Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 12

\[\boxed{\mathbf{12.}}\]

\[\textbf{а)}\ \sqrt[3]{x} = \sqrt{x - 4}\]

\[x - 4 \geq 0\]

\[x \geq 4.\]

\[x \geq 0.\]

\[M = \lbrack 4; + \infty).\]

\[\left( \sqrt[3]{x} \right)^{6} = \left( \sqrt{x - 4} \right)^{6}\]

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

\[x^{2} = x^{3} - 12x^{2} + 48x - 64\]

\[x^{3} - 13x^{2} + 48x - 64 = 0\]

\[Один\ из\ корней\ x = 8 > 4:\]

\[(x - 8)\left( x^{2} - 5x + 8 \right) = 0\]

\[x^{2} - 5x + 8 = 0\]

\[D = 25 - 32 = - 7 < 0\]

\[корней\ нет.\]

\[Ответ:x = 8.\]

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

\[2 - x \geq 0\]

\[x \leq 2.\]

\[x \geq 0.\]

\[M = \lbrack 0;2\rbrack.\]

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

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

\[x^{2} = 8 - 12x + 6x^{2} - x^{3}\]

\[x^{3} - 5x^{2} + 12x - 8 = 0\]

\[Один\ из\ корней\ x = 1:\]

\[(x - 1)\left( x^{2} - 4x + 8 \right) = 0\]

\[x^{2} - 4x + 8 = 0\]

\[D_{1} = 4 - 8 = - 4 < 0\]

\[нет\ корней.\]

\[Ответ:x = 1.\]

\(в)\ \sqrt[3]{x + 3} = \sqrt{x - 1}\)

\[1)\ x - 1 \geq 0\]

\[x \geq 1.\]

\[2)\ x + 3 \geq 0\]

\[x \geq - 3.\]

\[M = \lbrack 1; + \infty).\]

\[\left( \sqrt[3]{x + 3} \right)^{6} = \left( \sqrt{x - 1} \right)^{6}\]

\[(x + 3)^{2} = (x - 1)^{3}\]

\[x^{2} + 6x + 9 = x^{3} - 3x^{2} + 3x - 1\]

\[x^{3} - 4x^{2} - 3x - 10 = 0\]

\[Один\ из\ корней\ x = 5:\]

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

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

\[D = 1 - 8 = - 7 < 0\]

\[нет\ корней.\]

\[Ответ:x = 5.\]

\[\textbf{г)}\ \sqrt[3]{x + 2} = \sqrt{- x}\]

\[1) - x \geq 0\]

\[x \leq 0.\]

\[2)\ x + 2 \geq 0\]

\[x \geq - 2.\]

\[M = \lbrack - 2;0\rbrack.\]

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

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

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

\[x^{3} + x^{2} + 4x + 4 = 0\]

\[x^{2}(x + 1) + 4(x + 1) = 0\]

\[(x + 1)\left( x^{2} + 4 \right) = 0\]

\[x = - 1.\]

\[Ответ:x = - 1.\]