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How do you simplify ${{144}^{\dfrac{-1}{2}}}$ ?

Answer
VerifiedVerified
554.1k+ views
Hint: For answering this question we need to simplify the given expression ${{144}^{\dfrac{-1}{2}}}$ . To simplify this question we need to use our basic arithmetic concepts. That is to simplify the expression using the basic arithmetic simplifications.

Complete step by step answer:
Now considering from the question we have been given an expression \[{{144}^{\dfrac{-1}{2}}}\] .
So we need to simplify the expression using our basic arithmetic concepts.
We know that the square of the number $12$ is $144$ .
Let us use this in our expression and simplify it step by step.
By simplifying this we will have $\Rightarrow {{\left( {{12}^{2}} \right)}^{\dfrac{-1}{2}}}$ .
By further simplifying this using exponents for doing this we need to use the formulae ${{\left( {{a}^{x}} \right)}^{\dfrac{-1}{x}}}={{a}^{-1}}=\dfrac{1}{a}$ .
By using this we will have $\Rightarrow {{\left( {{12}^{2}} \right)}^{\dfrac{-1}{2}}}={{12}^{-1}}$ after that we will have $\Rightarrow \dfrac{1}{12}$ .

Hence we can conclude that the simplified form of the given expression ${{144}^{\dfrac{-1}{2}}}$ is $\dfrac{1}{12}$.

Note: We have to be very careful while answering questions of this type. These questions do not require much calculations so there is very less possibility of making mistakes in questions of this type. While answering questions of this type we should be sure with our calculations and concepts. Similarly we can solve any expression using the exponential formulae. Similarly we have many formulae like ${{a}^{x}}{{a}^{y}}={{a}^{x+y}}$ , ${{a}^{xy}}={{\left( {{a}^{x}} \right)}^{y}}$ , ${{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$ , $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ , $\dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}}$ , ${{a}^{x}}{{b}^{x}}={{\left( ab \right)}^{x}}$ ,${{a}^{0}}=1$ and ${{a}^{\dfrac{p}{q}}}=\sqrt[q]{{{a}^{p}}}$ . By combining some of these formulae we had obtained the above formulae which we had used in the solution of this question.
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