номер 115 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\left ( \frac{b-x}{b+x}-1 \right )\cdot \frac{b^{2}-x^{2}}{4x^{3}} \end{equation} 1)\begin{equation}\frac{b-x}{b+x}-1=\frac{b-x-b-x}{b+x}=-\frac{2x}{b+x} \end{equation} 2)\begin{equation}-\frac{2x}{b+x}\cdot \frac{b^{2}-x^{2}}{4x^{3}}= \end{equation} \begin{equation}=-\frac{2}{b+x}\cdot \frac{\left ( b-x \right )\left ( b+x \right )}{4x^{2}}= \end{equation} \begin{equation}=-\frac{b-x}{2x^{2}} \end{equation} 2)\begin{equation}\left ( a^{2}-b^{2} \right )\cdot \left ( \frac{a^{2}}{a-b}-b \right ) \end{equation} 1)\begin{equation}\frac{a^{2}}{a-b}-b=\frac{a^{2}-ab+b^{2}}{a-b} \end{equation} 2)\begin{equation}\left ( a^{2}-b^{2} \right )\cdot \frac{\left ( a^{2}-ab+b^{2} \right )}{a-b}= \end{equation} \begin{equation}=\frac{\left ( a-b \right )\left ( a+b \right )\cdot \left ( a^{2}-ab+b^{2} \right )}{a-b}= \end{equation} \begin{equation}=a^{3}+b^{3} \end{equation} 3)\begin{equation}\left ( x^{2}-y^{2} \right ):\left ( \frac{x^{2}}{x+y}+y \right ) \end{equation} 1)\begin{equation}\frac{x^{2}}{x+y}+y=\frac{x^{2}+xy+y^{2}}{x+y} \end{equation} 2)\begin{equation}\left ( x^{2}-y^{2} \right ):\frac{x^{2}+xy+y^{2}}{x+y}= \end{equation} \begin{equation}=\frac{\left ( x-y \right )\left ( x+y \right )\cdot \left ( x+y \right )}{x^{2}+xy+y^{2}}= \end{equation} \begin{equation}=\frac{\left ( x-y \right )\left ( x+y \right )^{2}}{x^{2}+xy+y^{2}} \end{equation} 2)\begin{equation}\left ( x-y \right )\left ( x+y \right )\cdot \frac{x^{2}+xy+y^{2}}{x+y}=x^{3}-y^{3} \end{equation} 4)\begin{equation}\left ( \frac{b}{a}+\frac{a}{b}+2 \right )\cdot \frac{ab}{b^{2}-a^{2}} \end{equation} 1)\begin{equation}\frac{b}{a}+\frac{a}{b}+2=\frac{b^{2}+a^{2}+2ab}{ab}= \end{equation} \begin{equation}=\frac{\left ( b+a \right )^{2}}{ab} \end{equation} 2)\begin{equation}\frac{\left ( b+a \right )^{2}}{ab}\cdot \frac{ab}{\left ( b-a \right )\left ( b+a \right )}= \end{equation} \begin{equation}=\frac{b+a}{b-a} \end{equation} 5)\begin{equation}\frac{\left ( p+q \right )^{2}}{9q^{2}-p^{2}}:\left ( \frac{p-3q}{p+3q}+2 \right ) \end{equation} 1)\begin{equation}\frac{p-3q}{p+3q}+2=\frac{p-3q+2p+6q}{p+3q}= \end{equation} \begin{equation}=\frac{3p-3q}{p+3q}=\frac{3\left ( p+q \right )}{p+3q} \end{equation} 2)\begin{equation}\frac{\left ( p+q \right )^{2}}{9q^{2}-p^{2}}:\frac{3\left ( p+q \right )}{p+3q}= \end{equation} \begin{equation}=\frac{\left ( p+q \right )^{2}}{\left ( 3q-p \right )\left ( 3q+p \right )}\cdot \frac{p+3q}{3\left ( p+q \right )}= \end{equation} \begin{equation}=\frac{p+q}{3\left ( 3q-p \right )} \end{equation} 6)\begin{equation}\left ( \frac{x}{y}+\frac{y}{x} -2\right ):\frac{\left ( 2x-2y \right )^{2}}{xy} \end{equation} 1)\begin{equation}\frac{x}{y}+\frac{y}{x}-2= \end{equation} \begin{equation}=\frac{x^{2}+y^{2}-2xy}{xy}=\frac{\left ( x-y \right )^{2}}{xy} \end{equation} 2)\begin{equation}\frac{\left ( x-y \right )^{2}}{xy}:\frac{4\left ( x-y \right )^{2}}{xy}= \end{equation} \begin{equation}=\frac{\left ( x-y \right )^{2}}{xy}\cdot \frac{xy}{4\left ( x-y \right )^{2}}=\frac{1}{4} \end{equation} 7)\begin{equation}\frac{a^{2}-ax+x^{2}}{\left ( ax \right )^{2}}:\left ( \frac{a^{2}}{x}+\frac{x^{2}}{a} \right ) \end{equation} 1)\begin{equation}\frac{a^{2}}{x}+\frac{x^{2}}{a}=\frac{a^{3}+x^{3}}{ax} \end{equation} 2)\begin{equation}\frac{a^{2}-ax+x^{2}}{\left ( ax \right )^{2}}:\frac{a^{3}+x^{3}}{ax}= \end{equation} \begin{equation}=\frac{a^{2}-ax+x^{2}}{\left ( ax \right )^{2}}\cdot \frac{ax}{a^{3}+x^{3}}= \end{equation} \begin{equation}=\frac{a^{2}-ax+x^{2}}{ax}\cdot \frac{1}{\left ( a+x \right )\left ( a^{2}-ax+x^{2} \right )}=\frac{1}{ax} \end{equation} 8)\begin{equation}\frac{b^{2}+5b+25}{10b}:\left ( \frac{b^{2}}{5}-\frac{25}{b} \right ) \end{equation} 1)\begin{equation}\frac{b^{2}}{5}-\frac{25}{b}=\frac{b^{3}-125}{5b}= \end{equation} \begin{equation}=\frac{b^{3}-5^{3}}{5b}= \end{equation} 2)\begin{equation}\frac{b^{2}+5b+25}{10b}:\frac{b^{3}-5^{3}}{5b}= \end{equation} \begin{equation}=\frac{b^{2}+5b+25}{10b}\cdot \frac{5b}{\left ( b-5 \right )\left ( b^{2}+5b+25 \right )}= \end{equation} \begin{equation}=\frac{1}{2\left ( b-5 \right )} \end{equation} 9)\begin{equation}\left ( a+b-\frac{4ab}{a+b} \right )\cdot \left ( a-b+\frac{4ab}{a-b} \right ) \end{equation} 1)\begin{equation}a+b-\frac{4ab}{a+b}=\frac{\left ( a+b \right )^{2}-4ab}{a+b}= \end{equation} \begin{equation}=\frac{a^{2}+2ab+b^{2}-4ab}{a+b}= \end{equation} \begin{equation}=\frac{a^{2}-2ab+b^{2}}{a+b}=\frac{\left ( a-b \right )^{2}}{a+b} \end{equation} 2)\begin{equation}a-b+\frac{4ab}{a-b}=\frac{\left ( a-b \right )^{2}+4ab}{a-b}= \end{equation} \begin{equation}=\frac{a^{2}-2ab+b^{2}+4ab}{a-b}= \end{equation} \begin{equation}=\frac{\left ( a^{2}+2ab+b^{2} \right )}{a-b}=\frac{\left ( a+b \right )^{2}}{a-b} \end{equation} 3)\begin{equation}\frac{\left ( a-b \right )^{2}}{a+b}\cdot \frac{\left ( a+b \right )^{2}}{a-b}= \end{equation} \begin{equation}=\left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2} \end{equation} 10)\begin{equation}\left ( \frac{a^{2}}{2a+1}+1 \right )\cdot \left ( \frac{4a^{2}-3}{a+1}+4 \right ) \end{equation} 1)\begin{equation}\frac{a^{2}}{2a+1}+1=\frac{a^{2}+2a+1}{2a+1}= \end{equation} \begin{equation}=\frac{\left ( a+1 \right )^{2}}{2a+1} \end{equation} 2)\begin{equation}\frac{4a^{2}-3}{a+1}+4=\frac{4a^{2}-3+4a+4}{a+1}= \end{equation} \begin{equation}=\frac{4a^{2}+4a+2}{a+1}=\frac{\left ( 2a+1 \right )^{2}}{a+1} \end{equation} 3)\begin{equation}\frac{\left ( a+1 \right )^{2}}{2a+1}\cdot \frac{\left ( 2a+1 \right )^{2}}{a+1}= \end{equation} \begin{equation}=\left ( a+1 \right )\left ( 2a+1 \right ) \end{equation} 11)\begin{equation}\left ( \frac{n^{2}+p^{2}}{p}-2p \right ):\left ( \frac{n^{2}+p^{2}}{p}-2n \right ) \end{equation} 1)\begin{equation}\frac{n^{2}+p^{2}}{p}-2p=\frac{n^{2}+p^{2}-2p^{2}}{p}= \end{equation} \begin{equation}=\frac{n^{2}-p^{2}}{p} \end{equation} 2)\begin{equation}\frac{n^{2}+p^{2}}{p}-2n=\frac{n^{2}+p^{2}-2np}{p}= \end{equation} \begin{equation}=\frac{\left ( n-p \right )^{2}}{p} \end{equation} 3)\begin{equation}\frac{n^{2}-p^{2}}{p}:\frac{\left ( n-p \right )^{2}}{p}= \end{equation} \begin{equation}=\frac{n^{2}-p^{2}}{p}\cdot \frac{p}{\left ( n-p \right )^{2}}= \end{equation} \begin{equation}=\frac{\left ( n+p \right )\left ( n-p \right )\cdot p}{p\cdot \left ( n-p \right )^{2}}= \end{equation} \begin{equation}=\frac{n+p}{n-p} \end{equation} 12)\begin{equation}\left ( \frac{b^{2}+c^{2}}{2b}+c \right ):\left ( \frac{b^{2}+c^{2}}{2b}-c \right ) \end{equation} 1)\begin{equation}\frac{b^{2}+c^{2}}{2b}+c=\frac{b^{2}+c^{2}+2bc}{2b}= \end{equation} \begin{equation}=\frac{\left ( b+c \right )^{2}}{2b} \end{equation} 2)\begin{equation}\frac{b^{2}+c^{2}}{2b}-c=\frac{b^{2}+c^{2}-2bc}{2b}= \end{equation} \begin{equation}=\frac{\left ( b-c \right )^{2}}{2b} \end{equation} 3)\begin{equation}\frac{\left ( b+c \right )^{2}}{2b}:\frac{\left ( b-c \right )^{2}}{2b}= \end{equation} \begin{equation}=\frac{\left ( b+c \right )^{2}}{2b}\cdot \frac{2b}{\left ( b-c \right )^{2}}= \end{equation} \begin{equation}=\left ( \frac{b+c}{b-c} \right )^{2} \end{equation}
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