Решебник по алгебре 8 класс Макарычев ФГОС Задание 640

 

\[\boxed{\text{640\ (640).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[bx + 2x = 3b + 6\]

\[x(b + 2) = 3(b + 2)\]

\[0 = 0\]

\[x - любое\ число\]

\[Ответ:x = 3\text{.\ }\]

\[\boxed{\text{640.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \frac{21}{x + 1} = \frac{16}{x - 2} - \frac{6}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ | \cdot x(x + 1)(x - 2)\]

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

\[x - 2 \neq 0,\ \ x \neq 2\]

\[x \neq 0\]

\[21x \cdot (x - 2) = 16x \cdot (x + 1) -\]

\[- 6 \cdot (x + 1)(x - 2)\]

\[21x^{2} - 42x = 16x^{2} + 16x -\]

\[- 6x^{2} + 12x - 6x + 12\]

\[11x^{2} - 64x - 12 = 0\]

\[D = 4096 + 528 = 4624 = 68^{2}\]

\[x_{1,2} = \frac{64 \pm 68}{22} = \frac{132}{22};\ - \frac{4}{22}\]

\[x_{1} = \frac{66}{11} = 6;\ \ x_{2} = - \frac{2}{11}\]

\[Ответ:x = \left\{ - \frac{2}{11};6 \right\}.\]

\[\textbf{б)}\frac{2}{y^{2} - 3y} - \frac{1}{y - 3} = \frac{5}{y^{3} - 9y}\]

\[\frac{2}{y(y - 3)} - \frac{1}{y - 3} =\]

\[= \frac{5}{y(y - 3)(y + 3)}\text{\ \ \ \ \ \ }\]

\[| \cdot y(y - 3)(y + 3)\]

\[y \neq 0\]

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

\[y + 3 \neq 0,\ \ y \neq - 3\]

\[2 \cdot (y + 3) - y \cdot (y + 3) = 5\]

\[2y + 6 - y^{2} - 3y = 5\]

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

\[D = 1 + 4 = 5\]

\[y_{1,2} = \frac{- 1 \pm \sqrt{5}}{2}\]

\[Ответ:y =\]

\[= \left\{ \frac{- 1 - \sqrt{5}}{2};\frac{- 1 + \sqrt{5}}{2} \right\}.\]

\[\textbf{в)}\frac{18}{4x^{2} + 4x + 1} - \frac{1}{2x^{2} - x} =\]

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

\[\frac{18}{(2x + 1)^{2}} - \frac{1}{x(2x - 1)} =\]

\[= \frac{6}{(2x - 1)(2x + 1)}\text{\ \ \ \ \ \ }\]

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

\[x \neq 0\]

\[2x - 1 \neq 0,\ \ x \neq \frac{1}{2}\]

\[2x + 1 \neq 0,\ \ x \neq - \frac{1}{2}\]

\[18x(2x - 1) - (2x + 1)^{2} =\]

\[= 6x(2x + 1)\]

\[36x^{2} - 18x - 4x^{2} - 4x - 1 =\]

\[= 12x^{2} + 6x\]

\[20x^{2} - 28x - 1 = 0\]

\[D = 784 + 80 = 864 = 144 \cdot 6\]

\[x_{1,2} = \frac{28 \pm \sqrt{144 \cdot 6}}{40} =\]

\[= \frac{28 \pm 12\sqrt{6}}{40} = \frac{7 \pm 3\sqrt{6}}{10}\]

\[Ответ:x = \left\{ \frac{7 - 3\sqrt{6}}{10};\frac{7 + 3\sqrt{6}}{10} \right\}.\]

\[\textbf{г)}\frac{3 \cdot \left( 4y^{2} + 10y - 7 \right)}{16y^{2} - 9} =\]

\[= \frac{3y - 7}{3 - 4y} + \frac{6y + 5}{3 + 4y}\]

\[\frac{3 \cdot \left( 4y^{2} + 10y - 7 \right)}{(4y - 3)(4y + 3)} =\]

\[= \frac{6y + 5}{4y + 3} - \frac{3y - 7}{4y - 3}\text{\ \ \ \ \ }\]

\[| \cdot (4y - 3)(4y + 3)\]

\[4y - 3 \neq 0,\ \ y \neq \frac{3}{4}\]

\[4y + 3 \neq 0,\ \ y \neq - \frac{3}{4}\]

\[3 \cdot \left( 4y^{2} + 10y - 7 \right) =\]

\[= (6y + 5)(4y - 3) -\]

\[- (3y - 7)(4y + 3)\]

\[12y^{2} + 30y - 21 = 24y^{2} -\]

\[- 18y + 20y - 15 -\]

\[- 12y^{2} - 9y + 28y + 21\]

\[9y = 27\]

\[y = 3\]

\[Ответ:y = 3\text{.\ }\]