Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 6

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

\[1)\ x^{2} - 5 \geq 0\]

\[\left( x + \sqrt{5} \right)\left( x - \sqrt{5} \right) \geq 0\]

\[x \leq - \sqrt{5};\ \ x \geq \sqrt{5}.\]

\[2)\ 5x + 9 \geq 0\]

\[5x \geq - 9\]

\[x \geq - 1,8.\]

\[M = \left\lbrack \sqrt{5};\ + \infty \right).\]

\[3)\ x^{2} - 5 = 5x + 9\]

\[x^{2} - 5x - 14 = 0\]

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = - 14\]

\[x_{1} = 7;\ \ \]

\[x_{2} = - 2\ (не\ подходит).\]

\[Ответ:x = 7.\]

\[\textbf{б)}\ \sqrt[4]{2x^{2} - 1} = \sqrt[4]{3x - 2}\]

\[1)\ 2x^{2} - 1 \geq 0\]

\[x^{2} - \frac{1}{2} \geq 0\]

\[\left( x + \sqrt{0,5} \right)\left( x - \sqrt{0,5} \right) \geq 0\]

\[x \leq - \sqrt{0,5};\ \ x \geq \sqrt{0,5}.\]

\[\sqrt{0,5} = \frac{\sqrt{2}}{2}.\]

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

\[x \geq \frac{2}{3}.\]

\[M = \left\lbrack \frac{\sqrt{2}}{2}; + \infty \right).\]

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

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

\[D = 9 - 8 = 1\]

\[x_{1} = \frac{3 + 1}{4} = 1;\]

\[x_{2} = \frac{3 - 1}{4} = 0,5\ (не\ подходит).\]

\[Ответ:x = 1.\]

\[\textbf{в)}\ \sqrt{3x^{2} - 13} = \sqrt{5x - 1}\]

\[1)\ 3x^{2} - 13 \geq 0\]

\[3\left( x^{2} - \frac{13}{3} \right) \geq 0\]

\[\left( x + \sqrt{\frac{13}{3}} \right)\left( x - \sqrt{\frac{13}{3}} \right) \geq 0\]

\[x \leq - \sqrt{\frac{13}{3}};\ \ x \geq \sqrt{\frac{13}{3}}.\]

\[2)\ 5x - 1 \geq 0\]

\[5x \geq 1\]

\[x \geq 0,2.\]

\[M = \left\lbrack \sqrt{\frac{13}{3}}; + \infty \right).\]

\[3)\ 3x^{2} - 13 = 5x - 1\]

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

\[D = 25 + 144 = 169\]

\[x_{1} = \frac{5 + 13}{6} = 3;\]

\[x_{2} = \frac{5 - 13}{6} =\]

\[= - \frac{4}{3}\ (не\ подходит).\]

\[Ответ:x = 3.\]

\[\textbf{г)}\ \sqrt{4x^{2} - 11} = \sqrt{13x + 31}\]

\[1)\ 4x^{2} - 11 \geq 0\]

\[4\left( x^{2} - \frac{11}{4} \right) \geq 0\]

\[\left( x + \frac{\sqrt{11}}{2} \right)\left( x - \frac{\sqrt{11}}{2} \right) \geq 0\]

\[x \leq - \frac{\sqrt{11}}{2};\ \ x \geq \frac{\sqrt{11}}{2}.\]

\[2)\ 13x + 31 \geq 0\]

\[13x \geq - 31\]

\[x \geq - \frac{31}{13}\]

\[x \geq - 2\frac{5}{13}.\]

\[M =\]

\[= \left\lbrack - 2\frac{5}{13}; - \frac{\sqrt{11}}{2} \right\rbrack \cup \left\lbrack \frac{\sqrt{11}}{2}; + \infty \right).\]

\[3)\ 4x^{2} - 11 = 13x + 31\]

\[4x^{2} - 13x - 42 = 0\]

\[D = 169 + 672 = 841\]

\[x_{1} = \frac{13 + 29}{8} = 5,25;\]

\[x_{2} = \frac{13 - 29}{8} = - 2.\]

\[Ответ:x = - 2;\ \ x = 5,25.\]