Решебник по алгебре 7 класс Мерзляк Задание 204

\[\boxed{\text{204\ (204).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ m^{5} \cdot m^{4} = m^{5 + 4} = m^{9}\]

\[2)\ x \cdot x^{2} = x^{1} \cdot x^{2} = x^{1 + 2} = x^{3}\]

\[3)\ a^{3} \cdot a^{3} = a^{3 + 3} = a^{6}\]

\[4)\ 6^{8} \cdot 6^{3} = 6^{8 + 3} = 6^{11}\]

\[5)\ y^{3} \cdot y^{5} \cdot y^{9} = y^{3 + 5 + 9} = y^{17}\]

\[6)\ c^{8} \cdot c^{9} \cdot c = c^{8 + 9 + 1} = c^{18}\]

\[7)\ (b - c)^{10} \cdot (b - c)^{6} =\]

\[= (b - c)^{10 + 6} = (b - c)^{16}\]

\[8)\ 11^{2} \cdot 11^{4} \cdot 11^{6} = 11^{2 + 4 + 6} =\]

\[= 11^{12}\]

\[9)\ x^{4} \cdot x \cdot x^{11} \cdot x^{2} = x^{4 + 1 + 11 + 2} =\]

\[= x^{18}\]

\[10)\ \left( \text{ab} \right)^{5} \cdot \left( \text{ab} \right)^{15} = \left( \text{ab} \right)^{5 + 15} =\]

\[= \left( \text{ab} \right)^{20}\]

\[11)\ (2x + 3y)^{6} \cdot (2x + 3y)^{14} =\]

\[= (2x + 3y)^{6 + 14} = (2x + 3y)^{20}\]

\[12)\ ( - xy)^{2} \cdot ( - xy)^{7} \cdot ( - xy)^{9} =\]

\[= ( - xy)^{2 + 7 + 9} = ( - xy)^{18}\]