ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 315

\[\boxed{\text{316.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ (a - 2)(a + 2)\left( a^{2} + 4 \right) =\]

\[= 25a^{2} - 16\]

\[\left( a^{2} - 4 \right)\left( a^{2} + 4 \right) = 25a^{2} - 16\]

\[a^{4} - 16 = 25a^{2} - 16\]

\[a^{4} - 25a^{2} = 0\]

\[a^{2}\left( a^{2} - 25 \right) = 0\]

\[a^{2}(a - 5)(a + 5) = 0\]

\[a_{1} = 0;\ \ a_{2} = 5;\ \ a_{3} = - 5.\]

\[Ответ:a = 0;\ \ a = \pm 5.\]

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

\[= 6x^{2} - 1\]

\[\left( x^{2} - 1 \right)\left( x^{2} + 1 \right) = 6x^{2} - 1\]

\[x^{4} - 1 = 6x^{2} - 1\]

\[x^{4} - 6x^{2} = 0\]

\[x^{2}\left( x^{2} - 6 \right) = 0\]

\[x^{2}\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right) = 0\]

\[x_{1} = 0;\ \ x_{2} = \sqrt{6};\ \ x_{3} = - \sqrt{6}.\]

\[Ответ:x = 0;\ \ x = \pm \sqrt{6}.\]