Question

$(\frac{ {x}^{ - 3} {y}^{4} }{9} ) {}^{ - 2} \times (\frac{3}{ {x}^{ - 2} {y}^{3} } ) {}^{ - 3}$
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Answer 1

Объяснение:

$(\frac{x^{-3}y^4}{9})^{-2}*(\frac{3}{x^{-2}y^3})^{-3}=(\frac{9}{x^{-3}y^4}})^2*(\frac{x^{-2}y^3}{3})^3=\frac{(3^2)^2}{x^{-6}y^8} *\frac{x^{-6}y^9}{3^3} =\frac{3^4}{x^{-6}y^8} *\frac{x^{-6}y^9}{3^3} =3y.$