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номер 102 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\frac{1}{n+k}+\frac{2k}{n^{2}-k^{2}}+\frac{1}{n-k} \end{equation} \begin{equation}\frac{n-k+2k+n+k}{n^{2}-k^{2}}=\frac{2n+2k}{n^{2}-k^{2}}= \end{equation} \begin{equation}=\frac{2\left ( n+k \right )}{\left ( n-k \right )\left ( n+k \right )}=\frac{2}{n-k} \end{equation} 2)\begin{equation}\frac{4a}{a^{2}-b^{2}}+\frac{3}{a-b}-\frac{2}{a+b} \end{equation} \begin{equation}\frac{4a}{a^{2}-b^{2}}+\frac{3}{a-b}-\frac{2}{a+b}= \end{equation} \begin{equation}=\frac{4a+3a+3b-2a+2b}{a^{2}-b^{2}}= \end{equation} \begin{equation}=\frac{5a+5b}{a^{2}-b^{2}}=\frac{5\left ( a+b \right )}{\left ( a-b \right )\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{5}{a-b} \end{equation} 3)\begin{equation}\frac{1}{3c+6}+\frac{3}{4c-8}-\frac{c+1}{c^{2}-4} \end{equation} \begin{equation}\frac{1}{3\left ( c+2 \right )}+\frac{3}{4\left ( c-2 \right )}-\frac{c+1}{c^{2}-4}= \end{equation} \begin{equation}=\frac{4c-8+9c+18-12c-12}{12\left ( c^{2}-4 \right )}= \end{equation} \begin{equation}=\frac{c-2}{12\left ( c-2 \right )\left ( c+2 \right )}=\frac{1}{12\left ( c+2 \right )} \end{equation} 4)\begin{equation}\frac{1}{x^{2}-2xy}+\frac{2}{4y^{2}-x^{2}}-\frac{1}{x^{2}+2xy} \end{equation} \begin{equation}\frac{1}{x\left ( x-2y \right )}+\frac{2}{4y^{2}-x^{2}}-\frac{1}{x\left ( x+2y \right )}= \end{equation} \begin{equation}=\frac{x+2y+2x-x+2y}{x\left ( x+2y \right )}= \end{equation} \begin{equation}=\frac{2x+4y}{x\left ( x+2y \right )}=\frac{2\left ( x+2y \right )}{x\left ( x+2y \right )}=\frac{2}{x} \end{equation} 5)\begin{equation}\frac{2x+3}{2x^{2}-3x}+\frac{2x-3}{2x^{2}+3x}-\frac{16x}{4x^{2}-9} \end{equation} \begin{equation}\frac{2x+3}{x\left ( 2x-3 \right )}+\frac{2x-3}{x\left ( 2x+3 \right )}-\frac{16x}{4x^{2}-9}= \end{equation} \begin{equation}=\frac{\left ( 2x+3 \right )^{2}+\left ( 2x-3 \right )^{2}-16x^{2}}{x\left ( 4x^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{4x^{2}+12x+9+4x^{2}-12x+9-16x^{2}}{x\left ( 4x^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{-8x^{2}+18}{x\left ( 4x^{2}-9 \right )}=-\frac{2\left ( 4x^{2}-9 \right )}{x\left ( 4x^{2}-9 \right )}=-\frac{2}{x} \end{equation} 6)\begin{equation}\frac{36a}{9a^{2}-25}-\frac{3a-5}{3a^{2}+5a}-\frac{3a+5}{3a^{2}+5a} \end{equation} \begin{equation}\frac{36a}{9a^{2}-25}-\frac{3a-5}{a\left ( 3a+5 \right )}-\frac{3a+5}{-a\left ( 3a+5 \right )}= \end{equation} \begin{equation}=\frac{36a^{2}-\left ( 3a-5 \right )^{2}-\left ( 3a+5 \right )\left ( 3a-5 \right )}{a\left ( 9a^{2}-25 \right )}= \end{equation} \begin{equation}=\frac{36a^{2}-\left ( 9a^{2}-30a+25 \right )-\left ( 9a^{2}-25 \right )}{a\left ( 9a^{2}-25 \right )}= \end{equation} \begin{equation}=\frac{36a^{2}-9a^{2}+30a-25-9a^{2}+25}{a\left ( 9a^{2}-25 \right )}= \end{equation} \begin{equation}=\frac{18a^{2}+30a}{a\left ( 9a^{2}-25 \right )}=\frac{6a\left ( 3a+5 \right )}{a\left ( 3a-5 \right )\left ( 3a+5 \right )}= \end{equation} \begin{equation}=\frac{6}{3a-5} \end{equation} 7)\begin{equation}\frac{4b}{a^{2}-b^{2}}+\frac{a-b}{a^{2}+ab}+\frac{a+b}{b^{2}-ab} \end{equation} \begin{equation}\frac{4b}{a^{2}-b^{2}}+\frac{a-b}{a\left ( a+b \right )}+\frac{a+b}{b\left ( b-a \right )}= \end{equation} \begin{equation}=\frac{4b}{a^{2}-b^{2}}+\frac{a-b}{a\left ( a+b \right )}-\frac{a+b}{b\left ( a-b \right )}= \end{equation} \begin{equation}=\frac{4ab^{2}+b\left ( a-b \right )^{2}-a\left ( a+b \right )^{2}}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{4ab^{2}+b\left ( a^{2} -2ab+b^{2}\right )-a\left ( a^{2}+2ab+b^{2} \right )}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{4ab^{2}+a^{2}b-2ab^{2}+b^{3}-a^{3}-2a^{2}b-ab^{2}}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{2ab^{2}-2a^{2}b+b^{3}-a^{3}}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{2ab\left ( b-a \right )+\left ( b-a \right )\left ( b^{2}+ab+a^{2} \right )}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( b-a \right )\left ( 2ab+b^{2}+ab+a^{2} \right )}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( b-a \right )\left ( a^{2}+3ab+b^{2} \right )}{ab\left ( a-b \right )\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{-a^{2}+3ab+b^{2}}{ab\left ( a+b \right )} \end{equation} 8)\begin{equation}\frac{k-3}{3k+9}+\frac{k+3}{k^{2}-3k}-\frac{12}{k^{2}-9} \end{equation} \begin{equation}\frac{k-3}{3\left ( k+3 \right )}+\frac{k+3}{k\left ( k-3 \right )}-\frac{12}{k^{2}-9}= \end{equation} \begin{equation}=\frac{k\left ( k-3 \right )^{2}+3\left ( k+3 \right )^{2}-36k}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{k\left ( k^{2}-6k+9 \right )+3\left ( k^{2}+6k+9 \right )-36k}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{k^{3}-6k^{2}+9k+3k^{2}+18k+27-36k}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{k^{3}+27-3k^{2}-9k}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{\left ( k^{3}+27 \right )-\left ( 3k^{2}+9k \right )}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{\left ( k+3 \right )\left ( k^{2}-3k+9 \right )-3k\left ( k+3 \right )}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{\left ( k+3 \right )\left ( k^{2}-3k+9-3k \right )}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{\left ( k+3 \right )\left ( k^{2}-6k+9 \right )}{3k\left ( k^{2}-9 \right )}= \end{equation} \begin{equation}=\frac{\left ( k+3 \right )\left ( k-3 \right )^{2}}{3k\left ( k-3 \right )\left ( k+3 \right )}=\frac{k-3}{3k} \end{equation} 9)\begin{equation}\frac{3a^{2}+6}{a^{3}+1}-\frac{3}{a^{2}-a+1}-\frac{1}{a+1} \end{equation} \begin{equation}\frac{3a^{2}+6-3a-3-a^{2}+a-1}{a^{3}+1}= \end{equation} \begin{equation}=\frac{2a^{2}-2a+2}{a^{3}+1}= \end{equation} \begin{equation}=\frac{2\left ( a^{2}-a+1 \right )}{\left ( a+1 \right )\left ( a^{2}-a+1 \right )}=\frac{2}{a+1} \end{equation} 10)\begin{equation}\frac{1}{b-3}+\frac{3b^{2}}{27-b^{3}}+\frac{9}{b^{2}+3b+9} \end{equation} \begin{equation}\frac{1}{b-3}-\frac{3b^{2}}{b^{3}-27}+\frac{9}{b^{2}+3b+9}= \end{equation} \begin{equation}=\frac{b^{2}+3b+9-3b^{2}+9b-27}{b^{3}-27}= \end{equation} \begin{equation}=\frac{-2b^{2}+12b-18}{b^{3}-27}=\frac{-2\left ( b^{2}-6b+9 \right )}{b^{3}-27}= \end{equation} \begin{equation}=\frac{-2\left ( b-3 \right )^{2}}{\left ( b-3 \right )\left ( b^{2}+3b+9 \right )}=-\frac{2\left ( b-3 \right )}{b^{2}+3b+9} \end{equation}