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МатематикаРешебник по алгебре 9 класс Макарычев Задание 375
Задание 375
\[\boxed{\text{375\ (375).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x^{3} + \frac{1}{x^{3}} = 22 \cdot \left( x + \frac{1}{x} \right)\]
\[Пусть\ \ t = x + \frac{1}{x} \neq 0;\ \ t^{3} =\]
\[= \left( x + \frac{1}{x} \right)^{3} = x^{3} + 3x + \frac{3}{x} + \frac{1}{x^{3}};\]
\[x^{3} + \frac{1}{x^{3}} = t^{3} - 3 \cdot \left( x + \frac{1}{x} \right) =\]
\[= t^{3} - 3t:\]
\[t^{3} - 3t = 22t\]
\[t^{3} - 25t = 0\]
\[t\left( t^{2} - 25 \right) = 0\]
\[t_{1} = 0;\ \ t_{2} = 5;\ \ t_{3} = - 5.\]
\[Так\ как\ t \neq 0 \Longrightarrow t = \pm 5.\]
\[1)\ x + \frac{1}{x} = 5\]
\[x^{2} - 5x + 1 = 0\]
\[D = 25 - 4 = 21\]
\[x_{1,2} = \frac{5 \pm \sqrt{21}}{2}.\]
\[2)\ x + \frac{1}{x} = - 5\ \ \]
\[x^{2} + 5x + 1 = 0\]
\[D = 25 - 4 = 21\]
\[x_{3,4} = \frac{- 5 \pm \sqrt{21}}{2}.\]
\[Ответ:x = \frac{5 \pm \sqrt{21}}{2};\ \]
\[x = \frac{- 5 \pm \sqrt{21}}{2}.\]
\[\textbf{б)}\ x^{3} - \frac{1}{x^{3}} = 19 \cdot \left( x - \frac{1}{x} \right)\]
\[Пусть\ \ t = x - \frac{1}{x};\ \ x^{3} - \frac{1}{x^{3}} =\]
\[= \left( x - \frac{1}{x} \right)\left( x^{2} + 1 + \frac{1}{x^{2}} \right);\]
\[x^{2} + \frac{1}{x^{2}} = t^{2} + 2;\]
\[x^{3} - \frac{1}{x^{3}} = t\left( t^{2} + 2 + 1 \right) =\]
\[= t\left( t^{2} + 3 \right):\]
\[t^{3} + 3t = 19t\]
\[t^{3} - 16t = 0\]
\[t\left( t^{2} - 16 \right) = 0\]
\[t(t - 4)(t + 4) = 0\]
\[t_{1} = 0;\ \ t_{2} = 4;\ \ t_{3} = - 4.\]
\[1)\ x - \frac{1}{x} = 0\ \]
\[x^{2} - 1 = 0\]
\[x^{2} = 1\]
\[x_{1,2} = \pm 1.\]
\[2)\ x - \frac{1}{x} = 4\ \ \]
\[x^{2} - 4x - 1 = 0,\]
\[D_{1} = 4 + 1 = 5\]
\[x_{3,4} = 2 \pm \sqrt{5}.\]
\[3)\ x - \frac{1}{x} = - 4\]
\[x^{2} + 4x - 1 = 0\ \]
\[D_{1} = 4 + 1 = 5\]
\[x_{5,6} = - 2 \pm \sqrt{5}.\]
\[Ответ:x = \pm 1;x = 2 \pm \sqrt{5};\ \ \]
\[x = - 2 \pm \sqrt{5}.\]
\[\boxed{\text{375.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[A(0,1;\ - 0,1);\ \ подставим:\]
\[\textbf{а)}\ x^{2} + y^{2} = 0,02\]
\[{0,1}^{2} + ( - {0,1)}^{2} = 0,01 + 0,01 =\]
\[= 0,02\]
\[Проходит.\]
\[\textbf{б)}\ x^{2} - y^{2} = 0\]
\[{0,1}^{2} - ( - 0,1)^{2} = 0,01 - 0,01 =\]
\[= 0.\]
\[Проходит.\]
