Решебник по алгебре 9 класс Макарычев Задание 274

 

\[\boxed{\text{274}\text{\ (274)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{3} + 7x^{2} - 6 = 0\]

\[Разложим\ левую\ часть\ уравнения\ на\ множители:\]

\[x^{3} + 7x^{2} - 6 = x^{3} + x^{2} + 6x^{2} -\]

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

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

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

\[(x + 1)\left( x^{2} + 6x - 6 \right) = 0\]

\[1)\ x + 1 = 0\]

\[x_{1} = - 1.\]

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

\[D_{1} = 9 + 6 = 15\]

\[x_{2,3} = - 3 \pm \sqrt{15}.\]

\[Ответ:x = - 1;x = - 3 \pm \sqrt{15}.\]

\[\textbf{б)}\ x^{3} + 4x^{2} - 5 = 0\]

\[Разложим\ левую\ часть\ уравнения\ на\ множители:\]

\[x^{3} + 4x^{2} - 5 = x^{3} - x^{2} + 5x^{2} -\]

\[- 5 = x^{2}(x - 1) + 5 \cdot \left( x^{2} - 1 \right) =\]

\[= x^{2}(x - 1) + 5 \cdot (x - 1)(x + 1) =\]

\[= (x - 1)\left( x^{2} + 5x + 5 \right).\]

\[1)\ x - 1 = 0\]

\[x_{1} = 1.\]

\[2)\ x^{2} + 5x + 5 = 0\]

\[D = 25 - 4 \cdot 5 = 25 - 20 = 5\]

\[x_{2,3} = \frac{- 5 \pm \sqrt{5}}{2}\]

\[Ответ:x = 1;\ \ x = \frac{- 5 \pm \sqrt{5}}{2}.\]

\[\boxed{\text{274.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ y = \sqrt{12x - 3x²}\]

\[12x - 3x^{2} \geq 0\]

\[x^{2} - 4x \leq 0\]

\[x(x - 4) \leq 0\]

\[x \in \lbrack 0;4\rbrack.\]

\[\textbf{б)}\ y = \frac{1}{\sqrt{2x^{2} - 12x + 18}}\]

\[2x^{2} - 12x + 18 > 0\]

\[x^{2} - 6x + 9 > 0\]

\[(x - 3)^{2} > 0\]

\[x \in ( - \infty;3) \cup (3;\ + \infty).\]