It being given that
Question:

It being given that $\sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{6}=2.449$ and $\sqrt{10}=3.162$, find to three places of decimal, the value of each of the following.

(i) $\frac{1}{\sqrt{6}+\sqrt{5}}$

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

(iii) $\frac{1}{4 \sqrt{3}-3 \sqrt{5}}$

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

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

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

Solution:

(i)

$\frac{1}{\sqrt{6}+\sqrt{5}}$

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

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

$=\frac{\sqrt{6}-\sqrt{5}}{6-5}$

$=\sqrt{6}-\sqrt{5}$

$=2.449-2.236$

$=0.213$

(ii)

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

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

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

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

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

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

$=3 \times(2.236-1.732)$

$=1.512$

(iii)

$\frac{1}{4 \sqrt{3}-3 \sqrt{5}}$

$=\frac{1}{4 \sqrt{3}-3 \sqrt{5}} \times \frac{4 \sqrt{3}+3 \sqrt{5}}{4 \sqrt{3}+3 \sqrt{5}}$

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

$=\frac{4 \sqrt{3}+3 \sqrt{5}}{48-45}$

$=\frac{4 \times 1.732+3 \times 2.236}{3}$

$=4.545$

(iv)

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

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

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

$=\frac{9+5+6 \sqrt{5}}{9-5}$

$=\frac{14+6 \sqrt{5}}{4}$

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

$=\frac{7+3 \times 2.236}{2}$

$=6.854$

(v)

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

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

$=\frac{2+\sqrt{3}+4 \sqrt{3}+6}{(2)^{2}-(\sqrt{3})^{2}}$

$=\frac{8+5 \sqrt{3}}{4-3}$

$=8+5 \sqrt{3}$

$=8+5 \times 1.732$

$=16.660$

(vi)

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

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

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

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

$=\frac{7+2 \sqrt{10}}{3}$

$=\frac{7+2 \times 3.162}{3}$

$=4.441$