11) \frac{x^3y^2+x^2y^3}{10(y-2x)} : \frac{3(2x-y)}{x+y} при х = -\frac{1}{9} иу =- 9

Ответ: -\(\frac{27}{20}\)

Разбираемся:

Упростим выражение:

\[\frac{x^3y^2+x^2y^3}{10(y-2x)} : \frac{3(2x-y)}{x+y} = \frac{x^2y^2(x+y)}{10(y-2x)} \cdot \frac{x+y}{3(2x-y)} = -\frac{x^2y^2(x+y)^2}{30(2x-y)^2}\]

Подставим значения х = -\(\frac{1}{9}\) и у = -9:

\[-\frac{\left(-\frac{1}{9}\right)^2 \cdot (-9)^2 \cdot \left(-\frac{1}{9} + (-9)\right)^2}{30 \cdot \left(2 \cdot \left(-\frac{1}{9}\right) - (-9)\right)^2} = -\frac{\frac{1}{81} \cdot 81 \cdot \left(-\frac{82}{9}\right)^2}{30 \cdot \left(-\frac{2}{9} + 9\right)^2} = -\frac{\left(\frac{82}{9}\right)^2}{30 \cdot \left(\frac{79}{9}\right)^2} = -\frac{82^2}{30 \cdot 79^2} = -\frac{6724}{30 \cdot 6241} = -\frac{6724}{187230} = -\frac{3362}{93615}\]

Ответ: -\(\frac{3362}{93615}\)