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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 7
Задание 7
\[\boxed{\mathbf{7.}}\]
\[\textbf{а)}\ \sqrt{x + 1} = x - 2\]
\[1)\ x + 1 \geq 0\]
\[x \geq - 1.\]
\[2)\ x - 2 \geq 0\]
\[x \geq 2.\]
\[M = \lbrack 2; + \infty).\]
\[3)\ x + 1 = (x - 2)^{2}\]
\[x + 1 = x^{2} - 4x + 4\]
\[x^{2} - 5x + 3 = 0\]
\[D = 25 - 12 = 13\]
\[x_{1} = \frac{5 + \sqrt{13}}{2};\]
\[x_{2} = \frac{5 - \sqrt{13}}{2}.\]
\[4)\ \sqrt{9} < \sqrt{13} < \sqrt{16}\]
\[3 < \sqrt{13} < 4\]
\[4 < \frac{5 + \sqrt{13}}{2} < 4,5.\]
\[- \sqrt{16} < - \sqrt{13} < - \sqrt{9}\]
\[- 4 < - \sqrt{13} < - 3\]
\[\frac{1}{2} < \frac{5 - \sqrt{13}}{2} < 1.\]
\[Ответ:x = \frac{5 + \sqrt{13}}{2}.\]
\[\textbf{б)}\ \sqrt{x - 1} = x - 4\]
\[1)\ x - 1 \geq 0\]
\[x \geq 1.\]
\[2)\ x - 4 \geq 0\]
\[x \geq 4.\]
\[M = \lbrack 4; + \infty).\]
\[3)\ x - 1 = (x - 4)^{2}\]
\[x - 1 = x^{2} - 8x + 16\]
\[x^{2} - 9x + 17 = 0\]
\[D = 81 - 68 = 13\]
\[x_{1} = \frac{9 + \sqrt{13}}{2};\]
\[x_{2} = \frac{9 - \sqrt{13}}{2}.\]
\[4)\ \sqrt{9} < \sqrt{13} < \sqrt{16}\]
\[3 < \sqrt{13} < 4\]
\[6 < \frac{9 + \sqrt{13}}{2} < 6,5.\]
\[- \sqrt{16} < - \sqrt{13} < - \sqrt{9}\]
\[- 4 < - \sqrt{13} < - 3\]
\[2,5 < \frac{9 - \sqrt{13}}{2} < 3.\]
\[Ответ:x = \frac{9 + \sqrt{13}}{2}.\]
\[\textbf{в)}\ \sqrt{x + 3} = x + 2\]
\[1)\ x + 3 \geq 0\]
\[x \geq - 3.\]
\[2)\ x + 2 \geq 0\]
\[x \geq - 2.\]
\[M = \lbrack - 2; + \infty).\]
\[3)\ x + 3 = (x + 2)^{2}\]
\[x + 3 = x^{2} + 4x + 4\]
\[x^{2} + 3x + 1 = 0\]
\[D = 9 - 4 = 5\]
\[x_{1} = \frac{- 3 + \sqrt{5}}{2};\]
\[x_{2} = \frac{- 3 - \sqrt{5}}{2}.\]
\[4)\ \sqrt{4} < \sqrt{5} < \sqrt{9}\]
\[2 < \sqrt{5} < 3\]
\[- 0,5 < \frac{- 3 + \sqrt{5}}{2} < 0.\]
\[- \sqrt{9} < - \sqrt{5} < - \sqrt{5}\]
\[- 3 < - \sqrt{5} < - 2\]
\[- 3 < \frac{- 3 - \sqrt{5}}{2} < - 2,5.\]
\[Ответ:x = \frac{- 3 + \sqrt{5}}{2}.\]
\[\textbf{г)}\ \sqrt{x} = x - 1\]
\[1)\ x \geq 0.\]
\[2)\ x - 1 \geq 0\]
\[x \geq 1.\]
\[M = \lbrack 1; + \infty).\]
\[3)\ x = (x - 1)^{2}\]
\[x = x^{2} - 2x + 1\]
\[x^{2} - 3x + 1 = 0\]
\[D = 9 - 4 = 5\]
\[x_{1} = \frac{3 + \sqrt{5}}{2};\]
\[x_{2} = \frac{3 - \sqrt{5}}{2}.\]
\[4)\ \sqrt{4} < \sqrt{5} < \sqrt{9}\]
\[2 < \sqrt{5} < 3\]
\[2,5 < \frac{3 + \sqrt{5}}{2} < 3.\]
\[- \sqrt{9} < - \sqrt{5} < - \sqrt{5}\]
\[- 3 < - \sqrt{5} < - 2\]
\[0 < \frac{3 - \sqrt{5}}{2} < 0,5.\]
\[Ответ:x = \frac{3 + \sqrt{5}}{2}.\]