Решебник по алгебре 7 класс Мерзляк ФГОС Задание 561

 

\[\boxed{\text{561\ (561).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Последовательность:\]

\[\ldots,\ a,\ b,\ c,\ d,\ 0,\ 1,\ 1,\ 2,\ 3,\ 5,\ 8,\ \ldots\]

\[d + 0 \Longrightarrow \ \ d = 1\ \ \]

\[d + c = 0;\ \ \ 1 + c = 0 \Longrightarrow \ \ c =\]

\[= - 1\]

\[d = c + b;\ \ 1 = - 1 + b \Longrightarrow \ \ b =\]

\[= 2\]

\[c = b + a;\ \ - 1 = 2 + a \Longrightarrow \ \ a =\]

\[= - 3\]

\[Ответ:\ a = - 3.\]

\[\boxed{\text{561.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (2x - 6)^{2} = (2 \cdot x - 2 \cdot 3)^{2} =\]

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

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

\[2)\ (5y + 5)^{2} = (5 \cdot y + 1 \cdot 5)^{2} =\]

\[= \left( 5 \cdot (y + 1) \right)^{2} =\]

\[= 5^{2} \cdot (y + 1)^{2} = 25 \cdot (y + 1)^{2}.\]

\[3)\ (36x + 30y)^{2} =\]

\[= (6 \cdot 6x + 6 \cdot 5y)^{2} =\]

\[= \left( 6 \cdot (6x + 5y) \right)^{2} =\]

\[= 6^{2} \cdot (6x + 5y)^{2} =\]

\[= 36 \cdot (6x + 5y)^{2}.\]

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

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

\[= 2^{4} \cdot (x + 2)^{4} = 16 \cdot (x + 2)^{4}.\]

\[5)\ (6x - 9y)^{3} =\]

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

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

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

\[= 27 \cdot (2x - 3y)^{3}.\]

\[6)\ \left( a^{2} + ab \right)^{2} =\]

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

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

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

\[7)\ ( - 7a - 14ab)^{2} =\]

\[= ( - 7a \cdot 1 - 7a \cdot 2b)^{2} =\]

\[= \left( - 7a \cdot (1 + 2b) \right)^{2} =\]

\[= ( - 7a)^{2} \cdot (1 + 2b)^{2} =\]

\[= 49a^{2} \cdot (1 + 2b)^{2}.\]

\[8)\ \left( 3c^{4} - 6c^{3} \right)^{4} =\]

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

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

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

\[= 81c^{12} \cdot (c - 2)^{4}\text{.\ }\]