Решебник по алгебре 7 класс Мерзляк ФГОС Задание 578

 

\[\boxed{\text{578\ (578).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ ( - x + 1)^{2} = (1 - x)^{2} =\]

\[= 1 - 2x + x^{2}\]

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

\[= \left( - 1 \cdot (m + 9) \right)^{2} =\]

\[= ( - 1)^{2} \cdot (m + 9)^{2} =\]

\[= m^{2} + 18m + 81\]

\[3)\ ( - 5a + 3b)^{2} = (3b - 5a)^{2} =\]

\[= 9b^{2} - 30ab + 25a^{2}\]

\[4)\ ( - 4x - 8y)^{2} =\]

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

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

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

\[5)\ ( - 0,7c - 10d)^{2} =\]

\[= \left( - 1 \cdot (0,7c + 10d) \right)^{2} =\]

\[= ( - 1)^{2} \cdot (0,7c + 10d)^{2} =\]

\[= 0,49c^{2} + 14cd + 100d^{2}\]

\[6)\ \left( - 4a^{2} + \frac{1}{8}\text{ab} \right)^{2} =\]

\[= \left( \frac{1}{8}ab - 4a^{2} \right)^{2} =\]

\[= \frac{1}{64}a^{2}b^{2} - a^{3}b + 16a^{4}.\]

\[\boxed{\text{578.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[2ax - 6x - 2a + 6 = 0\]

\[2ax - 6x = 2a - 6\]

\[2x \cdot (a - 3) = 2 \cdot (a - 3)\]

\[a - 3 = 0\]

\[\ \ \ \ \ \ \ \ a = 3\]

\[0 = 0 - бесконечно\ много.\]

\[Ответ:при\ a = 3.\]

\[- 4a - 6ax - 9a - 6 = 0\]

\[- 2x \cdot (2 + 3a) = 3 \cdot (3a + 2)\]

\[3a + 2 = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ a = - \frac{2}{3}\]

\[0 = 0 - бесконечно\ много.\]

\[Ответ:\ при\ a = - \frac{2}{3}\text{.\ \ }\]