Simplify \[\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}\].

Hint: A rational number is one that can be expressed as a combination of two integers with the denominator not being zero, whereas an irrational number cannot be expressed in fractions. Irrational numbers have non-terminating decimals, while rational numbers have terminating decimals.

Complete step by step solution:
A fraction is made up of two sections. The numerator is one, and the denominator is the other. Surd is a number that cannot be simplified any further, such as the square root or cube root. The surd may be in either the numerator or the denominator place.
Let us solve the question by rationalizing the denominator.
We must multiply both the numerator and the denominator by the denominator's conjugate to rationalise the denominator. Remember that all we have to do to find the conjugate is to change the sign between the two terms.
By eliminating the radicals from the denominators, rationalising a denominator makes it easier to understand what the quantity really is.
$\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}$
Let us rationalize the denominator
The conjugate of $\left( {7 + 2\sqrt {14} } \right)$ is $\left( {7 - 2\sqrt {14} } \right)$
Multiplying and dividing by the conjugate,
$\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}\,\, = \,\,\dfrac{{\left[ {\left( {5 - 3\sqrt {14} } \right)\left( {7 - 2\sqrt {14} } \right)} \right]}}{{\left[ {\left( {7 + 2\sqrt {14} } \right)\left( {7 - 2\sqrt {14} } \right)} \right]}}\,\,........................\left( 1 \right)$
We know that, $\left( {{a^2}\, - \,{b^2}} \right)\,\, = \,\,\left( {a + b} \right)\left( {a - b} \right)$
Using the above formula in the denominator part of equation $\left( 1 \right)$we get,
$\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}\,\, = \,\,\dfrac{{\left[ {\left( {5 - 3\sqrt {14} } \right)\left( {7 - 2\sqrt {14} } \right)} \right]}}{{\left[ {{{\left( 7 \right)}^2}\, - \,\,{{\left( {2\sqrt {14} } \right)}^2}} \right]}}\,\,$
On solving, the numerator and denominator we get,
$\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}\,\, = \,\,\dfrac{{\left[ {5\left( {7 - 2\sqrt {14} } \right) - 3\sqrt {14} \left( {7 - 2\sqrt {14} } \right)} \right]}}{{\left[ {49 - 56} \right]}}\,\,$
On further solving, we get
$\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}\,\, = \,\,\dfrac{{\left[ {35 - 10\sqrt {14} - 21\sqrt {14} + 84} \right]}}{{\left[ {49 - 56} \right]}}\,\,$
Simplifying the above equation, we get,
$\dfrac{{\left( {5 - 3\sqrt {14} } \right)}}{{\left( {7 + 2\sqrt {14} } \right)}}\,\,\, = \,\,\,\dfrac{{119 - 31\sqrt {14} }}{{ - 7}}\,\,\, = \,\,\,\,\dfrac{{31\sqrt {14} - 119}}{7}\,$

Note:
Rationalization is the method of eliminating imaginary numbers from an algebraic expression's denominator. It's the process of moving the radical (square root or cube root) from the fraction's bottom (denominator) to the top (numerator).