Решебник по алгебре 7 класс Мерзляк Задание 372

\[\boxed{\text{372\ (372).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (x - y) \cdot \ *\ = x^{2}y^{2} - x^{3}y\]

\[(x - y) \cdot \ *\ = - x^{3}y + x^{2}y^{2}\]

\[(x - y) \cdot \ *\ = - x^{2}y \cdot x + x^{2}y \cdot y\ \]

\[(x - y) \cdot \ * = - x^{2}y(x - y)\]

\[*\ = - x^{2}\text{y.}\]

\[2)\ \left( - 9x^{2} + \ * \right) \cdot y = \ * + y^{4}\]

\[- 9x^{2} \cdot y + *_{1} \cdot y = *_{2} + y^{4}\]

\[- 9x^{2} \cdot y = *_{2}\]

\[*_{1} \cdot y = y^{4}\]

\[*_{2} = - 9x^{2} \cdot y\]

\[*_{1} = \frac{y_{4}}{y} = y^{3}.\]

\[3)\ (1,4x - *) \cdot 3x = \ * - 0,6x^{3}\]

\[1,4x \cdot 3x - *_{1} \cdot 3x = *_{2} - 0,6x^{3}\]

\[4,2x^{2} - *_{1} \cdot 3x = *_{2} - 0,6x^{3}\]

\[4,2x^{2} = *_{2}\]

\[*_{1} \cdot 3x = 0,6x^{3}\]

\[*_{1} = \frac{0,6x^{3}}{3x} = 0,2x^{2}.\]

\[*_{2} = 4,2x^{2};\ \ \ \ \ *_{1} = 0,2x^{2}.\]

\[4)*\ \cdot \left( * - x^{2}y^{5} + 5y^{6} \right) =\]

\[= 8x^{3}y^{3} + 5x^{3}y^{8} - \ *\]

\[*_{1} \cdot *_{2} + *_{1} \cdot \left( - x^{2}y^{5} \right) + *_{1} \cdot 5y^{6} =\]

\[= 8x^{3}y^{3} + 5x^{3}y^{8} - *_{3}\]

\[1)*_{1} \cdot *_{2} = 8x^{3}y^{3}\]

\[2) - *_{1} \cdot x^{2}y^{5} = 5x^{3}y^{8}\]

\[3)*_{1} \cdot 5y^{6} = - *_{3}\]

\[*_{1} \cdot 5y^{6} = - *_{3}\]

\[*_{1} = \frac{5x^{2}y^{8}}{- x^{2}y^{5}} = - 5xy^{3}\]

\[*_{1} = - 5xy^{3}\]

\[*_{1} \cdot *_{2} = 8x^{3}y^{3}\]

\[- 5xy^{3} \cdot *_{2} = 8x^{3}y^{2}\]

\[*_{2} = \frac{8x^{3}y^{3}}{- 5xy^{3}} = - 1,6x^{2}\]

\[*_{3} = *_{1} \cdot 5y^{6}\]

\[*_{3} = - 5xy^{3} \cdot 5y^{6} = - 25xy^{9}.\]