Решебник по алгебре 7 класс Мерзляк ФГОС Задание 727

 

\[\boxed{\text{727\ (727).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\(1){\ (a - 1)}^{3} - 9 \cdot (a - 1) =\)

\[= (a - 1)(a - 4)(a + 2)\]

\[(a - 1)^{3} - 9 \cdot (a - 1) =\]

\[= (a - 1)\left( (a - 1)^{2} - 9 \right) =\]

\[= (a - 1)(a - 1 - 3)(a - 1 + 3) =\]

\[= (a - 1)(a - 4)(a + 2)\]

\[(a - 1)(a - 4)(a + 2) =\]

\[= (a - 1)(a - 4)(a + 2) \Longrightarrow\]

\[\Longrightarrow тождество.\]

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

\[= (x - 1)²(x + 1)²\]

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

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

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

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

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

\[\Longrightarrow тождество.\]

\[\boxed{\text{727.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Формула\ квадрата\ трехчлена:\]

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

\[= a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac.\]

\[Выведем:\]

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

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

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

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

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

\[= a^{2} + b^{2} + 16\ - 2ab - 8b + 8a.\]