Решебник по алгебре 7 класс Мерзляк Задание 571

\[\boxed{\text{571\ (571).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ a^{2} + (3a - b)^{2} =\]

\[= a^{2} + 9a^{2} - 6ab + b^{2} =\]

\[= 10a^{2} - 6ab + b^{2}\]

\[2)\ (4x + 5)^{2} - 40x =\]

\[= 16x^{2} + 40x + 25 - 40x =\]

\[= 16x^{2} + 25\]

\[3)\ 50a^{2} - (7a - 1)^{2} =\]

\[= 50a^{2} - 49a^{2} + 14a - 1 =\]

\[= a^{2} + 14a - 1\]

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

\[= c^{2} + 36 - c^{2} + 12c - 36 =\]

\[= 12c\]

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

\[= x^{2} - 4x + 4 + x^{2} + 10x =\]

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

\[6)\ 3m \cdot (m - 4) - (m + 2)^{2} =\]

\[= 3m^{2} - 12m - m^{2} - 4m - 4 =\]

\[= 2m^{2} - 16m - 4\]

\[7)\ (y - 9)^{2} + (4 - y)(y + 6) =\]

\[= - 20y + 105\]

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

\[= x^{2} - 16 - x^{2} + 2x - 1 =\]

\[= 2x - 17\]

\[9)\ (2a - 3b)^{2} + (3a + 2b)^{2} =\]

\[= 13a^{2} + 13b^{2}\]

\[10)\ (x - 5)^{2} - (x - 7)(x + 7) =\]

\[= x^{2} - 10x + 25 - x^{2} + 49 =\]

\[= 74 - 10x\]