Решебник по алгебре 7 класс Мерзляк ФГОС Задание 480

 

\[\boxed{\text{480\ (480).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 2a^{3} - 3a^{2} - 2ab + 3b =\]

\[= a^{2} \cdot (2a - 3) - b \cdot (2a - 3) =\]

\[= (2a - 3) \cdot \left( a^{2} - b \right)\]

\[если\ a = 0,5;\ \ \ \ \ b = 2,25:\]

\[(2 \cdot 0,5 - 3) \cdot \left( {0,5}^{2} - 2,25 \right) =\]

\[= (1 - 3) \cdot (0,25 - 2,25) =\]

\[= - 2 \cdot ( - 2) = 4.\]

\[2)\ xy + y^{2} - 12x - 12y =\]

\[= y \cdot (x + y) - 12 \cdot (x + y) =\]

\[= (x + y) \cdot (y - 12)\]

\[если\ x = 10,8;\ \ \ \ y = - 8,8:\]

\[(10,8 - 8,8) \cdot ( - 8,8 - 12) =\]

\[= 2 \cdot ( - 20,8) = - 41,6.\]

\[3)\ 27x^{3} - 36x^{2} + 6x - 8 =\]

\[= 9x^{2} \cdot (3x - 4) + 2 \cdot (3x - 4) =\]

\[= (3x - 4) \cdot \left( 9x^{2} + 2 \right)\]

\[при\ x = - 1\frac{1}{3} = - \frac{4}{3}:\]

\[\boxed{\text{480.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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