номер 104 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\frac{b}{a}-\frac{a}{a-b}+\frac{b^{2}}{a^{2}-ab}=-1 \end{equation} \begin{equation}\frac{b}{a}-\frac{a}{a-b}+\frac{b^{2}}{a\left ( a-b \right )}= \end{equation} \begin{equation}=\frac{ab-b^{2}-a^{2}+b^{2}}{a\left ( a-b \right )}= \end{equation} \begin{equation}=\frac{ab-a^{2}}{a\left ( a-b \right )}=\frac{a\left ( b-a \right )}{a\left ( a-b \right )}=-1 \end{equation}
-1 = -1
2)\begin{equation}\frac{36}{x^{2}-6x}-\frac{x}{x-6}+\frac{6}{x}=-1 \end{equation} \begin{equation}\frac{36}{x\left ( x-6 \right )}-\frac{x}{x-6}+\frac{6}{x}= \end{equation} \begin{equation}=\frac{36-x^{2}+6x-36}{x\left ( x-6 \right )}= \end{equation} \begin{equation}=\frac{-x^{2}+6x}{x\left ( x-6 \right )}=\frac{-x\left ( x-6 \right )}{x\left ( x-6 \right )}=-1 \end{equation}
-1 = -1
3)\begin{equation}\frac{ab\left ( x+y \right )}{xy\left ( a+b \right )}-\frac{ab-xy}{ax+bx}-\frac{ab+xy}{ay+by}=\frac{y-x}{a+b} \end{equation} \begin{equation}\frac{ab\left ( x+y \right )}{xy\left ( a+b \right )}-\frac{ab-xy}{x\left ( a+b \right )}-\frac{ab+xy}{y\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{ab\left ( x+y \right )-aby+xy^{2}-abx-x^{2}y}{xy\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{abx+aby-aby+xy^{2}-abx-x^{2}y}{xy\left ( a+b \right )}= \end{equation} \begin{equation}=\frac{xy^{2}-x^{2}y}{xy\left ( a+b \right )}=\frac{xy\left ( y-x \right )}{xy\left ( a+b \right )}=\frac{y-x}{a+b} \end{equation}
4)\begin{equation}\frac{ab+xy}{x^{2}+xy}-\frac{ab}{xy}-\frac{xy-ab}{xy+y^{2}}=\frac{y-x}{y+x} \end{equation} \begin{equation}\frac{ab+xy}{x\left ( x+y \right )}-\frac{ab}{xy}-\frac{xy-ab}{y\left ( x+y \right )}= \end{equation} \begin{equation}=\frac{aby+xy^{2}-abx-aby-x^{2}y+abx}{xy\left ( x+y \right )}= \end{equation} \begin{equation}=\frac{xy^{2}-x^{2}y}{xy\left ( x+y \right )}=\frac{xy\left ( y-x \right )}{xy\left ( x+y \right )}=\frac{y-x}{x+y} \end{equation} \begin{equation}\frac{y-x}{y+x}=\frac{y-x}{x+y} \end{equation}
5)\begin{equation}\frac{a}{c^{2}-cd}-\frac{b}{d^{2}-cd}-\frac{2\left ( a+b \right )}{c^{2}-d^{2}}=\frac{bc-ad}{cd\left ( c+d \right )} \end{equation} \begin{equation}\frac{a}{c\left ( c-d \right )}-\frac{b}{d\left ( d-c \right )}-\frac{2\left ( a+b \right )}{c^{2}-d^{2}}= \end{equation} \begin{equation}=\frac{a}{c\left ( c-d \right )}+\frac{b}{d\left ( c-d \right )}-\frac{2\left ( a+b \right )}{c^{2}-d^{2}}= \end{equation} \begin{equation}=\frac{adc+ad^{2}+bc^{2}+bcd-2acd-2bcd}{cd\left ( c^{2}-d^{2} \right )}= \end{equation} \begin{equation}=\frac{ad^{2}+bc^{2}-adc-bcd}{cd\left ( c^{2}-d^{2} \right )}= \end{equation} \begin{equation}=\frac{ad\left ( d-c \right )+bc\left ( c-d \right )}{cd\left ( c^{2}-d^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( c-d \right )\left ( bc-ad \right )}{cd\left ( c-d \right )\left ( c+d \right )}=\frac{bc-ad}{cd\left ( c+d \right )} \end{equation} \begin{equation}\frac{bc-ad}{cd\left ( c+d \right )}=\frac{bc-ad}{cd\left ( c+d \right )} \end{equation}
6)\begin{equation}\frac{k}{a^{2}+ab}+\frac{n}{b^{2}+ab}-\frac{2\left ( k-n \right )}{a^{2}-b^{2}}=\frac{an-bk}{ab\left ( a-b \right )} \end{equation} \begin{equation}\frac{k}{a\left ( a+b \right )}+\frac{n}{b\left ( b+a \right )}-\frac{2\left ( k-n \right )}{a^{2}-b^{2}}= \end{equation} \begin{equation}=\frac{bk\left ( a-b \right )+an\left ( a-b \right )-2ab\left ( k-n \right )}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{abk-b^{2}k+a^{2}n-anb-2abk+2abn}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{abn-abk-b^{2}k+a^{2}n}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( a^{2}n+abn \right )-\left ( abk+b^{2}k \right )}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{an\left ( a+b \right )-bk\left ( a+b \right )}{ab\left ( a^{2}-b^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( a+b \right )\left ( an-bk \right )}{ab\left ( a-b \right )\left ( a+b \right )}=\frac{an-bk}{ab\left ( a-b \right )} \end{equation} \begin{equation}\frac{an-bk}{ab\left ( a-b \right )}=\frac{an-bk}{ab\left ( a-b \right )} \end{equation}
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