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МатематикаРешебник по алгебре 9 класс Макарычев Задание 291
Задание 291
\[\boxed{\text{291}\text{\ (291)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{3x - 2^{\backslash x + 3}}{x - 1} - \frac{2x + 3^{\backslash x - 1}}{x + 3} =\]
\[= \frac{12x + 4}{x^{2} + 2x - 3}\]
\[x^{2} + 2x - 3 = (x + 3)(x - 1)\]
\[x_{1} + x_{2} = - 2;\ \ \ x_{1} \cdot x_{2} = - 3\]
\[x_{1} = - 3;\ \ \ x_{2} = 1.\]
\[\frac{x^{2} - 6x - 7}{(x - 1)(x + 3)} = 0\]
\[x^{2} - 6x - 7 = (x + 1)(x - 7)\]
\[D_{1} = 9 + 7 = 16\]
\[x_{1} = 3 + 4 = 7;\ \ \ x_{2} =\]
\[= 3 - 4 = - 1.\]
\[\frac{(x - 7)(x + 1)}{(x - 1)(x + 3)} = 0\]
\[(x - 7)(x + 1) = 0\]
\[x_{1} = 7;\ \ \ x_{2} = - 1.\]
\[Ответ:x = 7;\ \ x = - 1.\]
\[\textbf{б)}\ \frac{5x - 1^{\backslash x - 3}}{x + 7} - \frac{2x + 2^{\backslash x + 7}}{x - 3} +\]
\[+ \frac{63}{x^{2} + 4x - 21} = 0\]
\[x^{2} + 4x - 21 = (x + 7)(x - 3)\]
\[D_{1} = 4 + 21 = 25\]
\[x_{1} = - 2 + 5 = 3;\ \ \ x_{2} =\]
\[= - 2 - 5 = - 7.\]
\[\frac{3x^{2} - 32x + 52}{(x - 3)(x + 7)} = 0\]
\[3x^{2} - 32x + 52 = 0\]
\[D_{1} = 16^{2} - 3 \cdot 52 = 100\]
\[x_{1} = \frac{16 - 10}{3} = 2;\ \ \ \ \]
\[\ x_{2} = \frac{16 + 10}{3} = \frac{26}{3} = 8\frac{2}{3}.\]
\[Ответ:x = 2;\ \ x = 8\frac{2}{3}.\]
\[\textbf{в)}\frac{x}{x^{2} + 4x + 4} = \frac{4}{x^{2} - 4} -\]
\[- \frac{16}{x^{3} + 2x^{2} - 4x - 8}\]
\[\frac{x}{(x + 2)^{2}} = \frac{4}{(x - 2)(x + 2)} -\]
\[- \frac{16}{x^{2}(x + 2) - 4 \cdot (x + 2)}\]
\[\frac{x^{\backslash x^{2} - 2}}{(x + 2)^{2}} = \frac{4^{\backslash x + 2\ }}{(x - 2)(x + 2)} -\]
\[- \frac{16}{(x + 2)\left( x^{2} - 4 \right)}\]
\[\frac{x(x - 2) - 4 \cdot (x + 2) + 16}{(x + 2)^{2}(x - 2)} =\]
\[= 0;\ \ \ \ x \neq - 2;\ \ x \neq 2\]
\[\frac{x^{2} - 2x - 4x - 8 + 16}{(x + 2)^{2}(x - 2)} = 0\]
\[\frac{x^{2} - 6x + 8}{(x + 2)^{2}(x - 2)} = 0\]
\[x^{2} - 6x + 8 = (x - 2)(x - 4)\]
\[D_{1} = 9 - 8 = 1\]
\[x_{1} = 3 + 1 = 4;\ \ x_{2} = 3 - 2 = 2.\]
\[\frac{(x - 2)(x - 4)}{(x + 2)^{2}(x - 2)} = 0\]
\[\frac{x - 4}{(x + 2)^{2}} = 0\]
\[x - 4 = 0\]
\[x = 4.\]
\[Ответ:x = 4.\]
\[\boxed{\text{291.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[\textbf{а)}\ 2 \cdot (x - 18)(x - 19) > 0\]
\[x \in ( - \infty;18) \cup (19; + \infty)\text{.\ }\]
\[\textbf{б)} - 4 \cdot (x + 0,9)(x - 3,2) < 0\]
\[(x + 0,9)(x - 3,2) > 0\]
\[x \in ( - \infty;\ - 0,9) \cup (3,2;\ + \infty).\]
\[\textbf{в)}\ (7x + 21)(x - 8,5) \leq 0\]
\[7 \cdot (x + 3)(x - 8,5) \leq 0\]
\[x \in \lbrack - 3;8,5\rbrack.\]
\[\textbf{г)}\ (8 - x)(x - 0,3) \geq 0\]
\[(x - 0,3)(x - 8) \leq 0\]
\[x \in \lbrack 0,3;8\rbrack.\]
