Question

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






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Answer 1

Ответ:

${ (\frac{ {x}^{ - 2} {y}^{4} }{9} })^{ - 2} \times {( \frac{3}{ {x}^{ - 2} {y}^{3} }) }^{ - 3} = \frac{ {x}^{4} {y}^{ - 8} }{ {9}^{ - 2} } \times \frac{ {3}^{ - 3} }{ {x}^{6} {y}^{ - 9} } = \frac{ {x}^{4} {y}^{ - 8} \times \frac{1}{27} }{ {x}^{6} {y}^{ - 9} \times \frac{1}{81} } = {x}^{ 4 - 6} \times {y}^{ - 8 - ( - 9)} \times \frac{81}{27 } = {x}^{ - 2} \times {y}^{1} \times 3 = 3y {x}^{ - 2} = \frac{3y}{ {x}^{2} }$