Решебник по алгебре 9 класс Макарычев Задание 600

\[\boxed{\text{600\ (600).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x³ + 4x² - 32x = 0\]

\[x\left( x^{2} + 4x - 32 \right) = 0\]

\[1)\ x_{1} = 0;\]

\[2)\ x² + 4x - 32 = 0\ \ \]

\[по\ теореме\ Виета:\]

\[x_{2} = - 8,\ \ x_{3} = 4.\]

\[\textbf{б)}\ x³ - 10x^{2} + 4x - 40 = 0\]

\[x^{2}(x - 10) + 4 \cdot (x - 10) = 0\]

\[\left( x^{2} + 4 \right)(x - 10) = 0\]

\[x = 10.\]

\[Ответ:\ \ \ а) - 8;0;4;\ \ б)\ 10.\]

\[\boxed{\text{600.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{1} = 2,\ \ x_{5} = x_{1} \cdot q^{4} = 162,\]

\[\ \ 2q^{4} = 162,\ \ \]

\[q^{4} = 81 \Longrightarrow q = \pm 3;\]

\[1)\ q = 3,\ \ x_{1} = 2,\ \ \]

\[x_{2} = x_{1} \cdot q = 2 \cdot 3 = 6;\]

\[x_{3} = x_{1} \cdot q^{2} = 2 \cdot 3^{2} = 2 \cdot 9 =\]

\[= 18;\]

\[x_{4} = x_{1} \cdot q^{3} = 2 \cdot 3^{3} = 54;\ \ \ \]

\[\ x_{5} = 162.\]