Rationalize the denominator and simplify:

Solution:

(i) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$\sqrt{3}-\sqrt{2}$

$=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}$

As we know, $(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$

$=\frac{3-2 \sqrt{3} \sqrt{2}+2}{1}=5-2 \sqrt{6}$

(ii) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$7-4 \sqrt{3}$

$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{49-48}$

$=35-20 \sqrt{3}+14 \sqrt{3}-24=11-6 \sqrt{3}$

(iii) $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$3+2 \sqrt{2}$

$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{(3-2 \sqrt{2})(3+2 \sqrt{2})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{9-8}$

$=3+2 \sqrt{2}+3 \sqrt{2}+4=7+5 \sqrt{2}$

(iv) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$3 \sqrt{5}+2 \sqrt{6}$

$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5}-2 \sqrt{6})(3 \sqrt{5}+2 \sqrt{6})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{45-24}$

$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{21}$

$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{21}$

$=\frac{6 \sqrt{30}+24-15-2 \sqrt{30}}{21}$

$=\frac{4 \sqrt{30}+9}{21}$

(v) $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$\sqrt{48}-\sqrt{18}$

$=\frac{(4 \sqrt{3}+5 \sqrt{2})(\sqrt{48}-\sqrt{18})}{(\sqrt{48}+\sqrt{18})(\sqrt{48}-\sqrt{18})}$

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(4 \sqrt{3}+5 \sqrt{2})(\sqrt{48}-\sqrt{18})}{48-18}$

$=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}$

$=\frac{18+8 \sqrt{6}}{30}$

$=\frac{9+4 \sqrt{6}}{15}$

(vi) $\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$2 \sqrt{2}-3 \sqrt{3}$

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

$=\frac{(2 \sqrt{3}-\sqrt{5})(2 \sqrt{2}-3 \sqrt{3})}{8-27}$

$=\frac{(4 \sqrt{6}-2 \sqrt{10})-18+3 \sqrt{15})}{-19}$

$=\frac{(18-4 \sqrt{6}+2 \sqrt{10}-3 \sqrt{15})}{19}$