Решебник по алгебре 9 класс Макарычев Задание 353

 

\[\boxed{\text{353\ (353).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

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

\[Ответ:a = 0;\ \ a = \pm 5.\]

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

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

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

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

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

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

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

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

\[Ответ:x = 0;\ \ x = \pm \sqrt{6}.\]

\[\boxed{\text{353.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \left( x^{2} + 17 \right)(x - 6)(x + 2) < 0\]

\[x^{2} + 17 > 0\ при\ любом\ x\]

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

\[x \in ( - 2;6).\]

\[\textbf{б)}\ \left( 2x^{2} + 1 \right) \cdot x \cdot (x - 4) > 0\]

\[так\ как\ 2x^{2} + 1 > 0\]

\[\Longrightarrow x(x - 4) > 0\]

\[x \in ( - \infty;0) \cup (4; + \infty).\]

\[\textbf{в)}\ (x - 1)^{2}(x - 24) < 0\]

\[(x - 1)^{2} \geq 0\ при\ любом\ \]

\[значении\ x;\]

\[x - 24 < 0;\ \ \ x \neq 1.\]

\[x \in ( - \infty;1) \cup (1;24).\]

\[\textbf{г)}\ (x + 7)(x - 4)^{2}(x - 21) > 0\]

\[(x - 4)^{2} \geq 0\ при\ любом\ x;\]

\[(x + 7)(x - 21) > 0;\ \ \ \ x \neq 4.\]

\[x \in ( - \infty; - 7) \cup (21; + \infty).\]