Решебник по алгебре 9 класс Макарычев Задание 265

 

\[\boxed{\text{265\ (265).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 2x^{2} - 6x^{5} + 1 = 0\ \]

\[степень\ 5;\ \ \]

\[\textbf{б)}\ \ x^{6} - 4x^{5} + 1 = 0\]

\[степень\ 6;\ \ \]

\[\textbf{в)}\ \frac{1}{x}x^{5} = 0\]

\[степень\ 5;\]

\[\textbf{г)}\ (x + 8)(x - 7) = x^{2} + 8x -\]

\[- 7x - 56 = x^{2} + x - 56 = 0\]

\[степень\ 2;\]

\[\textbf{д)}\ \frac{x^{\backslash 2}}{2} - \frac{x}{4} = 5^{\backslash 4}\]

\[2x - x = 20\]

\[x = 20\]

\[степень\ 1;\ \]

\[\textbf{е)}\ 5x^{3} - 5x\left( x^{2} + 4 \right) = 17\]

\[5x^{3} - 5x^{3} - 20x - 17 = 0\]

\[- 20x - 17 = 0;\ \ \]

\[\ степень\ \ 1.\]

\[\boxed{\text{265.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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