ГДЗ по алгебре 9 класс Макарычев Задание 839

 

\[\boxed{\text{839\ (839).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{(n + 1)!}{n!} = \frac{n! \cdot (n + 1)}{n!} =\]

\[= n + 1.\]

\[\textbf{б)}\ \frac{n!}{(n + 2)!} =\]

\[= \frac{n!}{n!(n + 1)(n + 2)} =\]

\[= \frac{1}{(n + 1)(n + 2)}\]

\[\textbf{в)}\ \frac{(n + 3)!}{(n + 1)!} =\]

\[= \frac{(n + 1)!(n + 2)(n + 3)}{(n + 1)!} =\]

\[= (n + 2)(n + 3).\]

\[\textbf{г)}\ \frac{(n + 1)!(n + 3)}{(n + 4)!} =\]

\[= \frac{(n + 1)(n + 3)}{(n + 1)!(n + 2)(n + 3)(n + 4)} =\]

\[= \frac{1}{(n + 2)(n + 4)}.\]

\[\textbf{д)}\ \frac{(n + 11)!n}{(n + 10)!} =\]

\[= \frac{(n + 10)!(n + 11) \cdot n}{(n + 10)!} =\]

\[= (n + 11) \cdot n = n² + 11n.\]

\[\boxed{\text{839.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x^{2} - (a - 2)x - a - 1 =\]

\[= x^{2} - (a - 2) \cdot x - (a + 1),\]

\[по\ теореме\ Виета:\ \]

\[x_{1} + x_{2} = a - 2,\]

\[x_{1} \cdot x_{2} = - a - 1,\]

\[\left( x_{1} \right)^{2} + \left( x_{2} \right)^{2} =\]

\[= \left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} =\]

\[= a^{2} - 4a + 4 + 2a + 2,\]

\[a^{2} - 2a + 6 = 0,\]

\[a_{0} = - \frac{b}{2a} = 1.\]

\[Ответ:при\ a = 1.\]