The term “rational” in math is not used in the sense of “sane” or “sensible.” It is instead used to imply a ratio, or fraction.
Rational expression is a fraction whose numerator and denominator is a polynomial function in variables.
Example: $\frac{x + 2}{x^{2}-3}$ is a rational expression.
We know that division by zero is not allowed since it makes the value of rational expression undefined. So it is always assumed that the denominator of a rational expression is never zero. This condition puts restrictions to the possible values of the variable.
For example, in a rational expression $\frac{2x + 3 }{x - 4}$, $x$ can never take the value of 4 since $x = 4$ will make the denominator zero and the expression undefined.
We will see reducing rational expression to simplest form. Simplifying rational expression is quite similar to simplifying fractions.
First check for common factors (constants, variables or polynomials ) of numerator and denominator , cancel out the common terms. There are some common mistakes with theses problems.
In the expression $\frac{x}{(x - 4)}$ students may end up cancelling $x$ like $\frac{\not{x}}{\not{x} - 4}$ which is not valid since $x$ in the denominator is not times the whole number like $\frac{x}{4x}$.
In general, the rule is not to cancel something if there is + or on one side of the variable.

Example 1: $\frac{x^{2} + 2x}{6x + 3}$

First of all write the numerator and denominator as product of factors.
               $\frac{x(x + 2)}{3(x + 2)}$
The polynomial x+2 is common to both the numerator and denominator. We will cross them out.
               $\frac{x (\not {x} + \not {2})}{3 (\not {x} + \not {2})} = \frac{x}{3}$
         
We have to be careful while canceling out the common terms

Example 2: $\frac{x^{3} - x}{x^{2} - 5x + 4}$

Factorise the numerator and denominator.
              $\frac{x^{3} - x}{x^{2} - 5x + 4} = \frac{x(x^{2} - 1)}{(x - 4)(x - 1)}$

Let us factorize the numerator further.
              $\frac{x^{3} - x}{x^{2} - 5x + 4} = \frac{x(x - 1)(x + 1)}{(x - 4)(x - 1)}$

Here the binomial x-1 is a common factor, cancelling it we get
             $\frac{x^{3} - x}{x^{2} - 5x + 4} = \frac{x(x + 1)}{(x - 4)}$

Example 3: Adding and subtracting rational expression.

           Add $\frac{m - 3}{m^{2} + 5m + 6} + \frac{5}{m + 2}$

We treat the rational as fractions. While adding two rational expressions we need to have same denominators. Compute LCD (least common denominator) for the given expressions.

The denominators are $m^{2} + 5m + 6$ and $m + 2$.

To find common multiple let us first write factors for $m^{2} + 5m + 6$
                          $m^{2} + 5m + 6 = (m + 2)(m + 3)$
The common multiple would be $(m + 2)(m + 3)$.

We need to multiply the second expression by $(m+3) /(m+3)$.
So the rational expression would be

$\frac{m - 3}{m^{2} + 5m + 6} + \frac{5}{m + 2} \times \frac{(m + 3)}{(m + 3)}$

$= \frac{m - 3}{(m + 2)(m + 3)} + \frac{5m + 15}{(m + 2)(m + 3)}$

$= \frac{m - 3 + 5m + 15}{(m + 2)(m + 3)}$

$= \frac{6m + 12}{(m + 2)(m + 3)}$


Let us check if we can simplify it further.
$= \frac{6(m + 12)}{(m + 2)(m + 3)}$

Cancelling m+2 will give the final answer in reduced form.
$= \frac{6}{(m + 3)}$

Subtract :
$\frac{x}{x - 2} - \frac{1}{x - 5}$

Get LCD,
LCD = $(x-2)(x-5)$

    $\frac{x}{x - 2}\times \frac{x - 5}{x - 5} - \frac{1}{x - 5}\times \frac{x - 2}{x - 2}$

        $= \frac{x^{2} - 5x}{(x - 2)(x - 5)} - \frac{x - 2}{(x - 2)(x - 5)}$

        $= \frac{x^{2} - 5x - x + 2}{(x - 2)(x - 5)}$

        $= \frac{x^{2} - 6x + 2}{(x - 2)(x - 5)}$