Завдання № 511

№ 511 Алгебра = № 22.7 Математика

Виконайте множення:

1) $\frac{1}{4}$m²n(2,4mn – 2,8m²);
2) –$\frac{2}{5}$ab³(1,5ab – $\frac{5}{6}$b²);
3) (1$\frac{1}{2}$x²y – $\frac{9}{10}$xy⁴) ∙ $\frac{2}{3}$xy³;
4) (1,5a – $\frac{4}{7}$b) ∙ (–$\frac{1}{14}$a²b⁵).

Розв'язок:

1) $\frac{1}{4}$m²n(2,4mn – 2,8m²) = $\frac{1}{4}$m²n ∙ $\frac{24}{10}$mn – $\frac{1}{4}$m²n ∙ $\frac{28}{10}$m² = $\frac{3}{5}$m³n² – $\frac{7}{10}$m⁴n;

2) –$\frac{2}{5}$ab³(1,5ab – $\frac{5}{6}$b²) = –$\frac{2}{5}$ab³ ∙ $\frac{15}{10}$ab – (–$\frac{2}{5}$)ab³ ∙ $\frac{5}{6}$b² = –$\frac{3}{5}$a²b⁴ + $\frac{1}{3}$ab⁵;

3) (1$\frac{1}{2}$x²y – $\frac{9}{10}$xy⁴) ∙ $\frac{2}{3}$xy³ = $\frac{3}{2}$x²y ∙ $\frac{2}{3}$xy³ – $\frac{9}{10}$xy⁴ ∙ $\frac{2}{3}$xy³ = x³y⁴ – $\frac{3}{5}$x²y⁷;

4) (1,5a – $\frac{4}{7}$b) ∙ (–$\frac{1}{14}$a²b⁵) = $\frac{15}{10}$a ∙ (–$\frac{1}{14}$a²b⁵) – $\frac{4}{7}$b ∙ (–$\frac{1}{14}$a²b⁵) = –$\frac{3}{28}$a³b⁵ + $\frac{2}{49}$a²b⁶.