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номер 99 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\frac{9}{n^{2}-9n}+\frac{1}{9-n} \end{equation} \begin{equation}\frac{9}{n\left ( n-9 \right )}+\frac{1}{9-n}=\frac{9}{n\left ( n-9 \right )}-\frac{1}{n-9}= \end{equation} \begin{equation}=\frac{9-n}{n\left ( n-9 \right )}=-\frac{1}{n} \end{equation} 2)\begin{equation}\frac{4}{n+7}+\frac{4n}{7n+49} \end{equation} \begin{equation}\frac{4}{n+7}+\frac{4n}{7\left ( n+7 \right )}=\frac{28+4n}{7\left ( n+7 \right )}= \end{equation} \begin{equation}=\frac{4\left ( 7+n \right )}{7\left ( n+7 \right )}=\frac{4}{7} \end{equation} 3)\begin{equation}\frac{1}{a^{2}-5a}+\frac{1}{25-5a} \end{equation} \begin{equation}\frac{1}{a\left ( a-5 \right )}+\frac{1}{5\left ( 5-a \right )}= \end{equation} \begin{equation}=\frac{1}{a\left ( a-5 \right )}-\frac{1}{5\left ( a-5 \right )}= \end{equation} \begin{equation}=\frac{5-a}{5a\left ( a-5 \right )}=\frac{1}{5a} \end{equation} 4)\begin{equation}\frac{1}{x^{2}-3x}+\frac{1}{9-3x} \end{equation} \begin{equation}\frac{1}{x^{2}-3x}+\frac{1}{3\left ( 3-x \right )}= \end{equation} \begin{equation}=\frac{1}{x\left ( x-3 \right )}-\frac{1}{3\left ( x-3 \right )}= \end{equation} \begin{equation}=\frac{3-x}{3x\left ( x-3 \right )}=-\frac{1}{3x} \end{equation} 5)\begin{equation}\frac{1}{y^{2}+5y}+\frac{1}{25+5y} \end{equation} \begin{equation}\frac{1}{y\left ( y+5 \right )}+\frac{1}{5\left ( 5+y \right )}= \end{equation} \begin{equation}=\frac{5+y}{5y\left ( y+5 \right )}=\frac{1}{5y} \end{equation} 6)\begin{equation}\frac{1}{16-4c}+\frac{1}{c^{2}-4c} \end{equation} \begin{equation}\frac{1}{4\left ( 4-c \right )}+\frac{1}{c\left ( c-4 \right )}= \end{equation} \begin{equation}=\frac{1}{4\left ( 4+c \right )}-\frac{1}{c\left ( 4-c \right )}= \end{equation} \begin{equation}=\frac{c-4}{4c\left ( 4-c \right )}=-\frac{1}{4c} \end{equation} 7)\begin{equation}\frac{16a-15}{3a^{2}-2a}-\frac{13}{4-6a} \end{equation} \begin{equation}\frac{16a-15}{a\left (3a-2 \right )}-\frac{13}{2\left ( 2-3a \right )}= \end{equation} \begin{equation}=\frac{16a-15}{a\left ( 3a-2 \right )}+\frac{13}{2\left ( 3a-2 \right )}= \end{equation} \begin{equation}=\frac{32a-30+13a}{2a\left ( 3a-2 \right )}= \end{equation} \begin{equation}=\frac{45a-30}{2a\left ( 3a-2 \right )}= \end{equation} \begin{equation}=\frac{15\left ( 3a-2 \right )}{2a\left ( 3a-2 \right )}=\frac{15}{2a} \end{equation} 8)\begin{equation}\frac{11}{2x-5}+\frac{9x+5}{5x-2x^{2}} \end{equation} \begin{equation}\frac{11}{2x-5}+\frac{9x+5}{x\left ( 5-2x \right )}= \end{equation} \begin{equation}=\frac{11}{2x-5}-\frac{9x+5}{x\left ( 2x-5 \right )}= \end{equation} \begin{equation} =\frac{11x-9x-5}{x\left ( 2x-5 \right )}=\frac{2x-5}{x\left ( 2x-5 \right )}=\frac{1}{x} \end{equation} 9)\begin{equation}\frac{9b}{2a^{2}-3ab}-\frac{4a}{2ab-3b^{2}} \end{equation} \begin{equation}\frac{9b}{a\left ( 2a-3b \right )}-\frac{4a}{b\left ( 2a-3b \right )}=\frac{9b^{2}-4a^{2}}{ab\left ( 2a-3b \right )}= \end{equation} \begin{equation}=\frac{\left ( 3b-2a \right )\left ( 3b+2a \right )}{ab\left ( 2a-3b \right )}= \end{equation} \begin{equation}=-\frac{3b+2a}{ab} \end{equation} 10)\begin{equation}\frac{4x}{5y^{2}-2xy}-\frac{25y}{5xy-2x^{2}} \end{equation} \begin{equation}\frac{4x}{y\left ( 5y-2x \right )}-\frac{25y}{x\left ( 5y-2x \right )}= \end{equation} \begin{equation}=\frac{4x^{2}-25y^{2}}{xy\left ( 5y-2x \right )}= \end{equation} \begin{equation}=\frac{\left ( 2x-5y \right )\left ( 2x+5y \right )}{xy\left ( 5y-2x \right )}= \end{equation} \begin{equation}=-\frac{2x+5y}{xy} \end{equation} 11)\begin{equation}\frac{3}{a^{2}-4ac}-\frac{6}{a^{2}-16c^{2}} \end{equation} \begin{equation}\frac{3}{a\left ( a-4c \right )}-\frac{6}{\left ( a-4c \right )\left ( a+4c \right )}= \end{equation} \begin{equation}=\frac{3a+12c-6a}{a\left ( a-4c \right )\left ( a+4c \right )}= \end{equation} \begin{equation}=\frac{12c-3a}{a\left ( a-4c \right )\left ( a+4c \right )}= \end{equation} \begin{equation}=\frac{3\left ( 4c-a \right )}{a\left ( a-4c \right )\left ( a+4c \right )}=-\frac{3}{a\left ( a+4c \right )} \end{equation} 12)\begin{equation}\frac{12}{9a^{2}-16b^{2}}-\frac{2}{3a^{2}-4ab} \end{equation} \begin{equation}\frac{12}{9a^{2}-16b^{2}}-\frac{2}{a\left ( 3a-4b \right )}= \end{equation} \begin{equation}=\frac{12a-6a-8b}{a\left ( 9a^{2}-16b^{2} \right )}= \end{equation} \begin{equation}=\frac{6a-8b}{a\left ( 9a^{2}-16b^{2} \right )}= \end{equation} \begin{equation}=\frac{2\left ( 3a-4b \right )}{a\left ( 3a-4b \right )\left ( 3a+4b \right )}= \end{equation} \begin{equation}=\frac{2}{a\left ( 3a+4b \right )} \end{equation} 13)\begin{equation}\frac{1}{a^{2}-x^{2}}+\frac{1}{2ax+2x^{2}} \end{equation} \begin{equation}\frac{1}{a^{2}-x^{2}}+\frac{1}{2x\left ( a+x \right )}= \end{equation} \begin{equation}=\frac{2x+a-x}{2\left ( a^{2}-x^{2} \right )}= \end{equation} \begin{equation}=\frac{x+a}{2\left ( a-x \right )\left ( a+x \right )}=\frac{1}{2\left ( a-x \right )} \end{equation} 14)\begin{equation}\frac{4}{c^{2}-4p^{2}}-\frac{1}{cp-2p^{2}} \end{equation} \begin{equation}\frac{4}{c^{2}-4p^{2}}-\frac{1}{p\left ( c-2p \right )}= \end{equation} \begin{equation}=\frac{4p-c-2p}{p\left ( c^{2}-4p^{2} \right )}= \end{equation} \begin{equation}=\frac{2p-c}{p\left ( c-2p \right )\left ( c+2p \right )}= \end{equation} \begin{equation}=-\frac{1}{p\left ( c+2p \right )} \end{equation} 15)\begin{equation}\frac{1}{10xy+50y^{2}}+\frac{1}{x^{2}-25y^{2}} \end{equation} \begin{equation}\frac{1}{10y\left ( x+5y \right )}+\frac{1}{x^{2}-25y^{2}}= \end{equation} \begin{equation}=\frac{x+5y+10y}{10y\left ( x^{2}-25y^{2} \right )}=\frac{x+5y}{10y\left ( x^{2}-25y^{2} \right )}= \end{equation} \begin{equation}=\frac{x+5y}{10y\left ( x-5y \right )\left ( x+5y \right )}=\frac{1}{10y\left ( x-5y \right )} \end{equation} 16)\begin{equation}\frac{1}{12ac-8c^{2}}-\frac{1}{4c^{2}-9a^{2}} \end{equation} \begin{equation}\frac{1}{4c\left ( 3a-2c \right )}-\frac{1}{4c^{2}-9a^{2}}= \end{equation} \begin{equation}=\frac{2c+3a-4c}{4c\left ( 4c^{2}-9a^{2} \right )}= \end{equation} \begin{equation}=\frac{3a-2c}{4c\left ( 2c-3a \right )\left ( 2c+3a \right )}= \end{equation} \begin{equation}=-\frac{1}{4c\left ( 2c+3a \right )} \end{equation}