Question

помоги у меня щя урак алгебры
$( \frac{x { }^{ - 3}y {}^{4} {} }{9} ) {}^{ - 2} \times ( \frac{3}{x {}^{ - 2}y {}^{3} } ) {}^{ - 3}$
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Answer 1

Объяснение:

$( \frac{x {}^{ - 3}y {}^{4} }{9} ) {}^{ - 2} \times ( \frac{3}{x {}^{ - 2} y {}^{2} } ) {}^{ - 3 } = ( \frac{9}{x {}^{ - 3} y {}^{4} } ) {}^{2} \times ( \frac{x {}^{ - 2} y {}^{2} }{3} ) {}^{3} = \frac{81}{x {}^{ - 6} y {}^{8} } \times \frac{x {}^{ - 6} y {}^{6} }{27} = \frac{81}{y {}^{8} } \times \frac{y {}^{6} }{27} = \frac{3}{y {}^{8} } \times y {}^{6} = \frac{3}{y {}^{2} }$

Answer 2

Ответ:

3y

Объяснение:

$\displaystyle\bf\Bigg(\frac{x^{-3}y^4}{9} \Bigg)^{-2}\cdot\Bigg(\frac{3}{x^{-2}y^3} \Bigg)^{-3}=\Bigg(\frac{y^4}{9x^3} \Bigg)^{-2}\cdot\Bigg(\frac{3x^2}{y^3} \Bigg)^{-3}=\\\\\\=\Bigg(\frac{9x^3}{y^4}\Bigg)^2\cdot\Bigg(\frac{y^3}{3x^2} \Bigg)^3=\frac{9^2x^6}{y^8} \cdot\frac{y^9}{3^3x^6}=\frac{81y}{27} =3y$