The value of $[2 - 3 (2 - 3)^{- 1}]^{- 1}$ is __________.

\frac{- 1}{5}

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\frac{1}{5}

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- 5

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5

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Solution

The correct option is D $\frac{1}{5}$
$=$ $[2 - 3 (2 - 3)^{- 1}]^{- 1}$
$=$ $[2 - 3 (- 1)^{- 1}]^{- 1}$
$=$ $[2 - 3 (- 1)]^{- 1}$
$=$ $[2 + 3]^{- 1}$
$=$ $\frac{1}{5}$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} 32 & 24 & 16 8 & 3 & 5 4 & 5 & 3 \end{matrix} ∣ ∣ ∣ ∣ = λ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 2 2 & 3 & 5 1 & 5 & 3 \end{matrix} ∣ ∣ ∣ ∣

λ

\frac{1}{(216)^{- 2 / 3}} + \frac{1}{(256)^{- 3 / 4}} + \frac{1}{(32)^{- 1 / 5}}

The value of $\frac{1}{{(216)}^{- \frac{2}{3}}} + \frac{1}{{(256)}^{- \frac{3}{4}}} + \frac{1}{{(32)}^{- \frac{1}{5}}}$ is :

(\sqrt{3} + 1)^{2} - (\sqrt{3} - 1)^{2}

(2 + \sqrt{5})^{5} + (2 - \sqrt{5})^{5}

(81^{1 / 2} \times 8^{2 / 3} \times 32^{2 / 5})^{1 / 2}

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