ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1576

\[\boxed{\mathbf{1576}\mathbf{.}}\]

\[\frac{x^{3} + x^{2} - 4x - 4}{x^{3} + 6x^{2} + 5x - 12} > 0\]

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

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

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

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

\[x^{3} + 6x^{2} + 5x - 12 =\]

\[= \left( x^{3} + 7x^{2} + 12x \right) + \left( - x^{2} - 7x - 12 \right) =\]

\[= x\left( x^{2} + 7x + 12 \right) - 1\left( x^{2} + 7x + 12 \right) =\]

\[= (x - 1)\left( x^{2} + 7x + 12 \right) =\]

\[= (x - 1)\left( x^{2} + 3x + 4x + 12 \right) =\]

\[= (x - 1)\left( x(x + 3) + 4(x + 3) \right) =\]

\[= (x - 1)(x + 4)(x + 3).\]

\[Получим:\]

\[\frac{(x - 2)(x + 2)(x + 1)}{(x - 1)(x + 4)(x + 3)} > 0\]

\[(x + 4)(x + 3)(x + 2)(x + 1)(x - 1)(x - 2) > 0\]

\[x < - 4;\text{\ \ \ }\]

\[- 3 < x < - 2;\]

\[- 1 < x < 1;\ \]

\[x > 2.\]