How do you simplify \[\left[ \dfrac{1}{x+h} \right]-\dfrac{1}{x}\]?

Hint: The given term is in fractions. To solve the term you need to write the term with a common denominator and start solving the term. By writing the term with a common denominator the problem will be easy to solve the numerator in complex terms. To solve the denominator, choose the correct term.

Complete step by step answer:
The given term is \[\left[ \dfrac{1}{x+h} \right]-\dfrac{1}{x}\]
To simplify the term we need the simplified denominator
To calculate the denominator, we need to solve the denominator with one term to get the common denominator.
Let us multiply \[\left[ \dfrac{1}{x+h} \right]\] with \[\left( \dfrac{x}{x} \right)\] and multiply \[\dfrac{1}{x}\] with \[\left( \dfrac{x+h}{x+h} \right)\]
We use \[\left( \dfrac{x}{x} \right)\] and \[\dfrac{1}{x}\]to get common denominator so that the value of the equation will be the same
Now the let us calculate \[\left[ \dfrac{1}{x+h} \right]\] with \[\left( \dfrac{x}{x} \right)\] and \[\dfrac{1}{x}\]with \[\left( \dfrac{x+h}{x+h} \right)\]
And now we get the resultant equation with the common denominator
\[\Rightarrow \left( \dfrac{x}{x} \right)\left[ \dfrac{1}{x+h} \right]-\left( \dfrac{x+h}{x+h} \right)\left( \dfrac{1}{x} \right)\]
\[\Rightarrow \left[ \dfrac{x}{x\left( x+h \right)} \right]-\left[ \dfrac{x+h}{x\left( x+h \right)} \right]\]
By using subtraction lets combine the fractions and we get
\[\Rightarrow \dfrac{x-x+h}{x\left( x+h \right)}\]
Here in the numerator, the term $ x $ is common so that will be canceled and the term $ h $ will be leftover
\[\Rightarrow \dfrac{h}{x\left( x+h \right)}\]
In the denominator $ x $ will be multiplied with $ \left( x+h \right) $ and we get
\[\Rightarrow \dfrac{h}{{{x}^{2}}+hx}\]
\[\therefore \]\[\left[ \dfrac{1}{x+h} \right]-\dfrac{1}{x}\]=\[\dfrac{h}{x\left( x+h \right)}\]\[or\dfrac{h}{{{x}^{2}}+hx}\]
Hence both are the simplified terms of \[\left[ \dfrac{1}{x+h} \right]-\dfrac{1}{x}\].

Note:
 As the above equation is the subtraction of the mixed equation. We solved it by writing the denominator of both terms with a common denominator. Generally, there are many types of fractions like improper fractions, like fractions, unlike fractions equivalent fractions. Fractions are categorized depending upon the numerator and the denominator. Any type of fraction can be solved in various methods.