номер 101 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\frac{9}{4x^{2}+3xy}+\frac{12}{4xy+3y^{2}} \end{equation}
при x = -0,125  y = -0,24 \begin{equation}\frac{9}{x\left ( 4x+3y \right )}+\frac{12}{y\left ( 4x+3y \right )}= \end{equation} \begin{equation}=\frac{9y+12x}{xy\left ( 4x+3y \right )}=\frac{3\left ( 3y+4x \right )}{xy\left ( 4x+3y \right )}= \end{equation} \begin{equation}=\frac{3}{xy}=\frac{3}{\left ( -0,125 \right )\cdot \left ( -0,24 \right )}= \end{equation} \begin{equation} =\frac{3}{0,03}=100 \end{equation} 2)\begin{equation}\frac{12}{3a^{2}-2ab}-\frac{18}{3ab-2b^{2}} \end{equation} \begin{equation}при   a=-\frac{3}{35}  b=2\frac{4}{5} \end{equation} \begin{equation}\frac{12}{a\left ( 3a-2b \right )}-\frac{18}{b\left ( 3a-2b \right )}= \end{equation} \begin{equation}=\frac{12b-18a}{ab\left ( 3a-2b \right )}=\frac{6\left ( 2b-3a \right )}{ab\left ( 3a-2b \right )}= \end{equation} \begin{equation}=-\frac{6}{ab}=-\frac{6}{-\frac{3}{35}\cdot 2\frac{4}{5}}=\frac{6}{\frac{3}{35}\cdot \frac{14}{5}}= \end{equation} \begin{equation}=\frac{6}{\frac{6}{25}}=25 \end{equation}
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