номер 62 гдз 8 класс алгебра Муравин Муравина

1)\begin{equation}\frac{a-b}{4a^{4}}\cdot \frac{12a^{6}}{b^{2}-ab}= \end{equation} \begin{equation}=\frac{\left ( a-b \right )\cdot 12a^{6}}{4\cdot a^{4}\cdot b\left ( b-a \right )}=-\frac{3a^{2}}{b} \end{equation} 2) \begin{equation}\frac{x^{2}-y^{2}}{27x^{3}}\cdot \frac{18x^{2}}{y^{2}-xy}= \end{equation} \begin{equation}=\frac{\left ( x-y \right )\left ( x+y \right )\cdot 18x^{2}}{27\cdot x^{3}\cdot y\left ( y-x \right )}=-\frac{2\left ( x+y \right )}{3y} \end{equation} 3) \begin{equation}\frac{a^{2}-b^{2}}{25ab}:\frac{\left ( a-b \right )^{2}}{30a^{2}}= \end{equation} \begin{equation}=\frac{\left ( a-b \right )\left ( a+b \right )\cdot 30a^{2}}{25ab\cdot \left ( a-b \right )^{2}}=\frac{6a\left ( a+b \right )}{5\left ( a-b \right )} \end{equation} 4) \begin{equation}\frac{\left ( x-5 \right )}{48x^{2}y^{2}}:\frac{25-x^{2}}{36x^{3}y^{2}} \end{equation} \begin{equation}\frac{\left ( x-5 \right )\cdot 36x^{3}y^{2}}{48x^{2}y^{2}\cdot \left ( 5-x \right )\left ( 5+x \right )}=-\frac{3x}{4\left ( 5+x \right )} \end{equation} 5) \begin{equation}\frac{x^{3}+27}{16x^{3}}:\frac{x^{2}-3x+9}{8x^{2}} \end{equation} \begin{equation}\frac{\left ( x+3 \right )\left ( x^{2}-3x+9 \right )\cdot 8x^{2}}{16x^{3}\left ( x^{2}-3x+9 \right )}=\frac{x+3}{2x}\end{equation} 6) \begin{equation}\frac{a^{2}+ab+b^{2}}{6a+12b}\cdot \frac{a^{2}-4b^{2}}{a^{3}-b^{3}} \end{equation} \begin{equation}\frac{\left ( a^{2}+ab+b^{2} \right )\cdot \left ( a-2b \right )\left ( a+2b \right )}{6\left ( a+2b \right )\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )}= \end{equation} \begin{equation}=\frac{a-2b}{6\left ( a-b \right )} \end{equation} 7)\begin{equation}\frac{by-5y^{2}}{3bn}:\frac{b^{2}-5by}{45ny} \end{equation} \begin{equation}\frac{y\left ( b-5y \right )\cdot 45ny}{3bn\cdot b\left ( b-5y \right )}=\frac{15y^{2}}{b^{2}} \end{equation} 8) \begin{equation}\frac{b^{4}-c^{4}}{b^{3}+c^{3}}\cdot \frac{b^{2}-bc+c^{2}}{b^{2}+c^{2}} \end{equation} \begin{equation}\frac{\left ( b^{2}-c^{2} \right )\left ( b^{2}+c^{2} \right )\cdot \left ( b^{2}-bc+c^{2} \right )}{\left ( b+c \right )\left ( b^{2}-bc+c^{2} \right )\cdot \left ( b^{2}+c^{2} \right )}= \end{equation} \begin{equation}=\frac{\left ( b-c \right )\left ( b+c \right )}{b+c}=b-c \end{equation} 9) \begin{equation}\frac{y^{3}-27z^{3}}{6y-6z}:\frac{y-3z}{72z-72y} \end{equation} \begin{equation}\frac{\left ( y^{3}-27z^{3} \right )\left ( 72z-72y \right )}{\left ( 6y-6z \right )\cdot \left ( y-3z \right )}= \end{equation} \begin{equation}=\frac{\left ( y-3z \right )\left ( y^{2}+3yz+9z^{2} \right )\cdot 72\left ( z-y \right )}{6\left ( y-z \right )\left ( y-3z \right )}= \end{equation} \begin{equation}=-12\cdot \left ( y^{2}+3yz+9z^{2} \right ) \end{equation} 10)\begin{equation}\frac{75d-75n}{d^{3}+8n^{3}}\cdot \frac{d^{2}+2dn}{5d-5n} \end{equation} \begin{equation}\frac{75\left ( d-n \right )\cdot d\left ( d+2n \right )}{\left ( d+2n \right )\left ( d^{2}-2nd+4n^{2} \right )\cdot 5\left ( d-n \right )}= \end{equation} \begin{equation}=\frac{15d}{d^{2}-2nd+4n^{2}} \end{equation} 11)\begin{equation}\frac{5c^{4}}{4c^{2}+9+12c}:\frac{4c^{4}-9c^{2}}{8c^{3}+36c^{2}+54c+27} \end{equation} \begin{equation}\frac{5c^{4}\cdot \left ( \left ( 8c^{3}+27 \right )+18c\left ( 2c+3 \right ) \right )}{\left ( 2c+3 \right )^{2}\cdot \left ( 2c^{2}-3c \right )\left ( 2c^{2}+3c \right )}= \end{equation} \begin{equation}=\frac{5c^{4}\cdot \left ( 2c+3 \right )\left ( 2c^{2}-6c+9 \right )+18c\left ( 2c+3 \right )}{\left ( 2c+3 \right )^{2}\cdot c\left ( 2c-3 \right )\left ( 2c+3 \right )}= \end{equation} \begin{equation}=\frac{5c^{3}\cdot \left ( 2c+3 \right )\left ( 2c^{2}+12c+9 \right )}{\left ( 2c+3 \right )^{3}\cdot \left ( 2c-3 \right )}= \end{equation} \begin{equation}=\frac{5c^{3}\cdot \left ( 2c+3 \right )^{3}}{\left ( 2c+3 \right )^{3}\cdot \left ( 2c-3 \right )}=\frac{5c^{3}}{2c-3} \end{equation} 12)\begin{equation}\frac{64d^{3}-24d^{2}+12d-1}{5cd+3d}\cdot \frac{25c^{2}-9}{16d^{2}-8d+1} \end{equation} \begin{equation}\frac{\left ( \left ( 64d^{3}-1 \right )-12d\left ( 2d-1 \right )\right )\cdot \left ( 5c-3 \right )\left ( 5c+3 \right ) }{d\left ( 5c+3 \right )\cdot \left ( 16d^{2}-8d+1 \right )}= \end{equation} \begin{equation}=\frac{\left ( 4d-1 \right )\left ( 16d^{2}-8d+1 \right )\cdot \left ( 5c-3 \right )}{d\cdot \left ( 16d^{2}-8d+1 \right )}= \end{equation} \begin{equation}=\frac{\left ( 4d-1 \right )\left ( 5c-3 \right )}{d} \end{equation}
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