\[\boxed{\text{601\ (}\text{н}\text{).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[рис.\ 7:(a + b)(a + b) =\]
\[= a^{2} + \text{ab} + \text{ba} + b^{2} =\]
\[= a² + 2\text{ab} + b² = (a + b)^{2};\]
\[рис.\ 8:(a - b)(a - b) =\]
\[= a^{2} - ab - ba + b^{2} =\]
\[= a^{2} - 2ab + b^{2} = (a - b)²\]
\[\boxed{\text{601\ (}\text{c}\text{).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[рис.\ 5:(a + b)(a + b) =\]
\[= a^{2} + \text{ab} + \text{ba} + b^{2} =\]
\[= a² + 2\text{ab} + b² = (a + b)^{2};\]
\[рис.\ 6:(a - b)(a - b) =\]
\[= a^{2} - ab - ba + b^{2} =\]
\[= a^{2} - 2ab + b^{2} = (a - b)²\]
\[\boxed{\text{601.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ a^{n + 1} + a^{n} + a + 1 =\]
\[= a^{n} \cdot (a + 1) + (a + 1) =\]
\[= (a + 1) \cdot \left( a^{n} + 1 \right).\]
\[2)\ b^{n + 2} - b - 1 + b^{n + 1} =\]
\[= \left( b^{n + 2} + b^{n + 1} \right) - (b + 1) =\]
\[= b^{n + 1} \cdot (b + 1) - (b + 1) =\]
\[= (b + 1) \cdot \left( b^{n + 1} - 1 \right).\]
\[3)\ 3y^{n + 3} - 3y^{2} - 5 + 5y^{n + 1} =\]
\[= \left( 3y^{2} + 5 \right) \cdot \left( y^{n + 1} - 1 \right)\text{.\ }\]