Оцените это ГДЗ

Оценить: 100% - 1 голосов
100%

Общая оценка

номер 119 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\frac{3}{x+y}-\frac{3x-3y}{2x-3y}\cdot \left ( \frac{2x-3y}{x^{2}-y^{2}}-2x+3y \right )= \end{equation} \begin{equation}=3\left ( x-y \right ) \end{equation} 1)\begin{equation}\frac{2x-3y}{x^{2}-y^{2}}-\left ( 2x-3y \right )= \end{equation} \begin{equation}=\frac{\left ( 2x-3y \right )-\left ( 2x-3y \right )\cdot \left ( x^{2}-y^{2} \right )}{x^{2}-y^{2}}= \end{equation} \begin{equation}=\frac{\left ( 2x-3y \right )\left ( 1-x^{2}+y^{2} \right )}{x^{2}-y^{2}} \end{equation} 2)\begin{equation}\frac{\left ( 3x-3y \right )\left ( 2x-3y \right )\left ( 1-x^{2}+y^{2} \right )}{\left ( 2x-3y \right )\left ( x^{2}-y^{2} \right )}= \end{equation} \begin{equation}=\frac{3\cdot \left ( x-y \right )\cdot\left ( 2x-3y \right )\cdot \left ( 1-x^{2}+y^{2} \right ) }{\left ( 2x-3y \right )\cdot \left ( x-y \right )\left ( x+y \right )}= \end{equation} \begin{equation}=\frac{3\cdot \left ( 1-x^{2}+y^{2} \right )}{x+y} \end{equation} 3)\begin{equation}\frac{3}{x+y}-\frac{3\left ( 1-x^{2}+y^{2} \right )}{x+y}= \end{equation} \begin{equation}=\frac{3-3+3x^{2}-3y^{2}}{x+y}= \end{equation} \begin{equation}=\frac{3\left ( x^{2}-y^{2} \right )}{x+y}= \end{equation} \begin{equation}=\frac{3\cdot \left ( x-y \right )\left ( x+y \right )}{x+y}= \end{equation} \begin{equation}=3\left ( x-y \right ) \end{equation} \begin{equation}3\left ( x-y \right )=3\left ( x-y \right ) \end{equation} 2)\begin{equation}\left ( \frac{xy+y^{2}}{5x^{2}-5xy}+xy+y^{2} \right )\cdot \frac{5x}{x+y}-\frac{y}{x-y}=5xy \end{equation} 1)\begin{equation}\frac{xy+y^{2}}{5x^{2}-5xy}+xy+y^{2}= \end{equation} \begin{equation}=\frac{\left ( xy+y^{2} \right )+\left ( xy+y^{2} \right )\left ( 5x^{2}-5xy \right )}{\left ( 5x^{2}-5xy \right )}= \end{equation} \begin{equation}=\frac{\left ( xy+y^{2} \right )\left (1+5x^{2}-5xy \right )}{5x\left ( x-y \right )} \end{equation} 2)\begin{equation}\frac{\left ( xy+y \right )^{2}\cdot \left ( 1+5x^{2}-5xy \right )}{5x\left ( x-y \right )}\cdot \frac{5x}{x+y}= \end{equation} \begin{equation}=\frac{y\left ( x+y \right )\left ( 1+5x^{2}-5xy \right )\cdot 1}{x-y\left ( x+y \right )}= \end{equation} \begin{equation}=\frac{y\left ( 1+5x^{2}-5xy \right )}{x-y} \end{equation} 3)\begin{equation}\frac{y\left ( 1+5x^{2}-5xy \right )}{x-y}-\frac{y}{x-y}= \end{equation} \begin{equation}=\frac{y+5xy^{2}-5xy^{2}-y}{x-y}= \end{equation} \begin{equation}=\frac{5xy\left ( x-y \right )}{x-y}=5xy \end{equation} \begin{equation}5xy=5xy \end{equation}