Решебник по алгебре 9 класс Макарычев Задание 566

\[\boxed{\text{566}\text{\ (566)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[b_{n} = 2n^{2} + 3n\]

\[\textbf{а)}\ b_{5} = 2 \cdot 5² + 3 \cdot 5 = 50 + 15 =\]

\[= 65;\]

\[\textbf{б)}\ b_{10} = 2 \cdot 10² + 3 \cdot 10 =\]

\[= 200 + 30 = 230;\]

\[\textbf{в)}\ b_{50} = 2 \cdot 50² + 3 \cdot 50 = 5150.\]

\[\boxed{\text{566.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x³ + 4x² - 32x = 0\]

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

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

\[2)\ x² + 4x - 32 = 0\ \ \]

\[по\ теореме\ Виета:\]

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

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

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

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

\[x = 10.\]

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