Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1560

\[\boxed{\mathbf{1560}\mathbf{.}}\]

\[1)\ \sqrt{x^{2} + 4x + 4} - \sqrt{x^{2} - 6x + 9} = 6\]

\[\sqrt{(x + 2)^{2}} - \sqrt{(x - 3)^{2}} = 6\]

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

\[x + 2 \geq 0\]

\[x \geq - 2;\]

\[x - 3 \geq 0\]

\[x \geq 3.\]

\[x \geq 3:\]

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

\[0x + 5 = 6\]

\[0x = 1\]

\[корней\ нет.\]

\[- 2 \leq x < 3:\]

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

\[2x - 1 = 6\]

\[2x = 7\]

\[x = 3,5.\]

\[x < - 2:\]

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

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

\[0x - 5 = 6\]

\[0x = 11\]

\[корней\ нет.\]

\[Ответ:\ \ решений\ нет.\]

\[y = \sqrt[3]{8 - x};\text{\ \ }z = \sqrt[3]{27 + x}:\]

\[y^{3} + z^{3} = 8 - x + 27 + x = 35\]

\[(y + z)\left( y^{2} - yz + z^{2} \right) = 35\]

\[y^{2} - yz + z^{2} = 7.\]

\[Разделим\ первое\ на\ второе:\]

\[y + z = 5\]

\[z = 5 - y.\]

\[Подставим\ во\ 2\ уравнение:\]

\[y^{2} - y(5 - y) + (5 - y)^{2} = 7\]

\[y^{2} - 5y + y^{2} + 25 - 10y + y^{2} - 7 = 0\]

\[3y^{2} - 15y + 18 = 0\]

\[y^{2} - 5y + 6 = 0\]

\[D = 25 - 24 = 1\]

\[y_{1} = \frac{5 - 1}{2} = 2;\]

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

\[1)\ \sqrt[3]{8 - x} = 2\]

\[8 - x = 8\]

\[x = 0.\]

\[2)\ \sqrt[3]{8 - x} = 3\]

\[8 - x = 27\]

\[x = - 19.\]

\[Ответ:\ \ x_{1} = - 19;\ \ x_{2} = 0.\]

\[3)\ \sqrt[4]{8 - x} + \sqrt[4]{89 + x} = 5\]

\[y = \sqrt[4]{8 - x};\text{\ \ }z = \sqrt[4]{89 + x}:\]

\[y^{4} + z^{4} = 8 - x + 89 + x = 97\]

\[y + z = 5\]

\[z = 5 - y.\]

\[Подставим\ в\ первое\ уравнение:\]

\[y^{4} + (5 - y)^{4} = 97\]

\[y^{4} - 10y^{3} + 75y^{2} - 250y + 264 = 0\]

\[\left( y^{2} - 5y + 6 \right)\left( y^{2} - 5y + 44 \right) = 0\]

\[(y - 2)(y - 3)\left( y^{2} - 5y + 44 \right) = 0\]

\[D = 25 - 44 = - 19 < 0\]

\[y_{1} = 2\ \ и\ \ y_{2} = 3.\]

\[1)\ \sqrt[4]{8 - x} = 2\]

\[8 - x = 16\]

\[x = - 8.\]

\[2)\ \sqrt[4]{8 - x} = 3\]

\[8 - x = 81\]

\[x = - 73.\]

\[Ответ:\ \ x_{1} = - 73;\ \ x_{2} = - 8.\]