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МатематикаРешебник по алгебре 9 класс Макарычев Задание 294
Задание 294
\[\boxed{\text{294}\text{\ (294)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{1}{x - 4} + \frac{1}{x - 2} = \frac{1}{x + 4} + \frac{1}{x - 5}\]
\[\frac{1}{x - 4} - \frac{1}{x - 5} = \frac{1}{x + 4} - \frac{1}{x - 2}\]
\[ОДЗ:\ \ x \neq 4;5;\ - 4;2.\]
\[(x - 5)(x + 4)(x - 2) -\]
\[- (x - 4)(x + 4)(x - 2) =\]
\[= (x - 4)(x - 5)(x - 2) -\]
\[- (x - 4)(x - 5)(x + 4)\]
\[(x - 2)(x + 4)(x - 5 - x + 4) =\]
\[= (x - 4)(x - 5)(x - 2 - x - 4)\]
\[- (x - 2)(x + 4) =\]
\[= - 6 \cdot (x - 4)(x - 5)\]
\[- \left( x^{2} + 2x - 8 \right) =\]
\[= - 6 \cdot \left( x^{2} - 9x + 20 \right)\]
\[- x^{2} - 2x + 8 =\]
\[= - 6x^{2} + 54x - 120\]
\[5x^{2} - 56x + 128 = 0\]
\[D_{1} = 28^{2} - 5 \cdot 128 = 144\]
\[x_{1} = \frac{28 + 12}{5} = 8;\ \ \ \]
\[x_{2} = \frac{28 - 12}{5} = 3,2.\]
\[Ответ:x = 8;\ \ x = 3,2.\]
\[\textbf{б)}\ \frac{1}{x + 1} + \frac{1}{x + 3} = \frac{1}{x + 28} + \frac{1}{x}\]
\[\frac{1}{x + 1} - \frac{1}{x + 28} = \frac{1}{x} - \frac{1}{x + 3}\]
\[ОДЗ:\ \ \ x \neq - 1;28;0;\ - 3.\]
\[(x + 28) \cdot x \cdot (x + 3) -\]
\[- x(x + 1)(x + 3) =\]
\[= (x + 1)(x + 28)(x + 3) -\]
\[- x(x + 1)(x + 28)\]
\[x(x + 3)(x + 28 - x - 1) =\]
\[= (x + 1)(x + 28)(x + 3 - x)\]
\[27x(x + 3) = 3 \cdot (x + 1)(x + 28)\]
\[27x^{2} + 81x =\]
\[= 3 \cdot \left( x^{2} + 28x + x + 28 \right)\]
\[27x^{2} + 81x = 3x^{2} + 87x + 84\]
\[24x^{2} - 6x - 84 = 0\ \ \ \ |\ :6\]
\[4x^{2} - x - 14 = 0\]
\[D = 1 + 4 \cdot 4 \cdot 14 = 225\]
\[x_{1,2} = \frac{1 \pm 15}{8} = 2;\ - 1,75.\]
\[Ответ:x = 2;\ \ x = - 1,75.\]
\[\boxed{\text{294.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \frac{x - 5}{x + 6} < 0\]
\[(x - 5)(x + 6) < 0\]
\[(x + 6)(x - 5) < 0\]
\[x \in ( - 6;5).\]
\[\textbf{б)}\ \frac{1,4 - x}{x + 3,8} < 0\]
\[(1,4 - x)(x + 3,8) < 0\]
\[(x + 3,8)(x - 1,4) > 0\]
\[x \in ( - \infty;\ - 3,8) \cup (1,4;\ + \infty).\]
\[\textbf{в)}\ \frac{2x}{x - 1,6} > 0\]
\[2x(x - 1,6) > 0\]
\[x \in ( - \infty;0) \cup (1,6;\ + \infty).\]
\[\textbf{г)}\ \frac{5x - 1,5}{x - 4} > 0\]
\[(5x - 1,5)(x - 4) > 0\]
\[5 \cdot (x - 0,3)(x - 4) > 0\]
\[x \in ( - \infty;0,3) \cup (4;\ + \infty).\]
