Question

как сделать эту помогите??​

Answer 1

Ответ:

$\frac{ 2xy ^ { 2 } }{ 3x-y } \div \frac{ xy ^ { 3 } }{ 9x ^ { 2 } -y ^ { 2 } } \times \frac{ 2y }{ 3x+y } = \frac{2xy^{2}\left(9x^{2}-y^{2}\right)}{\left(3x-y\right)xy^{3}}\times \left(\frac{2y}{3x+y}\right) = \frac{2\left(9x^{2}-y^{2}\right)}{y\left(3x-y\right)}\times \left(\frac{2y}{3x+y}\right) = \frac{2\left(3x+y\right)\left(3x-y\right)}{y\left(3x-y\right)}\times \left(\frac{2y}{3x+y}\right) = \frac{2\left(3x+y\right)}{y}\times \left(\frac{2y}{3x+y}\right) = \frac{2\left(3x+y\right)\times 2y}{y\left(3x+y\right)} = 2\times 2 = 4$