Examine whether the following numbers are rational or irrational:

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

$=5^{2}-(\sqrt{5})^{2} \quad\left[(a-b)(a+b)=a^{2}-b^{2}\right]$

$=25-5$

$=20$, which is an integer

Hence, $(5-\sqrt{5})(5+\sqrt{5})$ is rational.

(ii) $(\sqrt{3}+2)^{2}$

$=(\sqrt{3})^{2}+2^{2}+2 \times \sqrt{3} \times 2 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=3+4+4 \sqrt{3}$

$=7+4 \sqrt{3}$

Since, the sum and product of rational numbers and an irrational number is always an irrational.

$\Rightarrow 7+4 \sqrt{3}$ is irrational.

Hence, $(\sqrt{3}+2)^{2}$ is irrational.

(iii) $\frac{2 \sqrt{13}}{3 \sqrt{52}-4 \sqrt{117}}$

$=\frac{2 \sqrt{13}}{3 \sqrt{13 \times 4}-4 \sqrt{13 \times 9}}$

$=\frac{2 \sqrt{13}}{\sqrt{13}(3 \sqrt{4}-4 \sqrt{9})}$

$=\frac{2}{(3 \times 2-4 \times 2)}$

$=\frac{2}{6-8}$

$=\frac{2}{-2}$

$=-1$, which is an integer

Hence, $\frac{2 \sqrt{13}}{3 \sqrt{52}-4 \sqrt{117}}$ is rational.

(iv) $\sqrt{8}+4 \sqrt{32}-6 \sqrt{2}$

$=2 \sqrt{2}+4 \times 4 \sqrt{2}-6 \sqrt{2}$

$=2 \sqrt{2}+16 \sqrt{2}-6 \sqrt{2}$

$=12 \sqrt{2}$

Since, the product of a rational number and an irrational number is always an irrational.

Hence, $\sqrt{8}+4 \sqrt{32}-6 \sqrt{2}$ is rational.