Show that: $\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{2\sqrt{2}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}=5$

Hint: In this question we have been given with a term for which we have to prove that the left-hand side is equal to the right-hand side. We will solve this question by first considering the left-hand side and then performing various operations on it and showing that the resultant value is equal to the value on the right-hand side. We will solve the terms in the roots by rationalizing the denominator of the fraction by multiplying it with the conjugate term and simplify the terms to get the required solution.


Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow \dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{2\sqrt{2}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}=5$
Consider the left-hand side of the expression as:
$=\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{2\sqrt{2}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}$
Now since there is no direct way to solve the given expression, we will rationalize the denominators in the given fractions. We can write the expression as:
$= \dfrac{1}{3-2\sqrt{2}}\times \dfrac{3+2\sqrt{2}}{3+2\sqrt{2}}-\dfrac{1}{2\sqrt{2}-\sqrt{7}}\times \dfrac{2\sqrt{2}+\sqrt{7}}{2\sqrt{2}+\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}\times \dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}\times \dfrac{\sqrt{5}+2}{\sqrt{5}+2}$
On simplifying the terms, we get:
$= \dfrac{3+2\sqrt{2}}{\left( 3-2\sqrt{2} \right)\left( 3+2\sqrt{2} \right)}-\dfrac{2\sqrt{2}+\sqrt{7}}{\left( 2\sqrt{2}-\sqrt{7} \right)\left( 2\sqrt{2}+\sqrt{7} \right)}+\dfrac{\sqrt{7}+\sqrt{6}}{\left( \sqrt{7}-\sqrt{6} \right)\left( \sqrt{7}+\sqrt{6} \right)}-\dfrac{\sqrt{6}+\sqrt{5}}{\left( \sqrt{6}-\sqrt{5} \right)\left( \sqrt{6}+\sqrt{5} \right)}+\dfrac{\sqrt{5}+2}{\left( \sqrt{5}-2 \right)\left( \sqrt{5}+2 \right)}$
Now we can see that we have the terms in the denominator of $\left( a-b \right)\left( a+b \right)$ therefore, we can use the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and simplify and write the expression as:
$= \dfrac{3+2\sqrt{2}}{{{3}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}}-\dfrac{2\sqrt{2}+\sqrt{7}}{{{\left( 2\sqrt{2} \right)}^{2}}-{{\left( \sqrt{7} \right)}^{2}}}+\dfrac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}-\dfrac{\sqrt{6}+\sqrt{5}}{{{\left( \sqrt{6} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}}+\dfrac{\sqrt{5}+2}{{{\left( \sqrt{5} \right)}^{2}}-{{2}^{2}}}$
On simplifying the terms, we get:
$=\dfrac{3+2\sqrt{2}}{9-8}-\dfrac{2\sqrt{2}+\sqrt{7}}{8-7}+\dfrac{\sqrt{7}+\sqrt{6}}{7-6}-\dfrac{\sqrt{6}+\sqrt{5}}{6-5}+\dfrac{\sqrt{5}+2}{5-4}$
On subtracting the terms, we get:
$=\dfrac{3+2\sqrt{2}}{1}-\dfrac{2\sqrt{2}+\sqrt{7}}{1}+\dfrac{\sqrt{7}+\sqrt{6}}{1}-\dfrac{\sqrt{6}+\sqrt{5}}{1}+\dfrac{\sqrt{5}+2}{1}$
Since the denominator is the same for all the fractions and it is $1$, we can discard the fraction and change the sign of the terms and write the terms as:
$= 3+2\sqrt{2}-2\sqrt{2}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2$
On cancelling the similar terms with the opposite signs, we get:
$= 3+2$
On adding, we get:
$= 5$, which is the right-hand side of the expression, hence proved.

Note: It is to be noted that rationalizing a fraction implies that multiplying a radical term with a term such that the radical part of the term is eliminated. Now since multiplying and dividing a term with the same term does not change its value, we use rationalization to remove the radical part. It is to be remembered that squared root can also be written in the form of an exponent where $\sqrt{a}={{a}^{\dfrac{1}{2}}}$.