ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 637

 

\[\boxed{\text{637\ (637).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ при\ x = 5 + 2\sqrt{6},\]

\[\ \ y = 5 - 2\sqrt{6}:\]

\[\frac{\text{xy}}{x + y} = \frac{\left( 5 + 2\sqrt{6} \right)\left( 5 - 2\sqrt{6} \right)}{5 + 2\sqrt{6} + 5 - 2\sqrt{6}} =\]

\[= \frac{25 - 4 \cdot 6}{10} = \frac{25 - 24}{10} =\]

\[= \frac{1}{10} = 0,1.\]

\[\textbf{б)}\ при\ x = \sqrt{11} + \sqrt{3},\]

\[\ \ y = \sqrt{11} - \sqrt{3}\]

\[\frac{x^{2} + y^{2}}{\text{xy}} =\]

\[= \frac{\left( \sqrt{11} + \sqrt{3} \right)^{2} + \left( \sqrt{11} - \sqrt{3} \right)^{2}}{\left( \sqrt{11} + \sqrt{3} \right)\left( \sqrt{11} - \sqrt{3} \right)} =\]

\[= \frac{28}{8} = \frac{14}{4} = \frac{7}{2} = 3,5.\ \]

\[\boxed{\text{637.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\frac{3y + 9}{3y - 1} + \frac{2y - 13}{2y + 5} =\]

\[= 2\ \ \ \ \ \ \ \ \ \ | \cdot (3y - 1)(2y + 5)\]

\[3y - 1 \neq 0,\ \ 3y \neq 1,\]

\[\ \ y \neq \frac{1}{3}\]

\[2y + 5 \neq 0,\ \ 2y \neq - 5,\]

\[\ \ y \neq - 2,5\]

\[(3y + 9)(2y + 5) +\]

\[+ (2y - 13)(3y - 1) =\]

\[= 2 \cdot (3y - 1)(2y + 5)\]

\[6y^{2} + 15y + 18y + 45 + 6y^{2} -\]

\[- 2y - 39y + 13 =\]

\[= 12y^{2} + 30y - 4y - 10\]

\[- 34y = - 68\]

\[y = 2\]

\[Ответ:y = 2.\]

\[\textbf{б)}\frac{5y + 13}{5y + 4} - \frac{4 - 6y}{3y - 1} =\]

\[= 3\ \ \ \ \ \ | \cdot (5y + 4)(3y - 1)\]

\[5y + 4 \neq 0,\ \ 5y \neq - 4,\]

\[\ \ y \neq - \frac{4}{5}\]

\[3y - 1 \neq 0,\ \ 3y \neq 1,\]

\[\ \ y \neq \frac{1}{3}\]

\[(5y + 13)(3y - 1) -\]

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

\[= 3 \cdot (5y + 4)(3y - 1)\]

\[15y^{2} - 5y + 39y - 13 - 20y -\]

\[- 16 + 30y^{2} + 24y =\]

\[= 45y^{2} - 15y + 36y - 12\]

\[17y = 17\]

\[y = 1\]

\[Ответ:y = 1.\]

\[\textbf{в)}\frac{y + 1}{y - 5} + \frac{10}{y + 5} =\]

\[= \frac{y + 1}{y - 5} \cdot \frac{10}{y + 5}\ \ \ \ \ \ \ | \cdot \left( y^{2} - 25 \right)\]

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

\[y + 5 \neq 0,\ \ y \neq - 5\]

\[(y + 1)(y + 5) + 10 \cdot (y - 5) =\]

\[= 10 \cdot (y + 1)\]

\[y^{2} + 5y + y + 5 + 10y - 50 =\]

\[= 10y + 10\]

\[y^{2} + 6y - 55 = 0\]

\[D = 36 + 220 = 256 = 16^{2}\]

\[y_{1,2} = \frac{- 6 \pm 16}{2}\]

\[y_{1} = 5 - не\ подходит\ по\ ОДЗ;\ \]

\[y_{2} = - 11\]

\[Ответ:y = - 11.\]

\[\textbf{г)}\frac{6}{y - 4} - \frac{y}{y + 2} =\]

\[= \frac{6}{y - 4} \cdot \frac{y}{y + 2}\text{\ \ \ \ \ \ }\]

\[| \cdot (y + 2)(y - 4)\]

\[y - 4 \neq 0,\ \ y \neq 4\]

\[y + 2 \neq 0,\ \ y \neq - 2\]

\[6 \cdot (y + 2) - y \cdot (y - 4) = 6y\]

\[6y + 12 - y^{2} + 4y = 6y\]

\[6y + 12 - y^{2} + 4y = 6y\ | \cdot ( - 1)\]

\[y^{2} - 4y - 12 = 0\]

\[D = 16 + 48 = 64\]

\[y_{1,2} = \frac{4 \pm 8}{2}\]

\[y_{1} = 6;\ \ y_{2} = - 2 - не\ \]

\[подходит\ по\ ОДЗ\]

\[Ответ:y = 6.\ \]