Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 835

\[\sqrt{(x - 3)^{2}} + \sqrt{(5 + x)^{2}} = 8\]

\[|x - 3| + |x + 5| = 8\]

\[x \geq 3:\]

\[(x - 3) + (x + 5) = 8\]

\[2x + 2 = 8\]

\[2x = 6\]

\[x = 3.\]

\[- 5 \leq x \leq 3:\]

\[(3 - x) + (x + 5) = 8\]

\[8 = 8\]

\[x \in R.\]

\[x \leq - 5:\]

\[(3 - x) - (x + 5) = 8\]

\[- 2x - 2 = 8\]

\[2x = - 10\]

\[x = - 5.\]

\[Ответ:\ - 5 \leq x \leq 3.\]

\[2)\ \sqrt{x^{2} + 4x + 4} - \sqrt{x^{2} - 6x + 9} = 5\]

\[\sqrt{(x + 2)^{2}} - \sqrt{(x - 3)^{2}} = 5\]

\[|x + 2| - |x - 3| = 5\]

\[x \geq 3:\]

\[(x + 2) - (x - 3) = 5\]

\[5 = 5\]

\[x \in R.\]

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

\[(x + 2) + (x - 3) = 5\]

\[2x - 1 = 5\]

\[2x = 6\]

\[x = 3.\]

\[x \leq - 2:\]

\[- (x + 2) + (x - 3) = 5\]

\[- 5 = 5\]

\[x \in \varnothing.\]

\[Ответ:\ \ x \geq 3.\]

\[a = \sqrt[3]{8 - x};\ b = \sqrt[3]{27 + x}:\]

\[a^{2} - ab + b^{2} = 7\]

\[a^{3} + b^{3} = (8 - x) + (27 + x) = 35\]

\[(a + b)\left( a^{2} - ab + b^{2} \right) = 35\]

\[7(a + b) = 35\]

\[a + b = 5\]

\[a = 5 - b.\]

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

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

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

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

\[D = 25 - 24 = 1\]

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

\[b_{2} = \frac{5 + 1}{2} = 3;\]

\[27 + x = 8\]

\[x_{1} = - 19.\]

\[27 + x = 27\]

\[x_{2} = 0\]

\[Ответ:\ - 19;\ 0.\]

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

\[a = \sqrt[4]{8 - x};\ b = \sqrt[4]{89 + x}:\]

\[a^{4} + b^{4} = (8 - x) + (89 + x) = 97\]

\[a - b = 5\]

\[a = 5 + b.\]

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

\[2b^{4} + 20b^{3} + 150b^{2} + 500b + 528 = 0\]

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

\[(b + 2)\left( b^{3} + 8b^{2} + 59b + 132 \right) = 0\]

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

\[D = 25 - 176 = - 151 < 0\]

\[x \in \varnothing.\]

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

\[x \in \varnothing.\]

\[Ответ:\ \ корней\ нет.\]