Evaluate ${{(\frac{4}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}}^{- 1}$

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Solution

${{(\frac{4}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}}^{- 1}$
$= {\frac{3}{4} - 4}^{- 1} = {\frac{3 - 16}{4}}^{- 1} = {(\frac{- 13}{4})}^{- 1} = \frac{- 4}{13}$ .

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