Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 13

\[\boxed{\mathbf{13.}}\]

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

\[1)\ 2 - x \geq 0\]

\[x \leq 2.\]

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

\[2x \leq 3\]

\[x \leq 1,5.\]

\[M = ( - \infty;1,5\rbrack.\]

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

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

\[9 - 12x + 4x^{2} =\]

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

\[x^{3} - 2x^{2} + 1 = 0\]

\[\left( x^{3} - x^{2} \right) - \left( x^{2} - 1 \right) = 0\]

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

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

\[x - 1 = 0\]

\[x = 1 < 1,5.\]

\[x^{2} - x - 1 = 0\]

\[D = 1 + 4 = 5\]

\[x_{1} = \frac{1 + \sqrt{5}}{2} > 1,5;\]

\[x_{2} = \frac{1 - \sqrt{5}}{2} < 1,5.\]

\[Ответ:x = 1;\ \ x = \frac{1 - \sqrt{5}}{2}\text{.\ }\]

\[\textbf{б)}\ \sqrt[3]{5 - 2x} = \sqrt{3 - x}\]

\[1)\ 3 - x \geq 0\]

\[x \leq 3.\]

\[2)\ 5 - 2x \geq 0\]

\[2x \leq 5\]

\[x \leq 2,5.\]

\[M = ( - \infty;2,5\rbrack.\]

\[3)\ \left( \sqrt[3]{5 - 2x} \right)^{6} = \left( \sqrt{3 - x} \right)^{6}\]

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

\[25 - 20x + 4x^{2} =\]

\[= 27 - 27x + 9x^{2} - x^{3}\]

\[x^{3} - 5x^{2} + 7x - 2 = 0\]

\[Один\ из\ корней\ x = 2:\]

\[(x - 2)\left( x^{2} - 3x + 1 \right) = 0\]

\[x^{2} - 3x + 1 = 0\]

\[D = 9 - 4 = 5\]

\[x_{1} = \frac{3 + \sqrt{5}}{2} > 2,5;\]

\[x_{2} = \frac{3 - \sqrt{5}}{2} < 2,5.\]

\[Ответ:x = 2;\ \ x = \frac{3 - \sqrt{5}}{2}.\]

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

\[1)\ x - 3 \geq 0\]

\[x \geq 3.\]

\[2)\ x + 1 \geq 0\]

\[x \geq - 1.\]

\[M = \lbrack 3; + \infty).\]

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

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

\[x^{2} + 2x + 1 =\]

\[= x^{3} - 9x^{2} + 27x - 27\]

\[x^{3} - 10x^{2} + 25x - 28 = 0\]

\[Один\ из\ корней\ x = 7:\]

\[(x - 7)\left( x^{2} - 3x + 4 \right) = 0\]

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

\[D = 9 - 16 = - 7 < 0\]

\[нет\ корней.\]

\[Ответ:x = 7.\]

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

\[1)\ x + 2 \geq 0\]

\[x \geq - 2.\]

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

\[2x \geq - 3\]

\[x \geq - 1,5.\]

\[M = \lbrack - 1,5; + \infty).\]

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

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

\[4x^{2} + 12x + 9 =\]

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

\[x^{3} + 2x^{2} - 1 = 0\]

\[\left( x^{3} + x^{2} \right) + \left( x^{2} - 1 \right) = 0\]

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

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

\[x + 1 = 0\]

\[x = - 1.\]

\[x^{2} + x - 1 = 0\]

\[D = 1 + 4 = 5\]

\[x_{1} = \frac{- 1 + \sqrt{5}}{2} > - 1,5;\]

\[x_{2} = \frac{- 1 - \sqrt{5}}{2} < - 1,5.\]

\[Ответ:x = - 1;\ \ x = \frac{- 1 + \sqrt{5}}{2}.\]