Решебник по алгебре 7 класс Мерзляк ФГОС Задание 593

 

\[\boxed{\text{593\ (593).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[Преобразуем\ левую\ часть\ \]

\[равенства:\]

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

\[= 2a^{2} + 2b^{2} = 2 \cdot \left( a^{2} + b^{2} \right)\]

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

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

\[Тождество\ доказано.\]

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

\[Преобразуем\ левую\ часть\ \]

\[равенства:\]

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

\[= 4ab\]

\[2ab + 2ab = 4ab\]

\[4ab = 4ab.\]

\[Тождество\ доказано.\]

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

\[Преобразуем\ правую\ часть\ \]

\[равенства:\]

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

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

\[Тождество\ доказано.\]

\[4)\ \left( a^{2} + b^{2} \right)\left( c^{2} + d^{2} \right) =\]

\[= (ac + bd)^{2} + (ad - bc)^{2}\]

\[Преобразуем\ обе\ части\ \]

\[равенства:\]

\[a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2} =\]

\[a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2} =\]

\[= a^{2}c^{2} + b^{2}d^{2} + a^{2}d^{2} + b^{2}c^{2}\]

\[a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2} =\]

\[= a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}.\]

\[Тождество\ доказано.\]

\[\boxed{\text{593.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 8c^{3} - 2c^{2} + 4c - 1 =\]

\[= \left( 8c^{3} + 4c \right) - \left( 2c^{2} + 1 \right) =\]

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

\[= \left( 2c^{2} + 1 \right) \cdot (4c - 1).\]

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

\[= (x^{2}y + xy^{2} + (x + y) =\]

\[= xy \cdot (x + y) + 1 \cdot (x + y) =\]

\[= (x + y) \cdot (xy + 1).\]

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

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

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

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

\[4)\ 8a^{2} - 2ab - 4ac + bc =\]

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

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

\[5)\ 2b^{3} - 7b^{2}c - 4b + 14c =\]

\[= \left( 2b^{3} - 4b \right) - \left( 7b^{2}c - 14c \right) =\]

\[= 2b \cdot \left( b^{2} - 2 \right) - 7c \cdot \left( b^{2} - 2 \right) =\]

\[= \left( b^{2} - 2 \right) \cdot (2b - 7c).\]

\[6)\ 6x^{5} + 4x^{2}y^{2} - 9x^{3}y - 6y^{3} =\]

\[= \left( 6x^{5} + 4x^{2}y^{2} \right) - \left( 9x^{3}y + 6y^{3} \right) =\]

\[= \left( 3x^{3} + 2y^{2} \right) \cdot \left( 2x^{2} - 3y \right)\text{.\ }\]