Simplify the following expression:

$(\sqrt{5}$ - $\sqrt{2})$ $(\sqrt{5}$ + $\sqrt{2})$

Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = $\dfrac{c}{d}$. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Rationalize the denominators of the following:

(i) $\dfrac{1}{\sqrt{7}}$

(ii) $\dfrac{1}{\sqrt{7} - \sqrt{6} }$

(iii)$\dfrac{1}{\sqrt{5} + \sqrt{2} }$

(iv)$\dfrac{1}{\sqrt{7} - 2 }$

Find :

(i) $\left(64^{\smash{\frac{1}{2}}} \right)$

(ii) $\left(32^{\smash{\frac{1}{5}}} \right)$

(iii) $\left(125^{\smash{\frac{1}{3}}} \right)$