Question

9х2 / 5у3:27х5/2у4×15/4у(х-1) .
$\frac{9x {}^{2} } {5y {}^{3} } \div \frac{27x {}^{5} }{2y {}^{4} } \times \frac{15}{4y(x - 1)}$
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Answer 1

Ответ:

Объяснение:

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