Решебник по алгебре 7 класс Мерзляк Задание 570

\[\boxed{\text{570\ (570).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ (a + 8)^{2} = a^{2} + 16a + 64\]

\[2)\ (b - 2)^{2} = b^{2} - 4b + 4\]

\[3)\ (7 + c)^{2} = 49 + 14c + c^{2}\]

\[4)\ (6 - d)^{2} = 36 - 12d + d^{2}\]

\[5)\ (2m + 1)^{2} = 4m^{2} + 4m + 1\]

\[6)\ (4x - 3)^{2} =\]

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

\[= 16x^{2} - 24x + 9\]

\[7)\ (5m - 4n)^{2} =\]

\[= (5m)^{2} - 2 \cdot 5m \cdot 4n + (4n)^{2} =\]

\[= 25m^{2} - 40mn + 16n^{2}\]

\[8)\ (10c + 7d)^{2} =\]

\[= (10c)^{2} + 2 \cdot 10c \cdot 7d + (7d)^{2} =\]

\[= 100c^{2} + 140cd + 49d^{2}\]

\[9)\ \left( 4x - \frac{1}{8}y \right)^{2} =\]

\[= (4x)^{2} - 2 \cdot 4 \cdot \frac{1}{4}xy + \left( \frac{1}{8}y \right)^{2} =\]

\[= 16x^{2} - xy + \frac{1}{64}y^{2}\]

\[10)\ (0,3a + 0,9b)^{2} =\]

\[= 0,09a^{2} + 0,54ab + 0,81b^{2}\]

\[11)\ \left( c^{2} - 6 \right)^{2} = c^{4} - 12c^{2} + 36\]

\[12)\ \left( 15 + k^{2} \right)^{2} =\]

\[= 225 + 30k^{2} + k^{4}\]

\[13)\ \left( m^{2} - 3n \right)^{2} =\]

\[= m^{4} - 6m^{2}n + 9n^{2}\]

\[14)\ \left( m^{4} - n^{3} \right)^{2} =\]

\[= m^{8} - 2m^{4}n^{3} + n^{6}\]

\[15)\ \left( 5a^{4} - 2a^{7} \right)^{2} =\]

\[= 25a^{8} - 20a^{11} + 4a^{14}\]