Express each one of the following with rational denominator:

Solution:

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

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

$3-\sqrt{2}$

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

As we know,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)=\frac{3-\sqrt{2}}{9-2}=\frac{3-\sqrt{2}}{7}$

(ii) $\frac{1}{\sqrt{6}-\sqrt{5}}$

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

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

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

As we know,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)=\frac{\sqrt{6}+\sqrt{2}}{6-2}=\frac{\sqrt{6}+\sqrt{2}}{4}$

(iii) $\frac{16}{\sqrt{41}-5}$

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

$\sqrt{41}+5$

$=\frac{16 \times(\sqrt{41}+5)}{(\sqrt{41}-5)(\sqrt{41}+5)}$

As we know,

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)=\frac{16 \sqrt{41}+80}{41-5}$

$=\frac{16 \sqrt{41}+80}{16}$

$=\frac{16(\sqrt{41}+5)}{16}$

$=\sqrt{41}+5$

(iv) $\frac{30}{5 \sqrt{3}-3 \sqrt{5}}$

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

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

$=\frac{30 \times(5 \sqrt{3}+3 \sqrt{5})}{(5 \sqrt{3}-3 \sqrt{5})(5 \sqrt{3}+3 \sqrt{5})}$

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

$=\frac{30 \times(5 \sqrt{3}+3 \sqrt{5})}{75-45}$

$=\frac{30 \times(5 \sqrt{3}+3 \sqrt{5})}{30}$

$=5 \sqrt{3}+3 \sqrt{5}$

(v) $\frac{1}{2 \sqrt{5}-\sqrt{3}}$

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

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

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

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

$=\frac{2 \sqrt{5}+\sqrt{3}}{20-3}=\frac{2 \sqrt{5}+\sqrt{3}}{17}$

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

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

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

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

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

$=\frac{(2 \sqrt{6}+3+2 \sqrt{2}+\sqrt{3})}{8-3}$

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

(vii) $\frac{6-4 \sqrt{2}}{6+4 \sqrt{2}}$

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

$6-4 \sqrt{2}$

$=\frac{(6-4 \sqrt{2})(6-4 \sqrt{2})}{(6+4 \sqrt{2})(6-4 \sqrt{2})}$

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

$=\frac{(6-4 \sqrt{2})^{2}}{36-32}$

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

$=\frac{36-48 \sqrt{2}+32}{4}$

$=\frac{68-48 \sqrt{2}}{4}$

$=\frac{4(17-12 \sqrt{2})}{4}$

$=17-12 \sqrt{2}$

(viii) $\frac{3 \sqrt{2}+1}{2 \sqrt{5}-3}$

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

$2 \sqrt{5}-3$

$=\frac{(3 \sqrt{2}+1) \times(2 \sqrt{5}-3)}{(2 \sqrt{5}-3)(2 \sqrt{5}-3)}$

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

$=\frac{6 \sqrt{10}-9 \sqrt{2}+2 \sqrt{5}-3}{(20-9)}$

$=\frac{6 \sqrt{10}-9 \sqrt{2}+2 \sqrt{5}-3}{11}$

(ix) $\frac{b^{2}}{\sqrt{\left(a^{2}+b^{2}\right)}+a}$

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

$\sqrt{\left(a^{2}+b^{2}\right)}-a$

$=\frac{b^{2}\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}{\left(\sqrt{\left(a^{2}+b^{2}\right)}+a\right)\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}$

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

$=\frac{b^{2}\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}{\left.\left(a^{2}+b^{2}\right)-a^{2}\right)}$

$=\frac{b^{2}\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}{b^{2}}$