Решебник по алгебре 9 класс Макарычев Задание 310

 

\[\boxed{\text{310\ (310).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Для\ того,\ чтобы\ уравнение\ \]

\[имело\ два\ корня,\ \]

\[необходимо:D > 0.\]

\[\textbf{а)}\ 3x^{2} + bx + 3 = 0\]

\[D = b^{2} - 4 \cdot 3 \cdot 3 = b^{2} - 36\]

\[b^{2} - 36 > 0\]

\[(b - 6)(b + 6) > 0;\]

\[b \in ( - \infty;\ - 6) \cup (6; + \infty).\]

\[\textbf{б)}\ x^{2} + 2bx + 15 = 0\]

\[D = b^{2} - 15\]

\[b^{2} - 15 > 0\]

\[\left( b - \sqrt{15} \right)\left( b + \sqrt{15} \right) > 0;\]

\[b \in \left( - \infty; - \sqrt{15} \right) \cup \left( \sqrt{15}; + \infty \right).\]

\[\boxed{\text{310.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

\[= 0\]

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

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

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

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

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

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

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

\[= 5929 - 5200 = 729\]

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

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

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

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

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

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

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

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

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

\[Ответ:0,2;0,5;2;5.\]