MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Exponents with Unlike Bases and Same Exponent

Simplify: [ ...

Question

Simplify:
${[{(2)}^{- 1} + {(4)}^{- 1} + {(3)}^{- 1}]}^{- 1}$

Open in App

Solution

${[{(2)}^{- 1} + {(4)}^{- 1} + {(3)}^{- 1}]}^{- 1}$
$= {(\frac{1}{2} + \frac{1}{4} + \frac{1}{3})}^{- 1}$
$= {(\frac{6 + 3 + 4}{12})}^{- 1}$
$= {(\frac{13}{12})}^{- 1}$
$= \frac{12}{13}$

Suggest Corrections

thumbs-up

0

Similar questions

{{(\frac{3}{4})}^{- 1} - {(\frac{1}{4})}^{- 1}}^{- 2}

{[{(\frac{2}{3})}^{- 1} \times {(\frac{3}{4})}^{- 1}]}^{- 1}

\frac{3}{2} \div {\frac{1}{2} + \frac{1}{3} + \frac{1}{6} \div (\frac{3}{4} - \frac{1}{2})}

(2^{- 1} + 3^{- 1})^{- 1} \div 4^{- 1}

(\frac{1}{2})^{- 1} + (\frac{1}{3})^{- 1} + (\frac{1}{4})^{- 1}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Laws of Exponents

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Exponents with Unlike Bases and Same Exponent

Standard VII Mathematics

Join BYJU'S Learning Program

CrossIcon