\[\boxed{\text{449\ (449).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{5} \cdot \left( x^{2} \right)^{3} = x^{5} \cdot x^{6} = x^{5 + 6} =\]
\[= x^{11}\]
\[\textbf{б)}\ \left( x^{3} \right)^{4} \cdot x^{8} = x^{12} \cdot x^{8} =\]
\[= x^{12 + 8} = x^{20}\]
\[\textbf{в)}\ \left( x^{4} \right)^{2} \cdot \left( x^{5} \right)^{3} = x^{8} \cdot x^{15} =\]
\[= x^{8 + 15} = x^{23}\]
\[\textbf{г)}\ \left( x^{2} \right)^{3} \cdot \left( x^{3} \right)^{5} = x^{6} \cdot x^{15} =\]
\[= x^{6 + 15} = x^{21}\]
\[\textbf{д)}\ \left( x^{3} \right)^{2} \cdot \left( x^{4} \right)^{5} = x^{6} \cdot x^{20} =\]
\[= x^{6 + 20} = x^{26}\]
\[\textbf{е)}\ \ \left( x^{7} \right)^{3} \cdot \left( x^{3} \right)^{4} = x^{21} \cdot x^{12} =\]
\[= x^{21 + 12} = x^{33}\]