Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 140

\[\boxed{\mathbf{140.}}\]

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

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

\[2x - 2 \geq 1\]

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

\[Ответ:\ \ равносильны.\]

\[2)\ (x - 1)(x + 2) < 0\ \ и\ \ \]

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

\[Решим\ первое\ неравенство:\]

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

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

\[- 2 < x < 1.\]

\[Решим\ второе\ неравенство:\]

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

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

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

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

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

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

\[- 2 < x < 1.\]

\[Ответ:\ \ равносильны.\]

\[3)\ (x - 2)(x + 1) < 3x + 3\ \ и\ \ \]

\[x - 2 < 3\]

\[Решим\ первое\ неравенство:\]

\[(x - 2)(x + 1) < 3x + 3\]

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

\[x^{2} - 4x - 5 < 0\]

\[D = 4^{2} + 4 \bullet 5 = 16 + 20 = 36\]

\[x_{1} = \frac{4 - 6}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{4 + 6}{2} = 5\]

\[(x + 1)(x - 5) < 0\]

\[- 1 < x < 5.\]

\[Решим\ второе\ неравенство:\]

\[x - 2 < 3\]

\[x < 3 + 2\]

\[x < 5.\]

\[Ответ:\ \ не\ равносильны.\]

\[4)\ x(x + 3) \geq 2x\ \ и\ \ \]

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

\[Решим\ первое\ неравенство:\]

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

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

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

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

\[x \leq - 1;\text{\ \ }x \geq 0.\]

\[Решим\ второе\ неравенство:\]

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

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

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

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

\[x = 0;\ \ x \geq - 1.\]

\[Ответ:\ \ не\ равносильны.\]