Find rational numbers a and b such that

Solution:

(i)

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

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

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

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

$=3-2 \sqrt{2}$

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

$\Rightarrow a=3, b=-2$

(ii)

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

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

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

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

$=\frac{9-4 \sqrt{5}}{-1}$

$=-9+4 \sqrt{5}$

$\therefore \frac{2-\sqrt{5}}{2+\sqrt{5}}=4 \sqrt{5}+(-9)=a \sqrt{5}+b$

$\Rightarrow a=4, b=-9$

(iii)

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

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

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

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

$=5+2 \sqrt{6}$

$\therefore \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2 \sqrt{6}=a+b \sqrt{6}$

$\Rightarrow a=5, b=2$

(iv)

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

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

$=\frac{35-20 \sqrt{3}+14 \sqrt{3}-24}{(7)^{2}-(4 \sqrt{3})^{2}}$

$=\frac{11-6 \sqrt{3}}{49-48}$

$=11-6 \sqrt{3}$

$\therefore \frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=11+(-6) \sqrt{3}=a+b \sqrt{3}$

$\Rightarrow a=11, b=-6$