Решебник по алгебре 9 класс Макарычев Задание 290

 

\[\boxed{\text{290}\text{\ (290)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{2^{\backslash x + 3}}{x - 2} - \frac{10^{\backslash x - 2}}{x + 3} =\]

\[= \frac{50}{x^{2} + x - 6} - 1^{\backslash x^{2} + x - 6}\]

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

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

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

\[\frac{x^{2} - 7x - 30}{(x - 2)(x + 3)} = 0\]

\[x^{2} - 7x - 30 = (x + 3)(x - 10)\]

\[x_{1} + x_{2} = 7;\ \ \ x_{1} \cdot x_{2} = - 30\]

\[x_{1} = 10;\ \ \ \ x_{2} = - 3.\]

\[\frac{(x - 10)(x + 3)}{(x - 2)(x + 3)} = 0\]

\[\frac{x - 10}{x - 2} = 0\]

\[x - 10 = 0\]

\[x = 10.\]

\[Ответ:x = 10.\]

\[\textbf{б)}\ \frac{x + 5^{\backslash x - 7}}{x - 1} + \frac{2x - 5^{\backslash x - 1}}{x - 7} -\]

\[- \frac{30 - 12x}{8x - x^{2} - 7} = 0\]

\[x^{2} - 8x + 7 = (x - 1)(x - 7)\]

\[D_{1} = 16 - 9 = 9\]

\[x_{1} = 4 + 3 = 7;\ \ x_{2} = 4 - 3 = 1.\]

\[= 0;\ \ \ \ x \neq 1;\ \ x \neq 7\]

\[\frac{3x^{2} - 21x}{(x - 1)(x - 7)} = 0\]

\[\frac{3x(x - 7)}{(x - 1)(x - 7)} = 0\]

\[\frac{3x}{x - 1} = 0\]

\[3x = 0\]

\[x = 0.\]

\[Ответ:x = 0.\]

\[\boxed{\text{290.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 5 \cdot (x - 13)(x + 24) < 0\]

\[5 \cdot (x + 24)(x - 13) < 0\]

\[x \in ( - 24;13).\]

\[\textbf{б)} - \left( x + \frac{1}{7} \right)\left( x + \frac{1}{3} \right) \geq 0\]

\[\left( x + \frac{1}{3} \right)\left( x + \frac{1}{7} \right) \leq 0\]

\[x \in \left\lbrack - \frac{1}{3};\ - \frac{1}{7} \right\rbrack.\]

\[\textbf{в)}\ (x + 12)(3 - x) > 0\]

\[(x + 12)(x - 3) < 0\]

\[x \in ( - 12;3).\]

\[\textbf{г)}\ (6 + x)(3x - 1) \leq 0\]

\[3 \cdot (x + 6)\left( x - \frac{1}{3} \right) \leq 0\]

\[x \in \left\lbrack - 6;\frac{1}{3} \right\rbrack.\]