Предмет: Математика, автор: aisanabralinova

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Ответы

Автор ответа: сок111213

$\frac{3 {x}^{2} }{5 {y}^{3} } \div \frac{9 {x}^{3} }{2 {y}^{2} } \times \frac{5y}{3(x - 1)} = \frac{3 {x}^{2} \times 2 {y}^{2} \times 5y}{5 {y}^{3} \times 9 {x}^{3} \times 3(x - 1) } = \\ = \frac{2}{9x(x - 1)} = \frac{2}{9 {x}^{2} - 9x }$

$\frac{5a(b - 1)}{3 {}^{2} d} \div \frac{5 {cd}^{2} }{9ab} \div \frac{ {a}^{2} (b - 1)}{ {c}^{3}d } = \\ = \frac{5a(b - 1) \times 9ab \times c {}^{3}d }{9d \times 5 {cd}^{2} \times {a}^{2} (b - 1)} = \frac{b {c}^{2} }{ {d}^{2} }$

$\frac{7 {p}^{4} }{10 {q}^{3} } \times \frac{5 {q}^{2}(p + 1) }{14 {p}^{2} } \div \frac{3p}{4 {q}^{4 } } = \frac{7p {}^{4} \times 5 {q}^{2} (p + 1) \times 4 {q}^{4} }{10 {q}^{3} \times 14 {p}^{2} \times 3p} = \\ = \frac{p {q}^{3}(p + 1) }{3} = \frac{ {p}^{2} {q}^{3} + p {q}^{3} }{3}$

$\frac{8 {x}^{2} {y}^{3} }{7ab {}^{2} } \div \frac{14x {y}^{2} }{7 {a}^{2} b} \div \frac{2 {x}^{2}(y + 2) }{ab} = \frac{8 {x}^{2} {y}^{3} \times 7 {a}^{2}b \times ab }{7 {ab}^{2} \times 14x {y}^{2} \times 2 {x}^{2}(y + 2) } = \\ = \frac{2 {a}^{2}y }{7x(y + 2)} = \frac{2 {a}^{2}y }{7xy + 14x}$