Решебник по алгебре 7 класс Мерзляк Задание 392

\[\boxed{\text{392\ (392).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ (a - 2) \cdot (b + 5) =\]

\[= ab + 5a - 2b - 10.\]

\[2)\ (m + n) \cdot (p - k) =\]

\[= mp - km + np - kn.\]

\[3)\ (x - 8) \cdot (x + 4) =\]

\[= x^{2} + 4x - 8x - 32 =\]

\[= x^{2} - 4x - 32.\]

\[4)\ (x - 10) \cdot (x - 9) =\]

\[= x^{2} - 9x - 10x + 90 =\]

\[= x^{2} - 19x + 90.\]

\[5)\ (c + 5) \cdot (c + 8) =\]

\[= c^{2} + 8c + 5c + 40 =\]

\[= c^{2} + 13c + 40.\]

\[6)\ (3y + 1) \cdot (4y - 6) =\]

\[= 12y^{2} - 18y + 4y - 6 =\]

\[= 12y^{2} - 14y - 6.\]

\[7)\ ( - 2m - 3) \cdot (5 - m) =\]

\[= - 10m + 2m^{2} - 15 + 3m =\]

\[= 2m^{2} - 7m - 15.\]

\[8)\ \left( 5x^{2} - x \right) \cdot \left( 6x^{2} + 4x \right) =\]

\[= 30x^{4} + 20x^{3} - 6x^{3} - 4x^{2} =\]

\[= 30x^{4} + 14x^{3} - 4x^{2}.\]

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

\[= - c^{4} - 4c^{3} - 3с - 12.\]

\[10)\ (x - 5) \cdot \left( x^{2} + 4x - 3 \right) =\]

\[= x^{3} - x^{2} - 23x + 15.\]

\[11)\ (2a + 3) \cdot \left( 4a^{2} - 4a + 3 \right) =\]

\[= 8a^{3} + 4a^{2} - 6a + 9.\]

\[12)\ a \cdot (5a - 4) \cdot (3a - 2) =\]

\[= \left( 5a^{2} - 4a \right) \cdot (3a - 2) =\]

\[= 15a^{3} - 10a^{2} - 12a^{2} + 8a =\]

\[= 15a^{3} - 22a^{2} + 8a.\ \]