\[\boxed{\text{355\ (355).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ 3x \cdot (2x + 5) = 6x^{2} + 15x.\]
\[2)\ 4x \cdot \left( x^{2} - 8x - 2 \right) =\]
\[= 4x^{3} - 32x^{2} - 8x.\]
\[3) - 2a \cdot \left( a^{2} + a - 3 \right) =\]
\[= - 2a^{3} - 2a^{2} + 6a.\]
\[4)\ 5b^{2} \cdot \left( 3b^{2} - 7b + 10 \right) =\]
\[= 15b^{4} - 35b^{3} + 50b^{2}.\]
\[5)\ mn \cdot \left( m^{2}n - n^{3} \right) =\]
\[= m^{3}n^{2} - mn^{4}.\]
\[6)\ 2ab \cdot \left( a^{3} - 3a^{2}b + b^{2} \right) =\]
\[= 2a^{4}b - 6a^{3}b^{2} + 2ab^{3}.\]
\[7)\ \left( 4y^{3} - 6y + 7 \right) \cdot \left( - 1,2y^{3} \right) =\]
\[= - 4,8y^{6} + 7,2y^{4} - 8,4y^{3}.\]
\[= 1,2x^{3}y^{3} - 2x^{3}y^{2} + 5,2x^{4}y^{4}.\]
\[= 4m^{2}n^{2} - 1,2mn^{3} + 6mn^{2}.\]
\[= c^{6}d - \frac{1}{3}c^{3}d^{8} - \frac{2}{7}cd^{11}\text{.\ }\]
\[\boxed{\text{355.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ 0,6a^{4}b^{3} \cdot 4a^{2}b =\]
\[= 2,4a^{6}b^{4}\ \ (К)\]
\[2) - 2,8a^{2}b^{5} \cdot 0,5a^{4}b^{6} =\]
\[= - 1,4a^{6}b^{11}\ \ (Р)\]
\[3)\ 1,6a^{2}b \cdot \left( - 0,25a^{3}b^{2} \right) =\]
\[= - 0,4a^{5}b^{3}\ \ (\ Ы)\]
\[4) - 0,7a^{6}b^{9} \cdot ( - 3ab) =\]
\[= 2,1a^{7}b^{10}\ \ \ \ (Л)\]
\[5)\ \frac{3}{16}a^{2}b^{4} \cdot \frac{8}{15}a^{4}b^{7} =\]
\[= \frac{1}{10}a^{6}b^{11} = 0,1a^{6}b^{11}\ \ \ \ (О)\]
\[6) - 2,6a^{3}b \cdot \frac{2}{13}a^{2}b^{3} =\]
\[= - 0,4a^{5}b^{4}\ \ \ (В)\]
\[Фамилия\ русского\ \]
\[математика:КРЫЛОВ\text{.\ }\]