Question

How do you simplify${\left( {\dfrac{9}{2}} \right)^{ - 1}}$?

Answer 1

Hint: In order to solve the above question ,replace the numerator with denominator and denominator with numerator and remove the exponent value ,you’ll get your required result.
Formula:
${x^a} = \dfrac{1}{{{x^{ - a}}}}\,$
$
  {\left( {{x^a}} \right)^b} = {x^{a \times b}} \\
  \dfrac{1}{{{x^a}}} = {x^{ - a}} \;
 $
${a^1} = a$

Complete step-by-step answer:
Given a rational number raised to the power negative value i.e.$ - 1$
In such case , to remove the negative sign from the number , replace the numerator with denominator and denominator with numerator and remove the negative sign from the exponent
$
   = {\left( {\dfrac{9}{2}} \right)^{ - 1}} \\
   = \dfrac{1}{{{{\left( {\dfrac{9}{2}} \right)}^{ (-) - 1}}}} \\
   = \dfrac{1}{{{{\left( {\dfrac{9}{2}} \right)}^1}}} \\
   = \dfrac{1}{{\left( {\dfrac{9}{2}} \right)}} \\
   = \dfrac{2}{9} \;
 $
Therefore ,our required answer is $\dfrac{2}{9}$
So, the correct answer is “$\dfrac{2}{9}$”.

Note: Alternatively we have another way to solve the above simplification problem
Bye using the some of the identities of exponent that are mentioned below:
$
  {\left( {{x^a}} \right)^b} = {x^{a \times b}} \\
  \dfrac{1}{{{x^a}}} = {x^{ - a}} \;
 $
Now, applying these identities on our question
$
   = {\left( {\dfrac{9}{2}} \right)^{ - 1}} \\
   = \left( {\dfrac{{{9^{ - 1}}}}{{{2^{ - 1}}}}} \right) \\
   = {9^{ - 1}} \times \dfrac{1}{{{2^{ - 1}}}} \\
   = \dfrac{1}{{{9^{ (-) - 1}}}} \times {2^{ (-) - 1}} \\
   = \dfrac{1}{{{9^1}}} \times {2^1} \\
   = \dfrac{1}{9} \times 2 \\
   = \dfrac{2}{9} \;
 $