ГДЗ по алгебре 9 класс Макарычев Задание 349

 

\[\boxed{\text{349\ (349).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ x^{3} - x + 3 = 0 \Longrightarrow среди\ \]

\[делителей\ 3\ нет\ целых\ корней;\]

\[2)\ x^{4} + x^{2} - 20 = 0\]

\[\left( x^{2} - 4 \right)\left( x^{2} + 5 \right) = 0\]

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

\[3)\ x^{4} + 5x^{2} + 4 = 0\]

\[\left( x^{2} + 1 \right)\left( x^{2} + 4 \right) = 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет;\]

\[4)\ x^{3} - 5x + 4 = 0\]

\[(x - 1)\left( x^{2} + x - 4 \right) = 0\]

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

\(D = 1 + 16 = 17\)

\[\Longrightarrow целый\ корень\ только\ \]

\[один:\ \ x = 1.\]

\[Ответ:\ \ 4).\]

\[\boxed{\text{349.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ (x + 1,2)(6 - x)(x - 4) > 0\]

\[(x + 1,2)(x - 4)(x - 6) < 0\]

\[x \in ( - \infty;\ - 1,2) \cup (4;6).\]

\[\textbf{б)}\ \left( \frac{1}{3} - x \right)(0,5 - x)\left( \frac{1}{7} - x \right) < 0\]

\[\left( x - \frac{1}{7} \right)\left( x - \frac{1}{3} \right)(x - 0,5) > 0\]

\[x \in \left( \frac{1}{7};\frac{1}{3} \right) \cup (0,5; + \infty).\]

\[\textbf{в)}\ (x + 0,6)(1,6 + x)(1,2 - x) > 0\]

\[(x + 1,6)(x + 0,6)(x - 1,2) < 0\]

\[x \in ( - \infty; - 1,6) \cup ( - 0,6;1,2).\]

\[\textbf{г)}\ (1,7 - x)(1,8 + x)(1,9 - x) < 0\]

\[(x + 1,8)(x - 1,7)(x - 1,9) < 0\]

\[x \in ( - \infty; - 1,8) \cup (1,7;1,9).\]