Решебник по алгебре 7 класс Мерзляк Задание 569

\[\boxed{\text{569\ (569).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

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

\[4)\ (5 - p)^{2} = 25 - 10p + p^{2}\]

\[5)\ (4 + k)^{2} = 16 + 8k + k^{2}\]

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

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

\[= 9a^{2} - 12a + 4\]

\[7)\ (7b + 6)^{2} =\]

\[= (7b)^{2} + 2 \cdot 7b \cdot 6 + 6^{2} =\]

\[= 49b^{2} + 84b + 36\]

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

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

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

\[9)\ (0,4m - 0,5n)^{2} =\]

\[= 0,16m^{2} - 0,4mn + 0,25n^{2}\]

\[11)\ (y - 13)^{2} = y^{2} - 26y + 169\]

\[12)\ (13 - y)^{2} = 169 - 26y + y^{2}\]

\[13)\ \left( b^{2} - 11 \right)^{2} =\]

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

\[= b^{4} - 22b^{2} + 121\]

\[14)\ \left( a^{2} + 4b \right)^{2} =\]

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

\[= a^{4} + 8a^{2}b + 16b^{2}\]

\[15)\ \left( x^{2} + y^{3} \right)^{2} =\]

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

\[= x^{4} + 2x^{2}y^{3} + y^{6}\]

\[16)\ \left( a^{3} - 4b \right)^{2} =\]

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

\[= a^{6} - 8a^{3}b + 16b^{2}\]

\[17)\ \left( a^{2} + a \right)^{2} = a^{4} + 2a^{3} + a^{2}\]

\[18)\ \left( 3b^{2} - 2b^{5} \right)^{2} =\]

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

\[= 9b^{4} - 12b^{7} + 4b^{10}\]