Решебник по алгебре 7 класс Мерзляк Задание 441

\[\boxed{\text{441\ (441).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 2x \cdot (a + b) + y \cdot (a + b) =\]

\[= (a + b) \cdot (2x + y).\]

\[2)\ (a - 4) - b \cdot (a - 4) =\]

\[= (a - 4) \cdot (1 - b).\]

\[3)\ 5a \cdot (m - n) + 7b \cdot (m - n) =\]

\[= (m - n) \cdot (5a + 7b).\]

\[4)\ 6x \cdot (4x + 1) - 11 \cdot (4x + 1) =\]

\[= (4x + 1) \cdot (6x - 11).\]

\[5)\ a \cdot (c - d) + b \cdot (d - c) =\]

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

\[6)\ x \cdot (x - 6) - 10 \cdot (6 - x) =\]

\[= (x - 6) \cdot (x + 10).\]

\[7)\ b \cdot (b - 20) + (20 - b) =\]

\[= (b - 20) \cdot (b - 1).\]

\[8)\ 6a \cdot (a - 3b) - 13b \cdot (3b - a) =\]

\[= (a - 3b) \cdot (6a + 13b).\]

\[9)\ (m - 9)^{2} - 3 \cdot (m - 9) =\]

\[= (m - 9) \cdot (m - 9 - 3) =\]

\[= (m - 9) \cdot (m - 12).\]

\[10)\ a \cdot (a + 5)^{2} + (a + 5) =\]

\[= (a + 5) \cdot \left( a \cdot (a + 5) + 1 \right) =\]

\[= (a + 5) \cdot \left( a^{2} + 5a + 1 \right).\]

\[11)\ \left( m^{2} - 3 \right) - n \cdot \left( m^{2} - 3 \right)^{2} =\]

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

\[= \left( m^{2} - 3 \right) \cdot \left( 1 - nm^{2} + 3n \right).\]

\[= (p - 12) \cdot \left( 8c + 7d \cdot (p - 12) \right) =\]

\[= (p - 12) \cdot (8c + 7dp - 84d)\text{.\ }\]