номер 116 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\left ( \frac{x+5}{x-5}-\frac{x^{2}-2x+7}{x^{2}-25} \right )\cdot \frac{x+5}{10x+15} \end{equation} 1)\begin{equation}\frac{x+5}{x-5}-\frac{x^{2}-2x+7}{x^{2}-25}= \end{equation} \begin{equation}=\frac{\left ( x+5 \right )^{2}-x^{2}+2x-7}{x^{2}-25}= \end{equation} \begin{equation}=\frac{x^{2}+10x+25-x^{2}+2x-7}{x^{2}-25}= \end{equation} \begin{equation}=\frac{12x+18}{x^{2}-25}=\frac{6\left ( 2x+3 \right )}{x^{2}-25} \end{equation} 2)\begin{equation}\frac{6\left ( 2x+3 \right )}{\left ( x-5 \right )\left ( x+5 \right )}\cdot \frac{x+5}{5\left ( 2x+3 \right )}=\frac{6}{5\left ( x-5 \right )} \end{equation} 2)\begin{equation}\frac{y^{2}+4y}{9y-18}\cdot \left ( \frac{y}{y-4}+\frac{y^{2}-2y+12}{16-y^{2}} \right ) \end{equation} 1)\begin{equation}\frac{y}{y-4}+\frac{y^{2}-2y+12}{16-y^{2}}= \end{equation} \begin{equation}=\frac{y}{y-4}-\frac{y^{2}-2y+12}{y^{2}-16}= \end{equation} \begin{equation}=\frac{y^{2}+4y-y^{2}+2y-12}{y^{2}-16}= \end{equation} \begin{equation}=\frac{6y-12}{y^{2}-4}=\frac{6\left ( y-2 \right )}{\left ( y-2 \right )\left ( y+2 \right )}= \end{equation} \begin{equation}=\frac{6}{y+2} \end{equation} 2)\begin{equation}\frac{y^{2}+4y}{9y-18}\cdot \frac{6}{y+2}= \end{equation} \begin{equation}=\frac{y\left ( y+4 \right )}{9\left ( y-2 \right )}\cdot \frac{6}{y+2}= \end{equation} \begin{equation}=\frac{2y\left ( y+4 \right )}{3\left ( y^{2}-4 \right )} \end{equation} 3)\begin{equation}\left ( \frac{2}{a+1}-\frac{1}{a-1} \right ):\left ( \frac{4}{a+1}+\frac{1}{a-4} \right ) \end{equation} 1)\begin{equation}\frac{2}{a+1}-\frac{1}{a-1}=\frac{2a-2-a-1}{\left ( a+1 \right )\left ( a-1 \right )}= \end{equation} \begin{equation}=\frac{a-3}{a^{2}-1} \end{equation} 2)\begin{equation}\frac{4}{a+1}+\frac{1}{a-4}= \end{equation} \begin{equation}=\frac{4a-16+a+1}{\left ( a+1 \right )\left ( a-4 \right )}= \end{equation} \begin{equation}=\frac{5a-15}{\left ( a+1 \right )\left ( a-4 \right )}=\frac{5\left ( a-3 \right )}{\left ( a+1 \right )\left ( a-4 \right )} \end{equation} 3)\begin{equation}\frac{a-3}{a^{2}-1}:\frac{5\left ( a-3 \right )}{\left ( a+1 \right )\left ( a-4 \right )}= \end{equation} \begin{equation}=\frac{a-3}{\left ( a+1 \right )\left ( a-1 \right )}:\frac{\left ( a+1 \right )\left ( a-4 \right )}{5\cdot \left ( a-3 \right )}= \end{equation} \begin{equation}=\frac{a-4}{5\left ( a-1 \right )} \end{equation} 4)\begin{equation}\left ( \frac{5}{b+3}-\frac{3}{b+5} \right ):\left ( \frac{2}{b-2}-\frac{1}{b+3} \right ) \end{equation} 1)\begin{equation}\frac{5}{b+3}-\frac{3}{b+5}=\frac{5b+25-3b-9}{\left ( b+3 \right )\left ( b+5 \right )}= \end{equation} \begin{equation}=\frac{2b+16}{\left ( b+3 \right )\left ( b+5 \right )}= \end{equation} \begin{equation}=\frac{2\left ( b+8 \right )}{\left ( b+3 \right )\left ( b+5 \right )} \end{equation} 2)\begin{equation}\frac{2}{b-2}-\frac{1}{b+3}=\frac{2b+6-b+2}{\left ( b-2 \right )\left ( b+3 \right )}= \end{equation} \begin{equation}=\frac{b+8}{\left ( b-2 \right )\left ( b+3 \right )} \end{equation} 3)\begin{equation}\frac{2\cdot \left ( b+8 \right )}{\left ( b+3 \right )\left ( b+5 \right )}:\frac{b+8}{\left ( b-2 \right )\left ( b+3 \right )}= \end{equation} \begin{equation}=\frac{2\left ( b+8 \right )}{\left ( b+3 \right )\left ( b+5 \right )}\cdot \frac{\left ( b-2 \right )\left ( b+3 \right )}{b+8}= \end{equation} \begin{equation}=\frac{2\left ( b-2 \right )}{\left ( b+5 \right )} \end{equation} 5)\begin{equation}\left ( \frac{2a+b}{a^{2}}+\frac{3\left ( a-b \right )}{ab}+\frac{a-4b}{b^{2}} \right )\cdot \end{equation} \begin{equation}\cdot \frac{ab}{\left ( a-b \right )^{2}} \end{equation} 1)\begin{equation}\frac{2a+b}{a^{2}}+\frac{3\left ( a-b \right )}{ab}+\frac{a-4b}{b^{2}}= \end{equation} \begin{equation}=\frac{2ab^{2}+b^{3}+3a^{2}b-3ab^{2}+a^{3}-4a^{2}b}{a^{2}b^{2}}= \end{equation} \begin{equation}=\frac{a^{3}+b^{3}-2a^{2}b+ab^{2}}{a^{2}b^{2}}= \end{equation} \begin{equation}=\frac{\left ( a^{3}+b^{3} \right )-ab\left ( a+b \right )}{a^{2}b^{2}}= \end{equation} \begin{equation}=\frac{\left ( a+b \right )\left ( a^{2}-ab+b^{2}-ab \right )}{a^{2}b^{2}}= \end{equation} \begin{equation}=\frac{\left ( a+b \right )\left ( a-b \right )^{2}}{a^{2}b^{2}} \end{equation} 2)\begin{equation}\frac{\left ( a+b \right )\left ( a-b \right )^{2}}{a^{2}b^{2}}\cdot \frac{ab}{\left ( a-b \right )^{2}}= \end{equation} \begin{equation}=\frac{a+b}{ab} \end{equation} 6)\begin{equation}\left ( \frac{2x-3}{x^{2}}-\frac{x+3}{x}+\frac{x+12}{9} \right )\cdot \frac{3x}{9-x^{2}} \end{equation} 1)\begin{equation}\frac{2x-3}{x^{2}}-\frac{x+3}{x}+\frac{x+12}{9}= \end{equation} \begin{equation}=\frac{18x-27-9x^{2}-27x+x^{3}+12x^{2}}{9x^{2}}= \end{equation} \begin{equation}=\frac{x^{3}+3x^{2}-9x-27}{9x^{2}}= \end{equation} \begin{equation}=\frac{\left ( x^{3}-27 \right )+\left ( 3x^{2}-9x\right )}{9x^{2}}= \end{equation} \begin{equation}=\frac{\left ( x-3 \right )\left ( x^{2}+3x+9 \right )+3x\left ( x-3 \right )}{9x^{2}}= \end{equation} \begin{equation}=\frac{\left ( x-3 \right )\left ( x^{2}+3x+9+3x \right )}{9x^{2}}= \end{equation} \begin{equation}=\frac{\left ( x-3 \right )\left ( x+3 \right )^{2}}{9x^{2}} \end{equation} 2)\begin{equation}\frac{\left ( x-3 \right )\left ( x+3 \right )^{2}}{9x^{2}}\cdot \frac{3x}{9-x^{2}}= \end{equation} \begin{equation}=\frac{-\left ( x-3 \right )\left ( x+3 \right )^{2}\cdot 3x}{9x^{2}\cdot \left ( x-3 \right )\left ( x+3 \right )}= \end{equation} \begin{equation}=-\frac{x+3}{3x} \end{equation} 7)\begin{equation}\left ( \frac{a-1}{a+4}-\frac{a-3}{a-4} \right )\cdot \frac{a+4}{3a-8}+\frac{a}{2a-8} \end{equation} 1)\begin{equation}\frac{a-1}{a+4}-\frac{a-3}{a-4}= \end{equation} \begin{equation}=\frac{\left ( a-4 \right )\left ( a-1 \right )-\left ( a+4 \right )\left ( a-3 \right )}{\left ( a+4 \right )\left ( a-4 \right )}= \end{equation} \begin{equation}=\frac{a^{2}-a-4a+4-a^{2}+3a-4a+12}{a^{2}-16}= \end{equation} \begin{equation}=\frac{16-6a}{a^{2}-16}=\frac{2\left ( 8-3a \right )}{a^{2}-16} \end{equation} 2)\begin{equation}\frac{2\left ( 8-3a \right )}{\left ( a-4 \right )\left ( a+4 \right )}\cdot \frac{a+4}{3a-8}=-\frac{2}{a-4} \end{equation} 3)\begin{equation}\frac{a}{2a-8}+\left ( -\frac{2}{a-4} \right )= \end{equation} \begin{equation}=\frac{a}{2\left ( a-4 \right )}-\frac{2}{a-4}= \end{equation} \begin{equation}=\frac{a-4}{2\left ( a-4 \right )}=\frac{1}{2} \end{equation} 8)\begin{equation}\left ( \frac{b+5}{b+3}-\frac{b-10}{b-3} \right ):\frac{6b+10}{b+3}-\frac{b}{2b-6} \end{equation} 1)\begin{equation}\frac{b+5}{b+3}-\frac{b-10}{b-3}= \end{equation} \begin{equation}=\frac{\left ( b-3 \right )\left ( b+5 \right )-\left ( b+3 \right )\left ( b-10 \right )}{b^{2}-9}= \end{equation} \begin{equation}=\frac{b^{2}+5b-3b-15-b^{2}+10b-3b+30}{b^{2}-9}= \end{equation} \begin{equation}=\frac{9b+15}{b^{2}-9}=\frac{3\left ( 3b+5 \right )}{b^{2}-9} \end{equation} 2)\begin{equation}\frac{3\cdot \left ( 3b+5\right )}{b^{2}-9}:\frac{6b+10}{b+3}= \end{equation} \begin{equation}=\frac{3\left ( 3b+5 \right )}{\left ( b-3 \right )\left ( b+3 \right )}\cdot \frac{b+3}{2\left ( 3b+5 \right )}= \end{equation} \begin{equation}=\frac{3}{2\left ( b-3 \right )} \end{equation} 3)\begin{equation}\frac{3}{2\left ( b-3 \right )}-\frac{b}{2b-6}=\frac{3}{2\left ( b-3 \right )}-\frac{b}{2\left ( b-3 \right )}= \end{equation} \begin{equation}=\frac{3-b}{2\left ( b-3 \right )}=-\frac{1}{2} \end{equation}
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