# Simplify:

Question:

Simplify:

(i) $\frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

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

(iii) $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$

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

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

Solution:

(i) $\frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}+\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

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

$3 \sqrt{2}-2 \sqrt{3}$ for $\frac{1}{3 \sqrt{2}+2 \sqrt{3}}$ and the rationalizing factor $\sqrt{3}+\sqrt{2}$ for $\frac{1}{\sqrt{3}-\sqrt{2}}$

$=\frac{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}{(3 \sqrt{2}+2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}+\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$

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

$=\frac{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}{18-12}+\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{3-2}$

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

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

$=\frac{(18-12 \sqrt{6}+12)}{6}+(6+2 \sqrt{6})$

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

$=5-2 \sqrt{6}+(6+2 \sqrt{6})=11$

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

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

$\sqrt{5}+\sqrt{3}$ for $\sqrt{15}-\sqrt{3}$ and the rationalizing factor $\sqrt{5}-\sqrt{3}$ for $\frac{1}{\sqrt{5}+\sqrt{3}}$

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

Now as we know, (a + b)(a - b)

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

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

$=\frac{8+2 \sqrt{15}+8-2 \sqrt{15}}{2}$

$=\frac{16}{2}=8$

(iii) $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$

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

$3-\sqrt{5}$ for $\frac{1}{3+\sqrt{5}}$ and the rationalizing factor $3+\sqrt{5}$ for $\frac{1}{3-\sqrt{5}}$

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

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

$=\frac{(7+3 \sqrt{5})(3-\sqrt{5})}{9-5}-\frac{(7-3 \sqrt{5})(3+\sqrt{5})}{9-5}$

$=\frac{(21-7 \sqrt{5}+9 \sqrt{5}-15)}{4}-\frac{(21+7 \sqrt{5}-9 \sqrt{5}-15)}{4}$

$=\frac{(6+2 \sqrt{5})}{4}-\frac{(6-2 \sqrt{5})}{4}$

$=\frac{4 \sqrt{5}}{4}$

$=\sqrt{5}$

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

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

$2-\sqrt{3}$ for $\frac{1}{2+\sqrt{3}}$, the rationalizing factor $\sqrt{5}+\sqrt{3}$ for $\frac{1}{\sqrt{5}-\sqrt{3}}$, and the rationalizing factor

$2+\sqrt{5}$ for $\frac{1}{2-\sqrt{5}}$

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

Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)=\frac{2-\sqrt{3}}{4-3}+\frac{2 \times(\sqrt{5}+\sqrt{3})}{5-3}+\frac{2+\sqrt{5}}{4-5}$

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

$=\frac{4-2 \sqrt{3}+2 \sqrt{5}+2 \sqrt{3}-4-2 \sqrt{5}}{2}=\frac{0}{2}=0$

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

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

$\sqrt{5}-\sqrt{3}$ for $\frac{1}{\sqrt{5}+\sqrt{3}}$, the rationalizing factor $\sqrt{3}-\sqrt{2}$ for $\frac{1}{\sqrt{3}+\sqrt{2}}$ and the rationalizing factor

$\sqrt{5}-\sqrt{2}$ for $\frac{1}{\sqrt{5}+\sqrt{2}}$

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

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

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

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

$=\frac{6 \sqrt{5}-6 \sqrt{3}+6 \sqrt{3}-6 \sqrt{2}-6 \sqrt{5}+6 \sqrt{2}}{3}=\frac{0}{3}=0$