ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 639

 

\[\boxed{\text{639\ (639).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x_{1} = \frac{\sqrt{3} - 1}{2};\ \ \ \ x_{2} = \frac{\sqrt{3} + 1}{2}\]

\[\left( x - x_{1} \right)\left( x - x_{2} \right) = 0\]

\[\left( x - \frac{\sqrt{3} - 1}{2} \right)\left( x - \frac{\sqrt{3} + 1}{2} \right) = 0\]

\[x^{2} - \frac{2\sqrt{3}}{2}x + \frac{2}{4} = 0\]

\[x^{2} - \sqrt{3}x + \frac{1}{2} = 0;\ \]

\[2x^{2} - \sqrt{3x} + 1 = 0 - искомое\ \]

\[уравнение.\ \]

\[\textbf{б)}\ x_{1} = 2 - \sqrt{3};\ \ x_{2} = \frac{1}{2 - \sqrt{3}}\]

\[\left( x - x_{1} \right)\left( x - x_{2} \right) = 0\]

\[\left( x - \left( 2 - \sqrt{3} \right) \right)\left( x - \frac{1}{2 - \sqrt{3}} \right) =\]

=\(0\)

\[x^{2} - \frac{8 - 4\sqrt{3}}{2 - \sqrt{3}}x + 1 = 0\]

\[x^{2} - \frac{4 \cdot \left( 2 - \sqrt{3} \right)}{2 - \sqrt{3}}x + 1 = 0\]

\[x^{2} - 4x + 1 = 0 - искомое\ \]

\[уравнение.\ \]

\[\boxed{\text{639.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \frac{10}{(x - 5)(x + 1)} + \frac{x}{x + 1} =\]

\[= \frac{3}{x - 5}\ \ \ \ \ \ | \cdot (x - 5)(x + 1)\]

\[x - 5 \neq 0,\ \ x \neq 5\]

\[x + 1 \neq 0,\ \ x \neq - 1\]

\[10 + x(x - 5) = 3(x + 1)\]

\[10 + x^{2} - 5x = 3x + 3\]

\[x^{2} - 8x + 7 = 0\]

\[D = 64 - 28 = 36\]

\[x_{1,2} = \frac{8 \pm 6}{2}\]

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

\[Ответ:x = \left\{ 1;7 \right\}.\]

\[\textbf{б)}\frac{17}{(x - 3)(x + 4)} - \frac{1}{x - 3} =\]

\[= \frac{x}{x + 4}\ \ \ \ \ \ \ \ \ | \cdot (x - 3)(x + 4)\]

\[x - 3 \neq 0,\ \ x \neq 3\]

\[x + 4 \neq 0,\ \ x \neq - 4\]

\[17 - (x + 4) = x(x - 3)\]

\[17 - x - 4 = x^{2} - 3x\]

\[x^{2} - 2x - 13 = 0\]

\[D = 4 + 52 = 56 = 4 \cdot 14\]

\[x_{1,2} = \frac{2 \pm \sqrt{4 \cdot 14}}{2} =\]

\[= \frac{2 \pm 2\sqrt{14}}{2} = 1 \pm \sqrt{14}\]

\[Ответ:x = \left\{ 1 - \sqrt{14};1 + \sqrt{14} \right\}.\]

\[\textbf{в)}\frac{4}{(x + 1)^{2}} - \frac{1}{(x - 1)^{2}} +\]

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

\[= 0\ \ \ \ \ \ \ | \cdot (x + 1)^{2}(x - 1)^{2}\]

\[x^{2} \neq 1,\ \ x \neq \pm 1\]

\[4 \cdot (x - 1)^{2} - (x + 1)^{2} +\]

\[+ \left( x^{2} - 1 \right) = 0\]

\[4 \cdot \left( x^{2} - 2x + 1 \right) -\]

\[- \left( x^{2} + 2x + 1 \right) + x^{2} - 1 = 0\]

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

\[- 2x - 1 + x^{2} - 1 = 0\]

\[4x^{2} - 10x + 2 = 0\ \ \ \ \ |\ :2\]

\[2x^{2} - 5x + 1 = 0\]

\[D = 25 - 8 = 17\]

\[x_{1,2} = \frac{5 \pm \sqrt{17}}{4}\]

\[Ответ:x = \left\{ \frac{5 - \sqrt{17}}{4};\frac{5 + \sqrt{17}}{4} \right\}.\]

\[\textbf{г)}\frac{4}{9x^{2} - 1} + \frac{1}{3x^{2} - x} =\]

\[= \frac{4}{9x^{2} - 6x + 1}\]

\[x \neq 0\]

\[3x + 1 \neq 0,\ \ x \neq - \frac{1}{3}\]

\[3x - 1 \neq 0,\ \ x \neq \frac{1}{3}\]

\[4x \cdot (3x - 1) +\]

\[+ (3x - 1)(3x + 1) =\]

\[= 4x \cdot (3x + 1)\]

\[12x^{2} - 4x + 9x^{2} + 3x -\]

\[- 3x - 1 = 12x^{2} + 4x\]

\[9x^{2} - 8x - 1 = 0\]

\[D = 64 + 36 = 100\]

\[x_{1,2} = \frac{8 \pm 10}{18}\]

\[x_{1} = 1;\ \ x_{2} = - \frac{1}{9}\]

\[Ответ:x = \left\{ - \frac{1}{9};1 \right\}\text{.\ }\]