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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 639
Задание 639
\[\boxed{\text{639\ (639).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x_{1} = \frac{\sqrt{3} - 1}{2};\ \ \ \ x_{2} = \frac{\sqrt{3} + 1}{2}\]
\[\left( x - x_{1} \right)\left( x - x_{2} \right) = 0\]
\[\left( x - \frac{\sqrt{3} - 1}{2} \right)\left( x - \frac{\sqrt{3} + 1}{2} \right) = 0\]
\[x^{2} - \frac{2\sqrt{3}}{2}x + \frac{2}{4} = 0\]
\[x^{2} - \sqrt{3}x + \frac{1}{2} = 0;\ \]
\[2x^{2} - \sqrt{3x} + 1 = 0 - искомое\ \]
\[уравнение.\ \]
\[\textbf{б)}\ x_{1} = 2 - \sqrt{3};\ \ x_{2} = \frac{1}{2 - \sqrt{3}}\]
\[\left( x - x_{1} \right)\left( x - x_{2} \right) = 0\]
\[\left( x - \left( 2 - \sqrt{3} \right) \right)\left( x - \frac{1}{2 - \sqrt{3}} \right) =\]
=\(0\)
\[x^{2} - \frac{8 - 4\sqrt{3}}{2 - \sqrt{3}}x + 1 = 0\]
\[x^{2} - \frac{4 \cdot \left( 2 - \sqrt{3} \right)}{2 - \sqrt{3}}x + 1 = 0\]
\[x^{2} - 4x + 1 = 0 - искомое\ \]
\[уравнение.\ \]
\[\boxed{\text{639.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ \frac{10}{(x - 5)(x + 1)} + \frac{x}{x + 1} =\]
\[= \frac{3}{x - 5}\ \ \ \ \ \ | \cdot (x - 5)(x + 1)\]
\[x - 5 \neq 0,\ \ x \neq 5\]
\[x + 1 \neq 0,\ \ x \neq - 1\]
\[10 + x(x - 5) = 3(x + 1)\]
\[10 + x^{2} - 5x = 3x + 3\]
\[x^{2} - 8x + 7 = 0\]
\[D = 64 - 28 = 36\]
\[x_{1,2} = \frac{8 \pm 6}{2}\]
\[x_{1} = 7;\ \ x_{2} = 1\]
\[Ответ:x = \left\{ 1;7 \right\}.\]
\[\textbf{б)}\frac{17}{(x - 3)(x + 4)} - \frac{1}{x - 3} =\]
\[= \frac{x}{x + 4}\ \ \ \ \ \ \ \ \ | \cdot (x - 3)(x + 4)\]
\[x - 3 \neq 0,\ \ x \neq 3\]
\[x + 4 \neq 0,\ \ x \neq - 4\]
\[17 - (x + 4) = x(x - 3)\]
\[17 - x - 4 = x^{2} - 3x\]
\[x^{2} - 2x - 13 = 0\]
\[D = 4 + 52 = 56 = 4 \cdot 14\]
\[x_{1,2} = \frac{2 \pm \sqrt{4 \cdot 14}}{2} =\]
\[= \frac{2 \pm 2\sqrt{14}}{2} = 1 \pm \sqrt{14}\]
\[Ответ:x = \left\{ 1 - \sqrt{14};1 + \sqrt{14} \right\}.\]
\[\textbf{в)}\frac{4}{(x + 1)^{2}} - \frac{1}{(x - 1)^{2}} +\]
\[+ \frac{1}{x^{2} - 1} =\]
\[= 0\ \ \ \ \ \ \ | \cdot (x + 1)^{2}(x - 1)^{2}\]
\[x^{2} \neq 1,\ \ x \neq \pm 1\]
\[4 \cdot (x - 1)^{2} - (x + 1)^{2} +\]
\[+ \left( x^{2} - 1 \right) = 0\]
\[4 \cdot \left( x^{2} - 2x + 1 \right) -\]
\[- \left( x^{2} + 2x + 1 \right) + x^{2} - 1 = 0\]
\[4x^{2} - 8x + 4 - x^{2} -\]
\[- 2x - 1 + x^{2} - 1 = 0\]
\[4x^{2} - 10x + 2 = 0\ \ \ \ \ |\ :2\]
\[2x^{2} - 5x + 1 = 0\]
\[D = 25 - 8 = 17\]
\[x_{1,2} = \frac{5 \pm \sqrt{17}}{4}\]
\[Ответ:x = \left\{ \frac{5 - \sqrt{17}}{4};\frac{5 + \sqrt{17}}{4} \right\}.\]
\[\textbf{г)}\frac{4}{9x^{2} - 1} + \frac{1}{3x^{2} - x} =\]
\[= \frac{4}{9x^{2} - 6x + 1}\]
\[x \neq 0\]
\[3x + 1 \neq 0,\ \ x \neq - \frac{1}{3}\]
\[3x - 1 \neq 0,\ \ x \neq \frac{1}{3}\]
\[4x \cdot (3x - 1) +\]
\[+ (3x - 1)(3x + 1) =\]
\[= 4x \cdot (3x + 1)\]
\[12x^{2} - 4x + 9x^{2} + 3x -\]
\[- 3x - 1 = 12x^{2} + 4x\]
\[9x^{2} - 8x - 1 = 0\]
\[D = 64 + 36 = 100\]
\[x_{1,2} = \frac{8 \pm 10}{18}\]
\[x_{1} = 1;\ \ x_{2} = - \frac{1}{9}\]
\[Ответ:x = \left\{ - \frac{1}{9};1 \right\}\text{.\ }\]
