ГДЗ по алгебре 7 класс Макарычев Задание 560

\[\boxed{\text{560\ (560).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( - x^{2}y^{2} \right)^{4} \cdot ( - xy)^{2} =\]

\[= x^{8}y^{8} \cdot x^{2}y^{2} = x^{10}y^{10}\]

\[\textbf{б)} - \left( \frac{1}{3}xy^{3} \right)^{2} \cdot ( - 3x)^{3} =\]

\[= \frac{1}{9}x^{2}y^{6} \cdot 27x^{3} = 3x^{5}y^{6}\ \]

\[\textbf{в)}\ \left( - 2x^{3}y^{2} \right)^{3} \cdot \left( - 2y^{2} \right)^{3} =\]

\[= 8x^{9}y^{6} \cdot 8y^{6} = 64x^{9}y^{12}\]

\[\textbf{г)}\ \left( \frac{1}{3}a^{2}b \right)^{3} \cdot \left( 9ab^{2} \right)^{2} =\]

\[= \frac{1}{27}a^{6}b^{3} \cdot 81a^{2}b^{4} = 3a^{8}b^{7}\]

\[\textbf{д)}\ \left( - 5a^{3}b \right)^{2} \cdot \left( \frac{1}{5}ab^{3} \right)^{3} =\]

\[= 25\ a^{6}b^{2} \cdot \frac{1}{125}a^{3}b^{9} = \frac{1}{5}a^{9}b^{11}\]

\[\textbf{е)}\ \left( - \frac{2}{7}ab^{4} \right)^{2} \cdot \left( - 3\frac{1}{2}a^{3}b \right)^{2} =\]

\[= \frac{4}{49}a^{2}b^{8} \cdot \frac{49}{4}a^{6}b^{2} = a^{8}b^{10}\]

\[\textbf{ж)}\ \left( x^{3}y \right)^{2} \cdot ( - 5xy)^{3} =\]

\[= x^{6} \cdot y^{2} \cdot ( - 125) \cdot x^{3}y^{3} =\]

\[= - 125x^{9}y^{5}\]

\[\textbf{з)}\ \left( \frac{1}{6}x^{2}y^{2} \right)^{2} \cdot \left( - 12x^{3}y^{5} \right)^{2} =\]

\[= \frac{1}{36}x^{4}y^{4} \cdot 144x^{6}y^{10} = 4x^{10}y^{14}\ \]