Решебник по алгебре 7 класс Мерзляк Задание 393

\[\boxed{\text{393\ (393).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ (a + b) \cdot (c - d) =\]

\[= ac - ad + bc - bd.\]

\[2)\ (x - 6) \cdot (x - 4) =\]

\[= x^{2} - 4x - 6x + 24 =\]

\[= x^{2} - 10x + 24.\]

\[3)\ (a - 3) \cdot (a + 7) =\]

\[= a^{2} + 7a - 3a - 21 =\]

\[= a^{2} + 4a - 21.\]

\[4)\ (11 - c) \cdot (c + 8) =\]

\[= 11c + 88 - c^{2} - 8c =\]

\[= - c^{2} + 3c + 88.\]

\[5)\ (d + 13) \cdot (2d - 1) =\]

\[= 2d^{2} - d + 26d - 13 =\]

\[= 2d^{2} + 25d - 13.\]

\[6)\ (3y - 5) \cdot (2y - 12) =\]

\[= 6y^{2} - 36y - 10y + 60 =\]

\[= 6y^{2} - 46y + 60.\]

\[7)\ \left( 2x^{2} - 3 \right) \cdot \left( x^{2} + 4 \right) =\]

\[= 2x^{4} + 8x^{2} - 3x^{2} - 12 =\]

\[= 2x^{4} + 5x^{2} - 12.\]

\[8)\ (x - 6) \cdot \left( x^{2} - 2x + 9 \right) =\]

\[= x^{3} - 8x^{2} + 21x - 54.\]

\[9)\ (5x - y) \cdot \left( 2x^{2} + xy - 3y^{2} \right) =\]

\[= 10x^{3} + 3y^{3} + 3x^{2}y - 16xy^{2}.\]

\[10)\ b \cdot (6b + 7) \cdot (3b - 4) =\]

\[= \left( 6b^{2} + 7b \right) \cdot (3b - 4) =\]

\[= 18b^{3} - 24b^{2} + 21b^{2} - 28b =\]

\[= 18b^{3} - 3b^{2} - 28b.\ \]