Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 8

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)}\ \sqrt{- 2x} = |x + 1|\]

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

\[x \leq 0.\]

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

\[x \in R.\]

\[M = ( - \infty;0\rbrack.\]

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

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

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

\[D_{1} = 4 - 1 = 3\]

\[x_{1} = - 2 + \sqrt{3} < 0;\]

\[x_{2} = - 2 - 2\sqrt{3} < 0.\]

\[Ответ:x = - 2 \pm \sqrt{3}.\]

\[\textbf{б)}\ \sqrt{4x} = |x - 2|\]

\[1)\ 4x \geq 0\]

\[x \geq 0.\]

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

\[x \in R.\]

\[M = \lbrack 0; + \infty).\]

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

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

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

\[D_{1} = 16 - 4 = 12 = \left( 2\sqrt{3} \right)^{2}\]

\[x_{1} = 4 + 2\sqrt{3} > 0;\]

\[x_{2} = 4 - 2\sqrt{3} > 0.\]

\[Ответ:x = 4 \pm 2\sqrt{3}.\]

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

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

\[x \leq 2.\]

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

\[x \in R.\]

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

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

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

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

\[D = 25 - 28 = - 3 < 0\]

\[нет\ корней.\]

\[Ответ:нет\ корней.\]

\[\textbf{г)}\ \sqrt{5 - x} = |x - 2|\]

\[1)\ 5 - x \geq 0\]

\[x \leq 5.\]

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

\[x \in R.\]

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

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

\[5 - x = x^{2} - 4x + 4\]

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

\[D = 9 + 4 = 13\]

\[x_{1} = \frac{3 + \sqrt{13}}{2} < 5;\]

\[x_{2} = \frac{3 - \sqrt{13}}{2} < 5.\]

\[Ответ:x = \frac{3 \pm \sqrt{13}}{2}.\]