ГДЗ по алгебре 7 класс Макарычев Задание 713

\[\boxed{\text{713\ (713).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ p^{2}q^{2} + pq - q^{3} - p^{3} =\]

\[= \left( qp - q^{3} \right) - \left( p^{3} - p^{2}q^{2} \right) =\]

\[= q\left( p - q^{2} \right) - p^{2}\left( p - q^{2} \right) =\]

\[= (p - q^{2})(q - p^{2})\]

\[если\ p = 0,5;\ \ q = - 0,5:\]

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

\[= (0,5 - 0,25)( - 0,5 - 0,25) =\]

\[= 0,25 \cdot ( - 0,75) = - 0,1875.\]

\[\textbf{б)}\ 3x^{3} - 2y^{3} - 6x^{2}y^{2} + xy =\]

\[= \left( 3x^{3} - 6x^{2}y^{2} \right) + \left( xy - 2y^{3} \right) =\]

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

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

\[если\ x = \frac{2}{3};\ y = \frac{1}{2}:\]

\[\left( \frac{2}{3} - 2 \cdot \left( \frac{1}{2} \right)^{2} \right)\left( 3 \cdot \left( \frac{2}{3} \right)^{2} + \frac{1}{2} \right) =\]

\[= \left( \frac{2}{3} - \frac{2}{4} \right)\left( \frac{12}{9} + \frac{1}{2} \right) =\]

\[= \left( \frac{2^{\backslash 2}}{3} - \frac{1^{\backslash 3}}{2} \right)\left( \frac{4^{\backslash 2}}{3} + \frac{1^{\backslash 3}}{2} \right) =\]

\[= \frac{4 - 3}{6} \cdot \frac{8 + 3}{6} = \frac{1}{6} \cdot \frac{11}{6} = \frac{11}{36}\]