ГДЗ по алгебре 9 класс Макарычев Задание 305

\[\boxed{\text{305\ (305).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2x^{2} + 3x - 5 \geq 0\]

\[2x^{2} + 3x - 5 = 0\]

\[D = 9 + 4 \cdot 2 \cdot 5 = 49\]

\[x_{1} = \frac{- 3 + 7}{4} = 1;\ \ \ \ \]

\[\ x_{2} = \frac{- 3 - 7}{4} = - 2,5;\]

\[2 \cdot (x + 2,5)(x - 1) \geq 0\]

\[x \in ( - \infty;\ - 2,5\rbrack \cup \lbrack 1; + \infty).\]

\[\textbf{б)} - 6x^{2} + 6x + 36 \geq 0\ |\ :( - 6)\]

\[x^{2} - x - 6 \leq 0\]

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

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 6\]

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

\[(x - 3)(x + 2) \leq 0\]

\[x \in \lbrack - 2;3\rbrack.\]

\[\textbf{в)} - x^{2} + 5 \leq 0\ \ \ \ \ \ \ \ \ \ \ |\ :( - 1)\]

\[x^{2} - 5 \geq 0\]

\[\left( x - \sqrt{5} \right)\left( x + \sqrt{5} \right) \geq 0\]

\[x \in \left( - \infty;\ - \sqrt{5} \right\rbrack \cup \left\lbrack \sqrt{5}; + \infty \right).\]