номер 117 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\left ( \frac{x^{2}+xy}{y^{2}}+1 \right ):\frac{x^{3}-y^{3}}{y^{3}}+\frac{x}{y-x} \end{equation} 1)\begin{equation}\frac{x^{2}+xy}{y^{2}}+1=\frac{x^{2}+xy+y^{2}}{y^{2}} \end{equation} 2)\begin{equation}\frac{x^{2}+xy+y^{2}}{y^{2}}:\frac{x^{3}-y^{3}}{y^{3}}= \end{equation} \begin{equation}=\frac{x^{2}+xy+y^{2}}{y^{2}}\cdot \frac{y^{3}}{\left ( x-y \right )\left ( x^{2}+xy+y^{2} \right )}= \end{equation} \begin{equation}=\frac{y}{x-y} \end{equation} 3)\begin{equation}\frac{y}{x-y}+\frac{x}{y-x}= \end{equation} \begin{equation}=\frac{y}{x-y}-\frac{x}{x-y}= \end{equation} \begin{equation}=\frac{y-x}{x-y}=-1 \end{equation} 2)\begin{equation}\left ( \frac{a^{2}+b^{2}}{ab}-1 \right ):\frac{a^{3}+b^{3}}{ab^{3}}-\frac{a^{2}}{a+b} \end{equation} 1)\begin{equation}\frac{a^{2}+b^{2}}{ab}-1=\frac{a^{2}+b^{2}-ab}{ab} \end{equation} 2)\begin{equation}\frac{a^{2}-ab+b^{2}}{ab}:\frac{a^{3}+b^{3}}{ab^{3}}= \end{equation} \begin{equation}=\frac{a^{2}-ab+b^{2}}{ab}\cdot \frac{ab^{3}}{\left ( a+b \right )\left ( a^{2}-ab+b^{2} \right )}= \end{equation} \begin{equation}=\frac{b^{2}}{a+b} \end{equation} 3)\begin{equation}\frac{b^{2}}{a+b}-\frac{a^{2}}{a+b}=\frac{b^{2}-a^{2}}{a+b}= \end{equation} \begin{equation}=\frac{-\left ( a+b \right )\left ( a-b \right )}{\left ( a+b \right )}=-\left ( a-b \right ) \end{equation} 3)\begin{equation}\frac{\left ( a+b \right )^{2}}{a}-\frac{a^{2}-b^{2}}{a^{2}+ab+b^{2}}\cdot \left ( \frac{a^{2}}{a+b}-\frac{b^{3}}{a^{2}+ab} \right ) \end{equation} 1)\begin{equation}\frac{a^{2}}{a+b}-\frac{b^{3}}{a^{2}+ab}=\frac{a^{2}}{a+b}-\frac{b^{3}}{a\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{a^{3}-b^{3}}{a\left ( a+b \right )} \end{equation} 2)\begin{equation}\frac{a^{2}-b^{2}}{a^{2}+ab+b^{2}}\cdot \frac{a^{3}-b^{3}}{a\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{\left ( a-b \right )\left ( a+b \right )\cdot \left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )}{\left ( a^{2}+ab+b^{2} \right )\cdot a\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{\left ( a-b \right )^{2}}{a} \end{equation} 3)\begin{equation}\frac{\left ( a+b \right )^{2}}{a}-\frac{\left ( a-b \right )^{2}}{a}= \end{equation} \begin{equation}=\frac{a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}}{a}= \end{equation} \begin{equation}=\frac{4ab}{a}=4b \end{equation} 4)\begin{equation}\left ( \frac{x^{2}}{x-y}-\frac{y^{3}}{xy-x^{2}} \right )\cdot \frac{x^{2}-y^{2}}{x^{2}-xy+y^{2}}-\frac{\left ( x-y \right )^{2}}{x} \end{equation} 1)\begin{equation}\frac{x^{2}}{x-y}-\frac{y^{3}}{xy-x^{2}}=\frac{x^{2}}{x-y}-\frac{y^{3}}{x\left ( y-x \right )}= \end{equation} \begin{equation}=\frac{x^{2}}{x-y}+\frac{y^{3}}{x\left ( x-y \right )}=\frac{x^{3}+y^{3}}{x\left ( x-y \right )} \end{equation} 2)\begin{equation}\frac{x^{3}+y^{3}}{x\left ( x-y \right )}\cdot \frac{x^{2}-y^{2}}{x^{2}-xy+y^{2}}= \end{equation} \begin{equation}=\frac{\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )\cdot \left ( x-y \right )\left ( x+y \right )}{x\left ( x-y \right )\cdot \left ( x^{2}-xy+y^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( x+y \right )^{2}}{x} \end{equation} 3)\begin{equation}\frac{\left ( x+y \right )^{2}}{x}-\frac{\left ( x-y \right )^{2}}{x}= \end{equation} \begin{equation}=\frac{x^{2}+2xy+y^{2}-x^{2}+2xy-y^{2}}{x}= \end{equation} \begin{equation}=\frac{4xy}{x}=4y \end{equation} 5)\begin{equation}\frac{a^{3}-8}{\left ( 2a-6 \right )^{2}}\cdot \frac{a-3}{\left ( a+2 \right )^{2}-2a}-\frac{7-2a}{4a-12} \end{equation} 1)\begin{equation}\frac{a^{3}-8}{\left ( 2a-6 \right )^{2}}\cdot \frac{a-3}{\left ( a+2 \right )^{2}-2a}= \end{equation} \begin{equation}=\frac{a^{3}-8}{4\left ( a-3 \right )^{2}}\cdot \frac{a-3}{a^{2}+2a+4}= \end{equation} \begin{equation}=\frac{\left ( a-2 \right )\left ( a^{2}+2a+4 \right )\cdot \left ( a-3 \right )}{4\left ( a-3 \right )^{2}\cdot \left ( a^{2}+2a+4 \right )}= \end{equation} \begin{equation}=\frac{a-2}{4\left ( a-3 \right )} \end{equation} 2)\begin{equation}\frac{a-2}{4a-12}-\frac{7-2a}{4a-12}=\frac{a-2-7+2a}{4a-12}= \end{equation} \begin{equation}=\frac{3a-9}{4a-12}=\frac{3\left ( a-3 \right )}{4\left ( a-3 \right )}=\frac{3}{4} \end{equation} 6)\begin{equation}\frac{b^{3}+27}{\left ( 3b+6 \right )^{2}}:\frac{\left ( b-3 \right )^{2}+3b}{b+2}+\frac{2b+3}{9b+18} \end{equation} 1)\begin{equation}\frac{b^{3}+27}{9\left ( b+2 \right )^{2}}:\frac{b^{2}-3b+9}{b+2}= \end{equation} \begin{equation}=\frac{\left ( b+3 \right )\left ( b^{2}-3b+9 \right )\cdot \left ( b+2 \right )}{9\left ( b+2 \right )^{2}\cdot \left ( b^{2}-3b+9 \right )}= \end{equation} \begin{equation}=\frac{b+3}{9\left ( b+2 \right )} \end{equation} 2)\begin{equation}\frac{b+3}{9b+18}+\frac{2b+3}{9b+18}= \end{equation} \begin{equation}=\frac{b+3+2b+3}{9b+18}=\frac{3b+6}{9b+18}= \end{equation} \begin{equation}=\frac{3\left ( b+2 \right )}{9\left ( b+2 \right )}=\frac{1}{3} \end{equation} 7)\begin{equation}\left ( \frac{6a+1}{a^{2}-6a}+\frac{6a-1}{a^{2}+6a} \right )\cdot \frac{a^{2}-36}{a^{2}+1} \end{equation} 1)\begin{equation}\frac{6a+1}{a^{2}-6a}+\frac{6a-1}{a^{2}+6a}= \end{equation} \begin{equation}=\frac{6a+1}{a\left ( a-6 \right )}+\frac{6a-1}{a\left ( a+6 \right )}= \end{equation} \begin{equation}=\frac{\left ( a+6 \right )\left ( 6a+1 \right )+\left ( a-6 \right )\left ( 6a-1 \right )}{a\left ( a^{2}-36 \right )}= \end{equation} \begin{equation}=\frac{6a^{2}+a+36a+6+6a^{2}-a-36a+6}{a\left ( a^{2}-36 \right )}= \end{equation} \begin{equation}=\frac{12a^{2}+12}{a\left ( a^{2}-36 \right )}=\frac{12\left ( a^{2}+1 \right )}{a\left ( a^{2}-36 \right )} \end{equation} 2)\begin{equation}\frac{12\cdot \left ( a^{2}+1 \right )\cdot \left ( a^{2}-36 \right )}{a\left ( a^{2}-36 \right )\cdot \left ( a^{2}+1 \right )}=\frac{12}{a} \end{equation} 8)\begin{equation}\left (\frac{5x+y}{x^{2}-5xy}+\frac{5x-y}{x^{2}+5xy}\right )\cdot \frac{x^{2}-25y^{2}}{x^{2}+y^{2}} \end{equation} 1)\begin{equation}\frac{5x+y}{x^{2}-5xy}+\frac{5x-y}{x^{2}+5xy}= \end{equation} \begin{equation}=\frac{5x+y}{x\left ( x-5y \right )}+\frac{5x-y}{x\left ( x+5y \right )}= \end{equation} \begin{equation}=\frac{\left ( 5x+y \right )\left ( x+5y \right )+\left ( 5x-y \right )\left ( x-5y \right )}{x\left ( x-5y \right )\left ( x+5y \right )}= \end{equation} \begin{equation}=\frac{5x^{2}+25xy+yx+5y^{2}+5x^{2}-25xy-xy+5y^{2}}{x\left ( x^{2}-25y^{2} \right )}= \end{equation} \begin{equation}=\frac{10x^{2}+10y^{2}}{x\left ( x^{2}-25y^{2} \right )}=\frac{10\left ( x^{2}+y^{2} \right )}{x\left ( x^{2}-25y^{2} \right )} \end{equation} 2)\begin{equation}\frac{10\left ( x^{2}+y^{2} \right )}{x\left ( x^{2}-25y^{2} \right )}\cdot \frac{x^{2}-25y^{2}}{x^{2}+y^{2}}=\frac{10}{x} \end{equation}