0\ \ и\]

\[\text{\ \ }x^{2} + x < 2;\]

\[Решим\ первое\ неравенство:\]

\[(x - 1)(x + 2) < 0;\]

\[(x + 2)(x - 1) < 0;\]

\[- 2 < x < 1;\]

\[Решим\ второе\ неравенство:\]

\[x^{2} + x < 2;\]

\[x^{2} + x - 2 < 0;\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[x_{1} = \frac{- 1 - 3}{2} = - 2\ \ и\ \]

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

\[(x + 2)(x - 1) < 0;\]

\[- 2 < x < 1;\]

\[Ответ:\ \ равносильны.\]

\[3)\ (x - 3)(x + 2) < 3x + 6\ \ и\ \]

\[\ x - 3 < 3;\]

\[Решим\ первое\ неравенство:\]

\[(x - 3)(x + 2) - 3 \cdot (x + 2) < 0\]

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

\[(x + 2)(x - 6) < 0\]

\[- 2 < x < 6.\]

\[Решим\ второе\ неравенство:\]

\[x - 3 < 3;\]

\[x < 3 + 3;\]

\[x < 6.\]

\[Ответ:\ \ не\ равносильны.\]

\[4)\ x(x + 3) \geq 2x\ \ и\]

\[\text{\ \ }x^{2}(x + 3) \geq 2x^{2};\]

\[Решим\ первое\ неравенство:\]

\[x(x + 3) \geq 2x;\]

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

\[x^{2} + x \geq 0;\]

\[(x + 1) \bullet x \geq 0;\]

\[x \leq - 1\ \ и\ \ x \geq 0;\]

\[Решим\ второе\ неравенство:\]

\[x^{2}(x + 3) \geq 2x^{2};\]

\[x^{2} \bullet (x + 3) - x^{2} \bullet 2 \geq 0;\]

\[x^{2} \bullet (x + 3 - 2) \geq 0;\]

\[x^{2} \bullet (x + 1) \geq 0;\]

\[x = 0\ \ и\ \ x \geq - 1;\]

\[Ответ:\ \ не\ равносильны.\]