Решебник по алгебре 9 класс Макарычев Задание 350

 

\[\boxed{\text{350\ (350).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[10x^{4} - 77x^{3} + 150x^{2} -\]

\[- 77x + 10 = 0\]

\[10x^{2} - 77x + 150 -\]

\[- \frac{77}{x} + \frac{10}{x^{2}} = 0\]

\[10 \cdot \left( x^{2} + \frac{1}{x^{2}} \right) - 77 \cdot \left( x + \frac{1}{x} \right) +\]

\[+ 150 = 0\]

\[Пусть\ t = x + \frac{1}{x};\ t^{2} =\]

\[= \left( x + \frac{1}{x} \right)^{2} = x^{2} + 2 + \frac{1}{x^{2}};\]

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

\[10 \cdot \left( t^{2} - 2 \right) - 77t + 150 = 0\]

\[10t^{2} - 20 - 77t + 150 = 0\]

\[10t^{2} - 77t + 130 = 0\]

\[D = 77^{2} - 4 \cdot 10 \cdot 130 =\]

\[= 5929 - 5200 = 729\]

\[t_{1,2} = \frac{77 \pm 27}{20} = 5,2;2,5.\]

\[1)\ x + \frac{1}{x} = 5,2\]

\[x^{2} - 5,2x + 1 = 0\]

\[D = 27,04 - 4 = 23,04\]

\[x_{1,2} = \frac{5,2 \pm 4,8}{2} = 5;\frac{1}{5}.\]

\[2)\ x + \frac{1}{x} = 2,5\]

\[x^{2} - 2,5x + 1 = 0\]

\[D = 6,25 - 4 = 2,25\]

\[x_{1,2} = \frac{2,5 \pm 1,5}{2} = 2;\frac{1}{2}.\]

\[Ответ:0,2;0,5;2;5.\]

\[\boxed{\text{350.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ (3x - 5)(x + 4)(2 - x) = 0\]

\[x_{1} = \frac{5}{3} = 1\frac{2}{3},\ \ x_{2} = - 4,\ \ \]

\[x_{3} = 2.\ \]

\[\textbf{б)}\ (3x - 5)(x + 4)(2 - x) > 0\]

\[\left( x - \frac{5}{3} \right)(x + 4)(x - 2) < 0\]

\[x \in ( - \infty;\ - 4) \cup \left( 1\frac{2}{3};2 \right).\]

\[\textbf{в)}\ (3x - 5)(x + 4)(2 - x) < 0\]

\[\left( x - \frac{5}{3} \right)(x + 4)(x - 2) > 0\]

\[x \in \left( - 4;1\frac{2}{3} \right) \cup (2; + \infty).\]