How do you simplify \[{{\left( \dfrac{3}{4} \right)}^{-2}}\]?

Hint: \[{{\left( \dfrac{a}{b} \right)}^{-n}}={{\left( \dfrac{b}{a} \right)}^{n}}=\dfrac{{{b}^{n}}}{{{a}^{n}}}\].
Using this exponential property, just compare the given question with this property and find the value of \[a,b,n\] and substitute this in the formula then we will get the fraction with square power then using the property of square that is multiply it with itself. Do perform this operation in numerator and denominator both and then you will get your final simplified answer.

Complete step by step solution:
Now using the exponential property \[{{\left( \dfrac{a}{b} \right)}^{-n}}={{\left( \dfrac{b}{a} \right)}^{n}}=\dfrac{{{b}^{n}}}{{{a}^{n}}}\]
Let’s compare this property with the given question
On comparing-
\[a=3,b=4,n={{2}^{{}}}\]
\[\Rightarrow {{\left( \dfrac{3}{4} \right)}^{-2}}\]
Now applying the property-
\[\Rightarrow {{\left( \dfrac{4}{3} \right)}^{2}}\]
Now using another exponential property that is
\[{{\left( \dfrac{c}{d} \right)}^{n}}=\dfrac{{{c}^{n}}}{{{d}^{n}}}\]
\[\Rightarrow \dfrac{{{4}^{2}}}{{{3}^{2}}}\]
Now squaring as we know that \[({{x}^{2}}=x\times x)\]
\[\Rightarrow \dfrac{4\times 4}{3\times 3}\]
\[\Rightarrow \dfrac{16}{9}\]

Hence, we have simplified the value of \[{{\left( \dfrac{3}{4} \right)}^{-2}}\]and that is \[\dfrac{16}{9}\].

Note: In these types of questions just apply the exponential property that we have studied in our class and further, the questions will automatically be solved. Just do what the question has given then you automatically move towards the solution.