Question

Pls answer the question on a paper
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Answer 1

Answer:

$\frac{x}{y} = ( \frac{4}{9} )^{ - 3} \div ( \frac{4}{9} )^{ - 2} \\ = ( \frac{9}{4} )^{ 3} \times ( \frac{4}{9} )^{ 2}\\ = ( \frac{9}{4} )^{ 3} \times ( \frac{9}{4} )^{ - 2} \\ = ( \frac{9}{4} )^{3 - 2} \\ = ( \frac{9}{4} )^{1} \\ \implies \: \frac{x}{y} =\frac{9}{4} \\ \implies \: \frac{y}{x} =\frac{4}{9} \\ now \\ (\frac{x}{y} )^{ - 1} + (\frac{y}{x} )^{ - 1} \\ = \frac{y}{x} + \frac{x}{y} \\ = \frac{4}{9} + \frac{9}{4} \\ = \frac{4 \times 4 + 9 \times 9}{9 \times 4} \\ = \frac{16 + 81}{36} \\ = \frac{97}{36} \\ thus \\ (\frac{x}{y} )^{ - 1} + (\frac{y}{x} )^{ - 1} = \frac{97}{36}$