ГДЗ по алгебре 9 класс Макарычев Задание 291

\[\boxed{\text{291}\text{\ (291)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{3x - 2^{\backslash x + 3}}{x - 1} - \frac{2x + 3^{\backslash x - 1}}{x + 3} =\]

\[= \frac{12x + 4}{x^{2} + 2x - 3}\]

\[x^{2} + 2x - 3 = (x + 3)(x - 1)\]

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

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

\[\frac{x^{2} - 6x - 7}{(x - 1)(x + 3)} = 0\]

\[x^{2} - 6x - 7 = (x + 1)(x - 7)\]

\[D_{1} = 9 + 7 = 16\]

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

\[= 3 - 4 = - 1.\]

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

\[(x - 7)(x + 1) = 0\]

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

\[Ответ:x = 7;\ \ x = - 1.\]

\[\textbf{б)}\ \frac{5x - 1^{\backslash x - 3}}{x + 7} - \frac{2x + 2^{\backslash x + 7}}{x - 3} +\]

\[+ \frac{63}{x^{2} + 4x - 21} = 0\]

\[x^{2} + 4x - 21 = (x + 7)(x - 3)\]

\[D_{1} = 4 + 21 = 25\]

\[x_{1} = - 2 + 5 = 3;\ \ \ x_{2} =\]

\[= - 2 - 5 = - 7.\]

\[\frac{3x^{2} - 32x + 52}{(x - 3)(x + 7)} = 0\]

\[3x^{2} - 32x + 52 = 0\]

\[D_{1} = 16^{2} - 3 \cdot 52 = 100\]

\[x_{1} = \frac{16 - 10}{3} = 2;\ \ \ \ \]

\[\ x_{2} = \frac{16 + 10}{3} = \frac{26}{3} = 8\frac{2}{3}.\]

\[Ответ:x = 2;\ \ x = 8\frac{2}{3}.\]

\[\textbf{в)}\frac{x}{x^{2} + 4x + 4} = \frac{4}{x^{2} - 4} -\]

\[- \frac{16}{x^{3} + 2x^{2} - 4x - 8}\]

\[\frac{x}{(x + 2)^{2}} = \frac{4}{(x - 2)(x + 2)} -\]

\[- \frac{16}{x^{2}(x + 2) - 4 \cdot (x + 2)}\]

\[\frac{x^{\backslash x^{2} - 2}}{(x + 2)^{2}} = \frac{4^{\backslash x + 2\ }}{(x - 2)(x + 2)} -\]

\[- \frac{16}{(x + 2)\left( x^{2} - 4 \right)}\]

\[\frac{x(x - 2) - 4 \cdot (x + 2) + 16}{(x + 2)^{2}(x - 2)} =\]

\[= 0;\ \ \ \ x \neq - 2;\ \ x \neq 2\]

\[\frac{x^{2} - 2x - 4x - 8 + 16}{(x + 2)^{2}(x - 2)} = 0\]

\[\frac{x^{2} - 6x + 8}{(x + 2)^{2}(x - 2)} = 0\]

\[x^{2} - 6x + 8 = (x - 2)(x - 4)\]

\[D_{1} = 9 - 8 = 1\]

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

\[\frac{(x - 2)(x - 4)}{(x + 2)^{2}(x - 2)} = 0\]

\[\frac{x - 4}{(x + 2)^{2}} = 0\]

\[x - 4 = 0\]

\[x = 4.\]

\[Ответ:x = 4.\]